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%\addtocounter{ex}{68} %this is the best thing in latex :)

\rem All sets are assumed to be topological spaces and all maps are assumed to be continuous unless stated otherwise. The symbol '=' will denote that two topological spaces are homeomorphic. The closed unit interval will be denoted by $I$ or $J$.

\ex Prove that being homotopic is an equivalence relation (on the set of continuous maps between topological spaces).
\begin{proof} Let $f,g,k:X\to Y$ be such that $f\sim g$, $g\sim k$, i.e. there exist maps $h,h':X\times I\to Y$ such that $h(x,0)=f(x)$, $h(x,1)=g(x)$, $h'(x,0)=g(x)$, $h'(x,1)=k(x)$.
\begin{itemize}
\item Reflexivity: the map $H_1:X\times I\to Y$ defined by $H_1(x,t):=f(x)$ for all $t\in I$ is a homotopy between $f$ and itself.
\item Symmetry: the map $H_2:X\times I\to Y$ defined by $H_2(x,t):=h(x,1-t)$ for all $t\in I$ is a homotopy between $g$ and $f$.
\item Transitivity: the map $H_3:X\times I\to Y$ defined by $H_3(x,t):=
\begin{cases}
h(x,2t)\quad \text{ for } 0\leq t \leq \frac{1}{2}\\
h'(x,2t)\quad \text{ for } \frac{1}{2} < t \leq 1
\end{cases}
$ is a homotopy between $f$ and $k$.
\end{itemize}
\end{proof}

\ex Let $\simeq$ be an equivalence relation on a topological space $X$. Prove that the map $f:X/\simeq \to Y$ is continuous iff $f\circ p:X\to Y$ is continuous, where $p:X\to X/\simeq$ is the canonical quotient projection.
\begin{proof}
The direction "$\Rightarrow$" follows from the facts that $p$ is continuous (in fact, the quotient topology is the final topology with respect to $p$) and the composition of continuous functions is again continuous. For "$\Leftarrow$", let $U\subseteq Y$ be open. Then $$p^{-1}(f^{-1}(U))=(f\circ p)^{-1}(U)$$ is open by continuity of $f\circ p$, so $f^{-1}(U)$ must also be open by the definition of quotient topology and we are done.
\end{proof}

\ex Show that $D^n/S^{n-1}=S^n$ using the map $f:D^n\to S^n$ given by $$f(x_1,\dots,x_n)=(2\sqrt{1-\norm{\textbf{x}}}\textbf{x},2\norm{\textbf{x}}^2-1).$$
\begin{proof}
It's easy to see that $f$ is continuous. Moreover, its restriction to the interior of $D^n$ gives a bijection to $S^n\setminus\{(0,\dots,0,1)\}$ (the inverse function is given by $(\textbf{y},z)\mapsto \frac{1}{\sqrt{\frac{1-z}{2}}}\textbf{y}$) and we have $f(S^{n-1})=\{(0,\dots,0,1)\}$, so we can define $f':D^n/S^{n-1}\to S^n$ by $f'([\textbf{x}])=f(\textbf{x})$. Then $f'$ is a bijection, and by the previous exercise it is continuous. Finally, both $D^n/S^{n-1}$ and $S^n$ are compact (Hausdorff) spaces (since both $D^n$ and $S^n$ are closed bounded subsets of $\R^n$ and $R^{n+1}$, respectively, and $S^{n-1}\subseteq D^n$ is closed), so $f'$ must be a homeomorphism (a general fact for continuous bijections between compact spaces).
\end{proof}

\ex Let $f:X\to Y$ and $M_f=X\times I\cup_{j\times 1}Y$. Moreover, let $\iota_X:X\to M_f$ be given by $x\mapsto (x,0)$, $\iota_Y:Y\to M_f$ be given by $y\mapsto [y]$ and $r:M_f\to Y$ be given by $r(y)=y$, $r(x,t)=f(x)$. Show that
\begin{enumerate}[i)]
\item $Y$ is a deformation retract of $M_f$,
\item $r\circ \iota_X=f$,
\item $\iota_Y\circ f\sim \iota_X$.
\end{enumerate}
\begin{proof}\mbox{}\newline
\begin{enumerate}[i)]
\item Geometrically, the deformation retraction is realized by pushing $X$ along $I$ towards $Y$.
\item We have $r\circ \iota_X(x)=r(x,0)=f(x)$ for all $x\in X$.
\item The required homotopy $h:X\times J \to M_f$ is given by $h(x,s)=[(x,s)]$.
\end{enumerate}
\end{proof}

\ex Show that the pair $(M_f,X)$ has the homotopic extension property (HEP), i.e. $\iota_X$ is a cofibration.
\begin{proof}
Let $g:I\times J\to\{0\}\times J\cup I\times\{0\}$ be any retraction such that $g(0,s)=(0,s)$ and $g(1,s)=(1,0)$. Then the map $r:M_f\times J\to X\times\{0\}\times J\cup M_f\times\{0\}$ defined by 
$r(x,t,s)=(x,g(t,s))$ and $r(y,s)=(y,0)$ is the required retraction.
\end{proof}

\ex The smash product between two based spaces is defined by $$(C,c_0)\wedge (D,d_0):=(C\times D)/\left(C\times \{d_0\}\cup \{c_0\}\times D\right).$$ Show that $X/A\wedge Y/B=(X\times Y)/(X\times B\cup A\times Y)$.
\begin{proof}
Let $p_1:X\times Y\to X/A\times Y/B$ be given by $p_1(x,y)=([x],[y])$ and $p_2:X/A\times Y/B\to X/A \wedge Y/B$ be given by $p_2(([x],[y]))=([[x]],[[y]])$. Then the composition $p_2\circ p_1$ is continuous and factors through $(X\times Y)/(X\times B\cup A\times Y)$, which implies that the canonical bijection between $(X\times Y)/(X\times B\cup A\times Y)$ and $X/A \wedge Y/B$ is continuous (using exercise 2). Using the definition of quotient topology several times, it can be shown that this bijection is also open, hence a homeomorphism.
\end{proof}

\ex Let $A=\{\frac{1}{n}\cup \{0\}\}\subseteq \R$. Show that $(I,A)$ does not have the HEP, i.e. the inclusion $A \hookrightarrow I$ is not a cofibration.
\begin{proof}
If $A\times J \cup I\times\{0\}$ was a retract of $I\times J$, the retraction would have to preserve connected subsets. But $I\times J$
 is locally connected while  $A\times J \cup I\times\{0\}$ is not, a contradiction.
\end{proof}

\ex Show that $(S^m,*)\wedge (S^n,*)\cong (S^{m+n},*)$.
\begin{proof} Using exercises 3 and 6 from the previous tutorial, we have 
\begin{equation*}
\begin{split}
(S^m,*)\wedge (S^n,*)&\cong (D^m/S^{m-1})\wedge(D^n/S^{n-1})\cong \\
&\cong(D^m\times D^n)/(S^{m-1}\times D^m\cup D^n\times S^{n-1})\cong\\
&\cong  (I^m\times I^n)/(\dd I^m\times I^n\cup \dd I^n\times I^m)=\\
&=I^{m+n}/(\dd (I^{m+n}))\cong D^{m+n}/\dd D^{m+n}\cong S^{m+n}.
\end{split}
\end{equation*}

\end{proof}

\ex Show that $\C P^n$ is a CW-complex.
\begin{proof}
Clearly $\C P^0$ is a point. Next, we have
\begin{equation*}
\begin{split}
\C P^n&=\C^{n+1}\setminus\{0\}/\{v\sim \lambda v, \lambda\in\C\setminus\{0\}\}\cong S^{2n+1}\setminus\{0\}/\{v\sim \lambda v, |\lambda|=1\}\cong\\
&\cong \{(w,\sqrt{1-|w^2|})\in \C^{n+1},w\in D^{2n}\}/\{w\sim \lambda w \text{ for }|w|=1\}\cong\\
&\cong (D^{2n}\cup S^{2n-1})/\{w\sim \lambda w \text{ for }w\in S^{2n-1}\}=D^{2n}\cup_f\C.
\end{split}
\end{equation*}
Now taking the canonical projection $S^{2n-1}\to\C P^{n-1}\cong S^{2n-1}/\sim$ as the attaching map $f$  yields a CW-complex with one cell in every even dimension till $2n$ and none in the odd ones.
\end{proof}

\ex From the lecture we know that $A:=\{\frac{1}{n},n\in\N\}\cup\{0\}$ as a subspace of $\R$ is not a CW-complex. Show that $X:=I\times\{0\}\cup A\times I$ is not a CW-complex either.
\begin{proof}
Suppose that $X$ is a CW-complex. Then it cannot contain cells of dimension $\geq 2$, because it becomes disconnected after removing any point. In fact, the space obtained after removing any point $(a,0)$ with $a\in A$ has more than two connected components (three for $a>0$, to be exact), so these points cannot lie inside a 1-cell. Therefore these points must form 0-cells, but we already know that $A$ does not have discrete topology, a contradiction.
\end{proof}

\ex Show that the Hawaiian earring given by $$X=\{(x,y)\in\R^2,(x-\frac{1}{n})^2+y^2=\frac{1}{n^2} \text{ for some } n\}$$ is not a CW-complex.
\begin{proof}
Suppose that $X$ is a CW-complex. Using similar arguments as in the previous exercise, we can see that $(0,0)$ must be a 0-cell and that $X$ must have either infinitely many 0-cells, or infinitely many 1-cells. But since $X$ is compact, exercise 5 implies that $X$ can have only finitely many cells, a contradiction.
\end{proof}

\ex Prove that every compact set $A$ in a CW-complex $X$ can have a nonempty intersection with only finitely many cells.
\begin{proof}
$X$ is comprised of cells that are indexed by elemnts of some set $J$. Let $B$ be a set containing exactly one point from each intersection $A\cap e^{\beta}, \beta\in J$. We need to show that $B$ is closed and discrete, which will imply that $B$ is compact (since $B\subseteq A$) and discrete, hence finite. We know that a set $C\subseteq X^{n}$ is closed iff both $C\cap X^{n-1}$ and $C\cap e_{\alpha}^n$ for each $\alpha\in J$ are closed, because $D^n\cup_f X^{n-1}$ is a pushout. Using induction, this implies that $C\subseteq X$ is closed iff $C\cap e_\alpha$ is closed for each $\alpha\in J$. Since $B\cap e_\alpha$ contains at most one point for any $\alpha\in J$ and $X$ is $T_1$ (even Hausdorff), this shows that $B$ is closed. Using the same argument, $B$ with any one point removed is closed. Therefore $B$ is also discrete and we are done.
\end{proof}

\ex Show that for a short exact sequence $0\to A\xrightarrow{f} B\xrightarrow{g}C\to 0$ of abelian groups (or more generally modules over a commutative ring) the following are equivalent:
\begin{enumerate}[(1)]
\item There exists $p:B\to A$ such that $pf=\id_A$.
\item There exists $q:C\to B$ such that $gq=\id_C$.
\item There exist $p:B\to A$ and $q:C\to B$ such that $fp+qg=\id_B$.
\end{enumerate}
(Another equivalent condition is $B\cong A\oplus C$, with $(p,g)$ and $f+q$ being the respective inverse isomorphisms.)
\begin{proof}\mbox{}\\
$(1)\implies(2)$ and $(3)$:\\
Since $g$ is surjective, for any $c\in C$ there is some $b\in B$ such that $g(b)=c$. Moreover, for any other $b'\in B$ such that also $g(b')=c$, we have $b-fp(b)=b'-fp(b')$, since $b-b'\in\ker g=\im f$, so that $b-b'=f(a)$ and $$fp(b-b')=fpf(a)=f(a)=b-b'.$$ This shows that we can correctly define $q(c):=b-fp(b)$ for any such $b$. Then we have $$gq(c)=g(b)-gfp(b)=g(b)=c$$ (since $gf=0$), which shows that $gq=\id_C$, and also
$qg(b)=b-fp(b)$, hence $fp+qg=\id_B$.\\\\
$(3)\implies(1)$ and $(2)$:\\
Applying $f$ from the right to the equation $fp+qg=\id_B$ yields $fpf=f$ (since $gf=0$), which together with the fact that $f$ is injective implies $pf=\id_A$. Similarly, applying $g$ from the left yields $gqg=g$, which together with the fact that $g$ is surjective implies $gq=\id_C$.
\end{proof}

\ex Let $0\to A_*\xrightarrow{f} B_*\xrightarrow{g}C_*\to 0$ be a short exact sequence of chain modules. We have defined the connecting homomorphism $\dd_*:H_n(C)\to H_{n-1}(A)$ by the formula $\dd_*[c]=[a]$, where $\dd c=0$, $f(a)=\dd b$ and $g(b)=c$. Show that this definition does not depend on $a$ nor $b$.
\begin{proof}
We have $\dd a=0$ iff $f(\dd a)=0$ (using injectivity of $f$) iff $0=\dd f(a)=\dd \dd b$, and the last condition is true.\\
Now let $b,b'\in B$ be such that $g(b)=g(b')=c$ with $a,a'\in A$ such that $f(a)=b, f(a')=b'$. Then $b-b'\in \ker g=\im f$, so $b-b'=f(\overline{a})$ for some $\overline{a}\in A$. Therefore $f(\dd \overline{a})=\dd b-\dd b'=f(a-a')$ and the injectivity of $f$ implies $\dd \overline{a}=a-a'$, hence $[0]=[\dd a']=[a]-[a']$ and we are done.
\end{proof} 


\ex Show $\dd\dd=0$. Use formula $\varepsilon_{n+1}^i\circ \varepsilon_n^j=\varepsilon_{n+1}^{j+1}\circ \varepsilon_n^i$, where $i<j$. The definition for $\sigma\in C_n(X)$,  $\sigma\colon\Delta^n\to X$, is 
$$\dd\sigma=\sum_{i=0}^n (-1)^i\,\sigma\circ \varepsilon_n^i.$$


\begin{proof} Easily workout
\begin{equation*}
\begin{split}
\dd(\dd\sigma)&=\dd \Big(\sum_{i=0}^{n+1} (-1)^i\,\sigma\circ \varepsilon_{n+1}^i\Big)=\sum_{i=0}^{n+1} (-1)^i\,\dd(\sigma\circ \varepsilon_{n+1}^{i})=\\
              &=\sum_{i=0}^{n+1}(-1)^i \Big(\sum_{j<i}(-1)^j \sigma \circ \varepsilon_{n+1}^{i}\circ\varepsilon_n^j + \sum_{j>i}(-1)^{j-1}\sigma\circ\varepsilon_{n+1}^i\circ \varepsilon_n^{j-1}\Big)=\\
              &=\sum_{i=1}^{n+1}\sum_{j<i} (-1)^i(-1)^j\sigma\circ\varepsilon_{n+1}^i\circ\varepsilon_n^j+\sum_{i=1}^{n}\sum_{j>i}(-1)^{j+i-1}\sigma\circ\varepsilon_{n+1}^i\circ\varepsilon_n^{j-1},
\end{split}
\end{equation*}
now, with proper reindex and shift, this yields $\varepsilon_{n+1}^i\circ\varepsilon_n^{j}=\varepsilon_{n+1}^{j+1}\circ\varepsilon_n^{i}=\varepsilon^j_{n+1}\circ\varepsilon_n^{i-1}$, both sums are of the same elements but with opposite signs. Hence, $\dd\dd=0$.
\end{proof}

\ex Simplicial homology of $\dd\Delta^2$.
\vspace{-0.314159265cm}
\begin{figure}[ht]
    \centering
    \includegraphics[scale=1.5]{algtopo3_triangle.jpg}
    %\caption{Triangle}
    %\caption{Awesome Image}
    %\label{fig:awesome_image}
\end{figure}
\vspace{-1.5cm}
\begin{proof}
Chain complex of this simplicial homology is $C_0=\Z[v_0,v_1,v_2]=\Z\oplus\Z\oplus\Z,$ $C_1=\Z[a_0,a_1,a_2]=\Z\oplus\Z\oplus\Z$. So
$$0\to C_1\stackrel{\dd}{\longrightarrow} C_0\to 0,$$
where we want to determine $\dd$ and we know $\dd a_0=v_2-v_1$, $\dd a_1=v_2-v_0$, $\dd a_2=v_1-v_0$. Using simple linear algebra, we study generators $\ker\dd$ and $\im\dd$:
\[
\left(
\begin{array}{ccc|ccc}
1 & 0 & 0 & 0  & -1 & 1 \\
0 & 1 & 0 & -1 & 0  & 1 \\
0 & 0 & 1 & -1 & 1  & 0
\end{array}
\right)
\sim 
\left(
\begin{array}{ccc|ccc}
1 & 0 & 0 & 0  & -1 & 1 \\
0 & 1 & 0 & -1 & 0  & 1 \\
1 & -1 & 1 & 0 & 0  & 0
\end{array}
\right),
\]
therefore $\ker\dd$ has a generator $a_0-a_1+a_2$ and $\im\dd$ has two generators $-v_1+v_2$ and $-v_0+v_2$. We get $H_0=\dfrac{\Z[v_0,v_1,v_2]}{\Z[-v_1+v_2,-v_0+v_2]}=\dfrac{\Z[-v_1+v_2,v_0+v_2,v_2]}{\Z[-v_1+v_2,-v_0+v_2]}=\Z[v_0]=\Z$ and $H_1=\ker\dd=\Z[a_0-a_1+a_2]=\Z$.
\end{proof}

\ex Simplicial complex, model of torus, compute differentials and homology.

\vspace{-0.5cm}
\begin{figure}[ht]
    \centering
    \includegraphics[scale=1.5]{algtopo3_torus.jpg}
    %\caption{Triangle}
    %\caption{Awesome Image}
    %\label{fig:awesome_image}
\end{figure}
\vspace{-1.12345cm}

\begin{proof}
Again, we get simplicial chain complex $C_*$ formed by free abelian groups generated by equivalence classes of simplicies. Note $a_1,a_2$ are actually one generator, same for $b_1,b_2$. All the vertices are also equivalent. We choose the orientation and fix it.\par 
Thus we get $C_0=\Z[v]=\Z, C_1=\Z[a,b,c]=\Z\oplus\Z\oplus\Z, C_2=\Z[e,f]=\Z\oplus\Z,C_3=0,$ and the following holds: $\dd a=0,\dd b=0,\dd c=0,$ as well as $\dd e=a+b-c,\dd f = c-a-b,\dd (e+f)=0,$ so we get $\ker\dd=\Z[e+f],\im\dd=\Z[a+b-c]$.\par 
Let $T$ be the torus. Then 
\begin{align*}
    H_2(T)&=\ker\dd_2=\Z[e+f]=\Z,\\
    H_1(T)&=\Z[a,b,c]/\Z[a+b-c]=\dfrac{\Z[a,b,a+b-c]}{\Z[a+b-c]}=\Z[a,b]=\Z\oplus\Z,\\
    H_0(T)&=\ker\dd_0=\Z.
\end{align*}

\end{proof}


\textbf{First criterion of homotopy equivalence} (Hatcher) \par\noindent
Let $(X,A)$ be a pair that satisfies HEP, i.e. $A\stackrel{i}{\hookrightarrow}X$ is a cofibration. Let $A$ be contractible in itself. Then $q\colon X\to X/A$ is a homotopy equivalence.


\ex $S^2\vee S^1 \simeq S^2/S^0$ (using First criterion)
\begin{proof}
%\textbf{OBRAZEK}\\

\begin{figure}[ht]
    \centering
    \includegraphics[scale=1.5]{algtopo3_threethings.jpg} %změnit scale
    %\caption{Triangle}
    %\caption{Awesome Image}
    %\label{fig:awesome_image}
\end{figure}
%\vspace{-1.1415926535cm}

In the picture (hopefully) above, $A$ is a segment as well as $B$, so contractible in itself. Clearly $S^2\vee S^1=X/B$ and $S^2/S^0=X/A$ and $X \simeq X/A$ and $X\simeq X/B$ by criterion, therefore $X/A\simeq X/B $ and we are done.
\end{proof}

\ex Let $i\colon A \hookrightarrow X$ is a cofibration, show $X/A\simeq X\cup CA=Ci$. (using First criterion)
\begin{proof}
\begin{figure}[ht]
    \centering
    \includegraphics[scale=1.414]{algtopo3_cone.jpg}
    %\caption{Stuff}
    %\caption{Awesome Image}
    %\label{fig:awesome_image}
\end{figure}
%\vspace{-1.1415926535cm}
%Consider mapping $Cf\colon A\times I \cup X/\sim, (a,1)\sim f(a), (a_{1},0)\sim(a_{2},0)$
We know $CA\hookrightarrow X\cup CA$ is a cofibration using homework 1, exercise 2, with $Y=CA$. Then by criterion $X\cup CA \simeq X\cup CA / CA$. Also $X/A$ is homeomorphic to $X\cup CA/CA$ (see picture above), which concludes the result.

\iffalse
$Cf\colon A\times I \cup X/\sim, (a,1)\sim f(a), (a_{1},0)\sim(a_{2},0)$ and $X\cup CA\to X\cup CA/CA$, have $X/A\to X\cup CA/CA$ continuous, and inverse is continuous, homeomorphic, $A$ is closed, take open set containing it (?)...
\fi
\end{proof}

\ex Prove the first criterion of homotopy equivalence. %(page 16) 

\begin{proof}
We take $h\colon A\times I\to A$, on $A\times\{0\}$ it is identity on $A$ and constant on $A\times\{1\}$.

\[\begin{tikzcd}
X\times I \arrow{r}{f} \arrow[swap]{d}{q\times\id_I} & X \arrow[swap]{d}{q}\\
X/A\times I \arrow{r}{\bar{f}} & X/A
\end{tikzcd}
\]
and find $g\colon X/A\to X$.
Define $\bar{f}(x,t)=f(x,t)$, $\bar{f}([x],t)=[f(x,t)]$. If we define $g\colon X/A\to A, [x]\mapsto f(x,1)$, then it is well defined. Now we want to show, that the compositions are homotopy equivalent to the identities.

$g\circ q\sim \id_X$: $g(q(x))=g([x])=f(x,1)$, just the way we defined it, so $f$ is the homotopy, as $f(-,0)=\id_X$ and $f(-,1)=g\circ q$, \par
$q\circ g\sim \id_{X/A}$: $q(g([x]))=q(f(x,1))=[f(x,1)]=\bar{f}([x],1)$ and $\id_{X/A}=\bar{f}([x],0)$, so in this case the map $\bar{f}$ is homotopy.
\end{proof}

\ex Application of the criterion: two types of suspensions, unreduced and reduced.\par \noindent
Unreduced suspension: $SX=X\times I/\sim$, where $(x_1,0)\sim(x_2,0)$, $(x_1,1)\sim(x_2,2)$.\\
Reduced suspension: $\Sigma X=SX/\{x_0\}\times I=(X,x_0)\wedge(S^1,s_0)$\\
(this might be a homework)\\
The criterion says, that if $\{x_0\}\hookrightarrow X$ is a cofibration, then $SX\simeq \Sigma X$.
$$I\simeq \{(x_0,t),t\in I\}\subseteq SX\longrightarrow SX/\{x_0,t),t\in I\}=\Sigma X$$


\ex There is a lemma, that says: Given the following diagram, where rows are long exact sequences and $m$ is iso,

\[\begin{tikzcd}
K_n \arrow{r}{f} \arrow[swap]{d}{k} & L_n \arrow{r}{g} \arrow[swap]{d}{l} & M_n \arrow{r}{h} \arrow[swap]{d}{m} & K_{n-1} \arrow{r}{} \arrow[swap]{d}{} & L_{n-1} \arrow{r}{} \arrow[swap]{d}{} & M_{n-1} \arrow[swap]{d}{} &\\
\overline{K}_n \arrow{r}{\bar{f}} & \overline{L}_n \arrow{r}{\bar{g}} & \overline{M}_n \arrow{r}{\bar{h}} & \overline{K}_{n-1} \arrow{r}{} & \overline{L}_{n-1} \arrow{r}{} & \overline{M}_{n-1}
\end{tikzcd}
\]
\noindent
we get a long exact sequence
$$K_n\stackrel{f\oplus k}{\longrightarrow} L_n \oplus \overline{K}_n\stackrel{l\oplus \bar{f}}{\longrightarrow}  \overline{L}_n\stackrel{h\circ m^{\!\!-1}\circ \bar{g}}{\longrightarrow} K_{n-1}\longrightarrow  \cdots$$
\noindent
This might be a homework.

\ex There is a lemma, that says: Given the following diagram, where rows are long exact sequences and $m$ is an iso

\[\begin{tikzcd}
K_n \arrow{r}{i} \arrow[swap]{d}{f} & L_n \arrow{r}{j} \arrow[swap]{d}{g} & M_n \arrow{r}{h} \arrow[swap]{d}{m} & K_{n-1} \arrow{r}{} \arrow[swap]{d}{} & L_{n-1} \arrow{r}{} \arrow[swap]{d}{} & M_{n-1} \arrow[swap]{d}{} &\\
\overline{K}_n \arrow{r}{\bar{i}} & \overline{L}_n \arrow{r}{\bar{j}} & \overline{M}_n \arrow{r}{} & \overline{K}_{n-1} \arrow{r}{} & \overline{L}_{n-1} \arrow{r}{} & \overline{M}_{n-1}
\end{tikzcd}
\]
\noindent
we get a long exact sequence
$$K_n\stackrel{(i,f)}{\longrightarrow} L_n \oplus \overline{K}_n\stackrel{g-\bar{i}}{\longrightarrow}  \overline{L}_n\stackrel{i\circ \text{iso}\circ \bar{j}}{\longrightarrow} K_{n-1}\longrightarrow  \cdots$$
\noindent
We can denote $\dd=h\circ m^{-1}\circ\bar{j}$.\\
Show exactness in $L_n\oplus\bar{K}_n$ and also in $\bar{L}_n$.

\begin{proof}
We have $(g-\bar{i})\circ(i,f)=\bar{i}f-gi=0$ obviously. For $x\in L_n,y\in \bar{K}_n$ we have $(g-\bar{i})(x,y)=0$, so $g(x)=\bar{i}(y)$. Now, let $x$ be such that $j(x)=0$, then there is $z\in K_n$ such that $i(z)=x$. Then, suppose $g(x)=a\in\bar{L}_n$, then by $m$ being iso we know $\bar{j}(a)=0$, so exists $y\in\bar{K}_n$ such that $\bar{i}(y)=a$. Since $f(z)$ and $y$ have the same image, their difference has a preimage, i.e. exists $b\in\bar{M}_{n+1}$ such that $b\mapsto y-f(z)$. By iso then there exists $c\mapsto z$, or denote $h(c)=z$. Now, all of this is much easier with a picture (that I don't draw). %I recommend you to attend the tutorial class.
Compute now:\\
$f(z+c)=f(z)+y-f(z)=y$ and $i(z+h(c))=i(z)=x,$ and we are done.\par 
Exactness in $\bar{L}_n$ is easier. It holds $\dd\circ(g-\bar{i})=0$, so take $x\in\ker\dd$ (also, $x\in\bar{L}_n$). Now, $x\mapsto a$, by iso there is $b$ in the upper row that maps to zero. Then there exists $y$ such that $y\mapsto b$. Now we can work with $x-g(y)$. There exists also $z$ such that, obviously, $z\mapsto x-g(y)\mapsto a-a =0$. Get $x=g(y)+\bar{i}(z)=g(y)-\bar{i}(-z)$, that is we needed to express $x$ as this difference, hence we are done.
\end{proof}

\ex There is a long exact sequence of the triple $(X,A,B)$, i.e. $(B\subseteq A\subseteq X)$:
$$\cdots\rightarrow H_n(A,B)\stackrel{i}{\longrightarrow}H_n(X,B)\stackrel{j_{X}}{\longrightarrow}H_n(X,A)\stackrel{D_{\ast}}{\longrightarrow}H_{n-1}(A,B)\rightarrow\cdots,$$
with $H_n(X,A)\stackrel{\dd_{\ast}}{\longrightarrow} H_{n-1}(A)\stackrel{j_A}{\longrightarrow}H_{n-1}(A,B)$. We get this sequence from a special short exact sequence of chain complexes. Show that it is exact and that the triangle commutes, that is $D_{\ast}=j_A\circ\dd_{\ast}$.

\begin{proof}
The chain complex of a pair is a quotient. We have
$$0\longrightarrow\dfrac{C_{\ast}(A)}{C_{\ast}(B)}\stackrel{i}{\longrightarrow}\dfrac{C_{\ast}(X)}{C_{\ast}(B)}\stackrel{j}{\longrightarrow}\dfrac{C_{\ast}(X)}{C_{\ast}(A)}\rightarrow 0.$$
Take $c\in C_{\ast}(A)$, then $ji[c]=j[ic]=[jic]=[c],$ but seen as a different class. So, $j\circ i=0$ and inclusion $\im\  i\subseteq \ker j$ holds. The other inclusion $\ker j\subseteq \im\  i$ is obvious.\par 
Analogous: $0\to C_{\ast}(A)\to C_{\ast}(X)\to \dfrac{C_{\ast}(X)}{C_{\ast}(A)}\to 0$.\par 
Now, show $D_{\ast}$ equals the composition:\\
Note, if you are familiar with the definition, it's clear.\\
Take a chain complex in $C_{\ast}(X)$ with boundary $0$, chain in $C_{\ast}(A)$, take preimage, boundary in $C_{\ast}(B)$, we have $[c]\in C_{n}(X,B)$ and its image  $[c]\in C_{n}(X,A)$, same representative, but different equivalence. Take $\dd c\in C_{n-1}(X,B)$, it has preimage $[\dd c]_{A,B}\in C_{n-1}(A,B)$. (Drawing diagram and chase helps.) The equality of $D_{\ast}$, that it is composition, holds, basically thanks to $j_A$ being inclusion.
\end{proof}

\ex Apply previous exercise to the triple $(D^k,S^{k-1},\ast)$, where $\ast$ is a point.
\begin{proof}
$$\cdots H_n(S^{k-1},\ast)\to H_n(D^k,\ast)\to H_n(D^k,S^{k-1})\stackrel{\varphi}{\rightarrow}H_{n-1}(S^{k-1},\ast)\to H_{n-1}(D^k,\ast)\to\cdots,$$
and, note $H_n(\ast)=\Z$ for $n=0$ and $H_n(\ast)=0$ otherwise.\par 
We also work with reduced homology groups: $\bar{H}_n(X)=H_n(X,x_0)$, $H_{\ast}(X)=\bar{H}_{\ast}(X)\oplus H_{\ast}(\ast)$. Since $D^k$ is contractible, $H_n(D^k,\ast)=\bar{H}_n(D^k)=0$ and $H_{n-1}(D^k,\ast)=0$, so we have
$$\cdots H_n(S^{k-1},\ast)\to 0 \to H_n(D^k,S^{k-1})\stackrel{\varphi}{\rightarrow}H_{n-1}(S^{k-1},\ast)\to 0\to\cdots,$$
where $\varphi$ is $iso$.\par
Reduced homology for pairs is the same as unreduced: $$H_n(D^k,S^{k-1})=\bar{H}_n(D^k,S^{k-1})\cong \bar{H}_{n-1}(S^{k-1}).$$ 
With this we know that $H_n(D^k,S^{k-1})=\bar{H}_n(D^k,S^{k-1})=\Z$ for $n=k$ and it is $0$ for $n\neq k$.\par 
Note also, that $(D^k,S^{k-1})\cong (\Delta^k,\dd \Delta^k)$, this might be useful later on.
\end{proof}

\ex Show that the chain in $C_k(\Delta^k,\dd\Delta^k)$ given by $\id\colon\Delta^k\to\Delta^k$ is the representative of the generator of $$H_k(\Delta^k,\dd\Delta^k)\cong\mathbb Z.$$ 
(Use induction and the long exact sequence for triple.)

\begin{proof}
First denote $\wedge^{k-1}$ boundary without interior of one face. Then work with the triple $\Delta^k,\dd \Delta^k,\wedge^{k-1}$, so the sequence needed is as follows:
$$0\to H_k(\Delta^k,\dd\Delta^k)=\Z\to H_{k-1}(\dd\Delta^k,\wedge^{k-1})\to 0,$$
where we have the zeroes because  $\Delta^k,\wedge^{k-1}$ are contractible to points. Use excision theorem, $H_{\ast}(X-C,A-C)\cong H_{\ast}(X,A)$, where $C$ is the boundary with bottom cut out (imagine upper part of the letter $\Delta$, i.e. triangle). We know that $H_{k-1}(\Delta^{k-1},\dd\Delta^{k-1})=\Z$ and this is isomorphic (by excision theorem) to $H_{k-1}(\dd\Delta^k,\wedge^{k-1})$.  So, everything is $\Z$. We want to show that images of $\id_{k-1}$ and $\id_{k}$ are the same, this is actually the inductive step.\par 
Suppose that the generator in $H_{k-1}(\Delta^{k-1},\dd\Delta^{k-1})$ is given by $\id_{k-1}\colon\Delta^{k-1}\to\Delta^{k-1}$. Then $\id_k\colon \Delta^k\to \Delta^k$ is cycle again, represents element $[\id_k]\in H_{k}(\Delta^{k},\dd\Delta^{k})$.\\ %Ask yourself, is this important?
The Beginning of the Induction \textit{(coming to theaters this summer)}:\par 
$$H_1([-1,1],\{-1,1\})\stackrel{\dd_{\ast}}{\to}H_0(\{-1,1\},\{1\}),$$
take $\id\colon [-1,1]\to [-1,1]$ as a chain complex, $1-(-1)=[-1]$, $[-1]$ generator.
\end{proof}

\ex Using the Mayor-Vietoris exact sequence compute the homology groups of the torus. (note: Vietoris died in 2002, aged 110, remarkable)
\begin{proof}
It goes as union, intersection, pair and we will want to determine the union. First draw two disks with holes (these glued together give a torus). Call one interior of the disk $A$ and the other $B$. We work with $X=A\cup B$, it is not a problem, that we need to work with $A,B$ open, as from the point of view of homology it doesn't matter.\par
The sequence is 
$$\cdots\to H_n(A\cap B)\to H_n(A)\oplus H_n(B)\to H_n(X)\to H_{n-1}(A\cap B)\to\cdots,$$
review: $X=A\cup B$ is torus, $A\cap B=S^1\sqcup S^1$ disjoint union, $H_n(A)=H_n(B)=\Z$ for $n=0,1$ and $0$ otherwise, $H_n(A\cap B)=H_n(A)\oplus H_n(B)=\Z\oplus\Z$ for $n=0,1$ and $0$ otherwise. We can therefore continue with this sequence:
$$H_2(A)\oplus H_2(B)\to H_2(X)\to H_1(A\cap B)\stackrel{f}{\to}H_1(A)\oplus H_1(B)\to H_1(X)\to$$ 
$$\to H_0(A\cap B)\stackrel{g}{\to}H_0(A)\oplus H_0(B)\to H_0(X)\to 0,$$
and we can rewrite it as
$$0\to H_2(X)\to \Z\oplus\Z\stackrel{f}{\to}\Z\oplus\Z\to H_1(X)\to$$ 
$$\to \Z\oplus\Z\stackrel{g}{\to}\Z\oplus\Z\to \Z\to 0,$$
where we use the fact, that torus is connected, so $H_0(X)=\Z$. Now we want to compute $H_2(X)$ and $H_1(X).$\par 
We know $H_2(X)$ is $\ker f$, $\Z\oplus\Z\to \Z\oplus\Z$, $(a,b)\mapsto (a+b,a+b)$, then $(a,-a)$ is in the kernel, so $H_2(X)=\Z[a,-a]=\Z$.\par 
For the $H_1(X)$ group use the fact, that $\ker g$ is $\Z$ (it has same idea, basically). Now consider the sequence
$$0\to \Z\hookrightarrow H_1(X)\to \ker g = \Z \to 0,$$
which splits, so $H_1(X)=\Z\oplus\Z.$ We are done. \par 
Sphere with two handels might be a homework.

\end{proof}

\ex Prove that $S^n$ has a nonzero vector field if and only if $n$ is odd.
\begin{proof}
First note, that we have $v\colon S^n\to \R^{n+1}$ such that $v(x)\perp x$. Consider the case of $S^1$ and $(x_0,x_1)\mapsto (x_1,-x_0)$. Take $(x_0,x_1,x_2,x_3,\dots,x_{2n+1})\in S^{2n+1}\subseteq \R^{2n}$. Then we get $(x_1,-x_0,x_3,-x_2,\dots, x_{2n+1},-x_{2n})$ as image and there is nothing more obvious than that the product is zero, i.e. it's perpendicular.\par 
Note, for $S^1\subseteq \C$ it is $z\mapsto ez$, where $e$ is the complex unit, usually denoted as $i$.\par 
Now we want to prove that if $S^n$ has a nonzero vector field, then $n$ is odd. We use the fact that $\de(\id)=1$ and $\de(-\id)=(-1)^{n+1}$. Take $v\colon S^n\to S^n$. If we show $\id\sim -\id $, then $1=(-1)^{n+1}\Rightarrow n$ is odd. The homotopy is $h(x,t)$ we are looking for is $h(x,t)=cos(t)x+sin(t)v(x)$, where $t\in[0,\pi]$. Also note $\parallel\! h(x,t)\!\parallel=cos^2t+sin^2t=1$. We are done.
\end{proof}

\ex Prove $\de(f)=\sum_{i=1}^k \de(f/x_i)$.

\begin{proof}
We have $f\colon S^n\to S^n$ and $y\in S^n$ has a neighborhood (nbhd) $V$. Denote $f^{-1}(y)=\{x_1,\dots,x_k\}$ and $U_i$ nbhd's of $x_i$, $U_i\cap U_j=\emptyset$. Let degree of $f/x_i$ be $n_i$. We have $f\colon (U_i,U_i-x_i)\to (V,V-y)$ and $(f/U_i)_*\colon H_n(U_i,U_i-x_i)\cong \Z \to H_n(V,V-y)\cong \Z$, so $z\mapsto n_i z$. Consider the following diagram, its evolution (i.e. drawing order) was revealed in the tutorial class:
\[
\xymatrix{
 &\bigoplus H_n(U_i,U_i-x_i) \ar[r]^{(f/U_i)_*} \ar[ld]_{\cong} \ar[d]^k & H_n(V,V-y) \ar[d]^h\\
H_n(S^n,S^n-x_i) & H_n(S^n,S^n-f^{-1}(y)) \ar[r] &H_n(S^n,S^n-y)\\
 &H_n(S^n) \ar[lu]^{\cong} \ar[u]^{j_*} \ar[r]^{f_*} & H_n(S^n) \ar[u]^{j_{y*}}
}
\] 

We have $j_{y*}$ iso and $h$ is iso by excision theorem.\par 
So, we have $H_n(S^n,S^n-f(y))\stackrel{\cong}{\leftarrow} \bigoplus H_n(U_i,U_i-x_i)\cong H_n(\bigsqcup_1^k(U_i,U_i-x_i))$. Iso between $H_n(S^n,S^n-f(y))$ and $H_n(\bigsqcup_1^k(U_i,U_i-x_i))$ by excision. So $k$ is also iso.\par 
We can take $j_{y*}=h\circ \sum(f/U_i)_* \circ k^{-1}\circ j_*$ and for generator $1\in H_n(S^n)$ we have $((j_y)_*\circ f_*)(1)=\de(f)\cdot 1$. We are done.
\[
\xymatrix{
(1,\dots,1) \ar@{->}[r]  & \sum deg(f/x_i) \ar@{->}[d]\\
1 \ar@{|->}[u] &\sum \de(f/x_i)}
\]

\end{proof}



\ex Compute homology groups of $\mathbb{RP}^n$ using CW-structure.

\begin{proof}
Let $X$ be the $CW$-complex, we have $C^{CW}_n=\bigoplus \Z[e^n_{\alpha}]$, $H_n(C^{CW}_*,d)=H_n(X)$. Also, $d[e^n_{\alpha}]=\sum_{\beta}d^{\beta}_{\alpha}[e^{n-1}_{\beta}]$, where $\alpha^{\beta}_{\alpha}$ is the degree of the map 
$$S^{n-1}\stackrel{f_{\alpha}}{\longrightarrow}X^{n-1}\to X^{n-1}/(X^{n-1}-e^{n-1}_{\beta})\cong S^{n-1}.$$
We know that $\mathbb{RP}^n=e^0\cup e^1 \cup \dots \cup e^n$, cell in every dimension. Attaching map is
$$f\colon S^i\to (\mathbb{RP})^{(i)}=\mathbb{RP}^i=S^i/(x\sim -x).$$
Then have 
$$0\to \Z \to \Z \to \cdots \to\Z\to\Z\to\Z\to 0,$$
where $\Z$ appears $n+1$ times. Now, $S^i\stackrel{f}{\to}\mathbb{RP}^i\to \mathbb{RP}^i/\mathbb{RP}^{i-1}\cong S^i$, hence we can apply the attaching map. Class $[y]$ has two preimages $y,-y$, there are nbhd's, $deg(f/y)=\pm 1$. Note that $f/-y=f/y\circ (-\id)$. Now we can compute the degree: $\de(f)=\de(f/y)+\de(f/-y)=\pm(1+(-1)^{i+1})$. It is $\pm 2$ if $i$ is odd and $0$ otherwise. So, we can add to the sequence above:
$$0\to \Z \to \Z \to \cdots \to\Z\stackrel{2\cdot}{\to}\Z \stackrel{0}{\to}\Z\to 0,$$
and compute for $X=\mathbb{RP}^n$, $n$ even: $H_0(X)=\Z,H_1(X)=\Z_2,H_2(X)=0,H_i(X)=0$ for $i$ even, $H_i(X)=\Z_2$ for $i$ odd. The homology is zero for $i>0$. For $n$ even the only difference is $H_n(X)=\Z$.\par 
(Note: this is because for even they are not orientable, for odd it is oriented space.)
\end{proof}

\ex Compute homology groups of oriented two dimensional surfaces. What about nonorientable?

\begin{proof}
Denote $M_g$ surface of genus $g$ (it is the same as sphere with $g$ handles, i.e. $M_1$ is torus and $M_2$ is double torus (homework 4). The $CW$-model is $e^0\cup e^1_1 \cup\dots\cup e^1_{2g}\cup e^2$ and we have 
$$0\to\Z\to \bigoplus_1^{2g}\Z \to \Z\to 0.$$
Second differential is zero, $e^0-e^0=0$. The first one is zero as well (glue the model, the arrows go with $+$ and then $-$). Then we get $H_0=\Z,H_1=\bigoplus_1^{2g}\Z,H_2=\Z$.\par 
For nonorientable surfaces, $N_g$ is modelled by one $2$-dimensional disc which has boundary composed with $g$ segments every of which repeats twice with the same orientation. So we have one cell in dimensions $2$ and $0$ and $g$ cells in dimension one. We get (quite similarly)
$$0\to\Z\to \bigoplus_1^{g}\Z \to \Z\to 0.$$
This equality holds: $d[e^2]=2[e^1_1]+2[e^1_2]+\dots,$ so $H_2=0,H_0=\Z$ and the only interesting case is 
$$H_1=\dfrac{\Z[e^1_1,\dots,e^1_g]}{\Z[2e^1_1+\cdots +2e^1_g]}=\Z_2+\bigoplus_1^{g-1}\Z.$$
\end{proof}

\ex Have $f\colon S^n\to S^n$ map of degree $k$. (such map always exists). Let $X=D^{n+1}\cup_{f}S^n$ and compute homology of $X$ and the projection $p\colon X\to X/S^n$ in homology.

\begin{proof}
Easy, $X=e^0\cup e^n\cup e^{n+1}$ and $0\to\Z\stackrel{k\cdot}{\to}\Z\to0\to\dots\to0\to\Z\to0$, also
$$S^n\to X^{(n)}\to X^{(n)}/(X^{(n)}-e^n)=S^n.$$
We get $H_{n+1}(X)=0,H_n=\Z_k,H_0(X)=\Z$. Note, $X/S^n\cong S^{n+1}$. So, for the $p_*$ we have 
$$p_*\colon H_{n+1}(X)=0\stackrel{0}{\to}H_{n+1}(S^{n+1})=\Z$$
and 
$$p_*\colon H_n(X)=\Z_k\stackrel{0}{\to}H_n(S^{n+1})=0$$
and at $H_0$ it is identity $\Z\to\Z$.\par 
\end{proof}

\ex Prove the following equalities (assuming some conditions):
\begin{align*}
 \bar{H}_{*} (X \vee Y) &= \bar{H}_{*}(X) \oplus \bar{H}_{*}(Y)\\
 \bar{H}_{*} (\bigvee_{i = 1}^n X_i) &= \bigoplus_{i=1}^n \bar{H}_{*}(X_i)\\
 \bar{H}_{*} (\bigvee_{i = 1}^\infty X_i) &= \bigoplus_{i=1}^\infty \bar{H}_{*}(X_i)
\end{align*}

\begin{proof}
Denote $z$ the distinguished point of $X \vee Y$. For the pair $(X \vee Y, X)$ we have the following long exact sequence%
\[ \dots \rightarrow \bar{H}_{i+1} (X\vee Y,X) \xrightarrow{0} \bar{H}_i (X) \rightarrow \bar{H}_{i}(X \vee Y) \rightarrow \bar{H}_{i}(X \vee Y, X) \xrightarrow{0} \bar{H}_{i-1}(X) \rightarrow \cdots.\]
Thus we have short exact sequence, which splits, because we have continuous (cts) map $\id \vee \operatorname{const}_z \colon X \vee Y \rightarrow X$ (which maps $Y$ to $z$). Thus we have $\bar{H}_{*} (X \vee Y) \cong \bar{H}_{*} (X) \oplus \bar{H}_{*} (X \vee Y, X)$. Now it remains to prove $\bar{H}_{*} (X \vee Y, X) \cong \bar{H}_i (Y)$. If $X \vee Y$ is a CW-complex and $X$ its subcomplex, it is known that $\bar{H}_i (X \vee Y, X) \cong \bar{H}_i (X \vee Y / X) = \bar{H}_i (Y)$. More generally, let $U$ be some (sufficiently small) neighborhood of $z$ in $X$. From excision theorem we have:%
\[\bar{H}_i (X \vee Y, X) \cong \bar{H}_i \big(X \vee Y \smallsetminus (X \smallsetminus U), X\smallsetminus (X\smallsetminus U)\big) = \bar{H}_i (U \vee Y, U).\] Because $U$ should be\footnote{It is true at least for $X$ locally contractible. It is not true generally.} contractible, $\bar{H}_i (U \vee Y, U) \cong \bar{H}_i (Y, z) = \bar{H}_i (Y)$.\par
The second equality we get from the first by induction.\par
Let us prove the third equality. Denote $Y_n = X_1 \vee X_2 \vee \dots \vee X_n$ and $Y = \bigvee_{n=1}^\infty Y_n$ and denote $z$ the distinguished point of $Y$ and $Y_n$ for every $n$. We have the following diagram (where each arrow is an inclusion):%
\[\xymatrix @=.5cm {%
 C_{*}(Y_1, z) \ar[dr] \ar[r] & C_{*}(Y_2, z) \ar[d] \ar[r] & \cdots \ar[dl]\\
 & C_{*} (Y, z)}
\]
Since $\Delta^k$ is compact, every continuous (cts) map $\Delta^k \rightarrow Y$ has image in some $Y_n$, thus it is easy to prove $C_{*} (Y,z) = \operatorname{colim} C_{*}(Y_n,z)$, thus%
\[\bar{H}_{*} (Y) = \operatorname{colim} \bar{H}_{*}(Y_n) = \operatorname{colim} \bigoplus_{i=1}^n X_i = \bigoplus_{i=1}^\infty X_i. \qedhere\]
\end{proof}


Let $X$ be a topological space with finitely generated homological groups and let $H_i(X) = 0$ for each sufficiently large $i$. Every finitely generated abelian group can be written as $\underbrace{\Z \oplus \Z \oplus \dots \oplus \Z}_{k-\text{times}} \oplus \tor{}$, where $\tor{}$ denote torsion part of the group. The number $k$ is called the \emph{rank} of the group.\par
Euler characteristic $\chi$ of $X$ is defined by:
\[\chi(X) = \sum_{i=0}^\infty (-1)^i \rk H_i(X)\]
\par \noindent
\textnormal{\textbf{Example.}}\ We know $H_i(S^n) = \begin{cases} \Z, & i=0,n\\ 0, &\text{otherwise.} \end{cases}$ Thus $\chi (S^n) = 1 - (-1)^{n}$.

\ex Let $(C_{*}, \partial)$ be a chain complex with homology $H_{*} (C_{*})$. Prove that $\chi (X) = \chi(C_{*})$, where
\[ \chi(C_{*}) = \sum_{i=0}^\infty (-1)^i \rk C_i. \]

\begin{proof}
We have two short exact sequences:
\begin{gather*}
 0 \rightarrow Z_i \hookrightarrow C_i \xrightarrow{\partial} B_{i-1} \rightarrow 0\\
 0 \rightarrow B_i \hookrightarrow Z_i \rightarrow Z_i / B_i = H_i \rightarrow 0,
\end{gather*}
where $C_i$, cycles $Z_i$ and boundaries $B_i$ are free abelian groups, thus $\rk C_i = \rk Z_i + \rk B_{i-1}$ and $\rk H_i = \rk Z_i - \rk B_i$. Thus we have
\begin{align*}
 \chi (C_{*}) &= \sum_{i=0}^\infty (-1)^i \rk Z_i + \sum_{i=0}^\infty (-1)^i \rk B_{i-1}\\
 &= \sum_{i=0}^\infty (-1)^i \rk Z_i - \sum_{i=0}^\infty (-1)^i \rk B_{i} = \chi (X). \qedhere
\end{align*}
\end{proof}

Let $X$ be a topological space with finitely generated homological groups and let $H_i(X) = 0$ for every sufficiently large $i$. Let $f \colon X \rightarrow X$ be a continuous map. Map $f$ induces homomorphism on the chain complex $f_{*} \colon C_{*}(X) \rightarrow C_{*}(X)$ and on the homologiy groups $H_{*}f \colon H_{*} (X) \rightarrow H_{*} (X)$, where $H_{*}f (\tor H_{*}(X)) \subseteq \tor H_{*}(X)$. Thus it induces homomorphism
\[ H_{*}f \colon H_{*} (X)/ \tor H_{*}(X) \rightarrow H_{*} (X)/ \tor H_{*}(X).\]
Since $H_{*} (X)/ \tor H_{*}(X) \cong \underbrace{\Z \oplus \Z \oplus \dots \oplus \Z}_{\rk H_{*}(X)}$, map $H_{*}f$ can be written as a matrix, thus we can compute its trace. So we can define the Lefschetz number of a map $f$:
\[ L(f) = \sum_{i=0}^\infty (-1)^i \tr H_if .\]\par
Similarly to the case of the Euler characteristic, it can be proved that\footnote{$f_i \colon C_{i}(X) \rightarrow C_{i}(X)$}
\[ \sum_{i=0}^\infty (-1)^i \tr H_if = \sum_{i=0}^\infty (-1)^i \tr f_i .\]
\par \noindent
\textnormal{\textbf{Theorem.}}\ If $L(f) \neq 0$, then f has a fixed point.

\ex Use the theorem above to show, that every cts map $f$ on $D^n$ and $\R P^n$ where $n$ is even has a fixed point.

\begin{proof}
 We know that that $H_i(D^n) = \begin{cases} \Z, & i=0\\ 0, &\text{otherwise.} \end{cases}$ Because $H_0 f \colon H_0(D^n) \cong \Z \rightarrow \Z \cong H_0(D^n)$ can be only the identity, we have $L(f)=1$, thus $f$ has a fixed point.\par
 Since $H_i(\R P^n) = \begin{cases} \Z, & i=0,\\ \Z/2, & i<n, i \text{ odd};\\ 0, &\text{otherwise}, \end{cases}$ and $\Z/2$ is torsion, we have $L(f)=1$ as in the previous case.
\end{proof}

\ex Let $M$ be a smooth compact manifold. Prove, that there is a nonzero vector field on $M$ if and only if $\chi (M) = 0$.

\begin{proof}
We will prove only implication ${\Rightarrow}$. Let $v$ be a nonzero vector field on $M$. Define a map $X \colon [0,1] \times M \rightarrow M$ which satisfies $\dot{X}(t,x) = v(X(t,x))$ for every $x \in M$ and $X(0,x) = x$. There exists $t_0$ such that $X(t_0, x) \neq x$. Denote $f(x) = X(t_0,x)$, thus $f$ has no fixed point, thus $L(f)=0$. Because $f$ is homotopic to $\id$ and $\tr H_i\id = \rk H_i(M)$, we get from homotopy invariance $0 = L(f) = L(\id) = \chi(M)$.
\end{proof}

\ex Use $\Z/2$ coefficients to show, that every cts map $f \colon S^n \rightarrow S^n$ satisfying $f(-x) = -f(x)$ has an odd degree.

\begin{proof}
The map $f$ induces a map $g \colon \R P^n \rightarrow \R P^n$, since $f(\{x,-x\}) \subseteq \big\{f(x), -f(x)\big\}$. We have the short exact sequence\footnote{$2 \sigma = 0$ because of the $\Z/2$ coefficient.}
\[ \xymatrix @C=.5cm @R=.1cm {%
 & \sigma \ar@{|->}[r] & \sigma_1 + \sigma_2 \ar@{|->}[r] & 2 \sigma = 0 & \\
 0 \ar[r] & C_{*}(\R P^n, \Z/2) \ar[r] & C_{*}(S^n, \Z/2) \ar[r] & C_{*}(\R P^n, \Z/2) \ar[r] & 0,}
\]
where $\sigma \colon \Delta^i \rightarrow \R P^n$ is an arbitrary element of $C_{*}(\R P^n)$, $\sigma_1, \sigma_2$ are its preimages of a projection:
\[ \xymatrix @R=.5cm @C=1.5cm {%
 & S^n \ar[d]\\
 \Delta^i \ar[ur]^{\sigma_1, \sigma_2} \ar[r]_\sigma & \R P^n}\]
From the short exact sequence we get the long exact sequence
\[\xymatrix @=1cm{%
 H_i(\R P^n; \Z/2) \ar[r] \ar[d]^{g_{*}} & H_i(S^n; \Z/2) \ar[r] \ar[d]^{f_{*}} & H_i(\R P^n; \Z/2) \ar[r] \ar[d]^{g_{*}} & H_{i-1}(\R P^n; \Z/2) \ar[d]^{g_{*}} \ar[r] & 0\\
 H_i(\R P^n; \Z/2) \ar[r] & H_i(S^n; \Z/2) \ar[r] & H_i(\R P^n; \Z/2) \ar[r] & H_{i-1}(\R P^n; \Z/2) \ar[r] & 0}
\]
Because $H_0(\R P^n; \Z/2)=\Z/2$ and $g_{0}$ on $H_0(\R P^n;\Z/2)$ is an isomorphism, we can show by induction, that $H_i(\R P^n, \Z/2)=\Z/2$ and $g_{i}$ is an isomorphism for every $i \leq n-1$. An induction step is shown on the following diagram (three isomorphisms imply the fourth):
\[\xymatrix @=1cm{%
 0 \ar[r] & \Z/2 \ar[r]^{\cong} \ar@{-->}[d]^{\cong} & \Z/2 \ar[d]^{\cong}\\
 0 \ar[r] & \Z/2 \ar[r]_{\cong} & \Z/2}
\]
For $i=n$ we have the following situation (the vertical isomorphisms were proved by induction):
\[\xymatrix @=1cm{%
 \Z/2 \ar[r]^{\cong} \ar[d]^{\cong} & \Z/2 \ar[r]^{0} \ar[d]^{?} & \Z/2 \ar[r]^{\cong} \ar[d]^{\cong} & \Z/2 \ar[r] \ar[d]^{\cong} & 0\\
 \Z/2 \ar[r]_{\cong} & \Z/2 \ar[r]^0 & \Z/2 \ar[r]_{\cong} & \Z/2 \ar[r] & 0}
\]
Thus $f_{*}$ (the arrow marked by $?$) has to be an isomorphism for $H_n$, thus it maps $[1]_2$ to $[1]_2$, hence $f$ has degree $1 \mod 2$.
\end{proof}

\ex Use theorem "$f\colon S^n\to S^n$ s.t. $f(-x)=-f(x) \Rightarrow \operatorname{deg}(f)$ is odd" to prove that for $g\colon S^n \to \R^n$ there exists $x\in S^n\colon g(x)=g(-x)$.

\begin{proof}
We will use proof by contradiction. Suppose that $g(x)-g(-x)$ is always non-zero and define $f(x)=\dfrac{g(x)-g(-x)}{||g(x)-g(-x)||}$. Then obviously $f(-x)=-f(x)$. Since $f\colon S^n\to S^{n-1}\subset S^n$ is odd , the map $f$ has odd degree, but $f(S^n)\subsetneq S^n$ which implies $\operatorname{deg}(f)=0$ ($f$ is homotopic to a constant map). A contradiction.
\end{proof}

\ex Let $\varphi\in C^k(X;R), \psi\in C^l(Y;R)$. Prove $\delta(\varphi\cup \psi)=\delta\varphi\cup\psi+(-1)^{k}\varphi\cup\delta\psi$. Use $\tau=[e_0,\dots,e_{k+l+1}]\in C_{k+l+1}(X)$.

\begin{proof}

Easily work out 
\begin{align*}
    \delta(\varphi\cup\psi)(\tau)&=(\varphi\cup\psi)(\delta\tau)=(\varphi\cup\psi)\Big(\sum_{i=0}^{k+l+1}(-1)^i \tau/[e_0,\dots,\hat{e_i},\dots,e_{k+l+1}]\Big)=\\
=&\sum_{i=0}^k (-1)^i \varphi(\tau/[e_0,\dots,\hat{e_i},\dots,e_{k+1}])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])+\\
+&\sum_{i=k+1}^{k+l+1} (-1)^i\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_k,\dots,\hat{e_i},\dots,e_{k+l+1}]).
\end{align*}
Now, the right hand side of the formula, the first part gives
\begin{align*}
    (\delta\varphi\cup\psi)(\tau)&=\delta\varphi(\tau/[e_0,\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=\\
                                 &=\varphi(\delta\tau/[e_0,\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=\\
                                 &=\sum_{i=0}^{k+1}(-1)^i \varphi(\delta\tau/[e_0,\dots,\hat{e_i},\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}]).
\end{align*}
The second part is
\begin{align*}
    (-1)^k(\varphi\cup\delta\psi)(\tau)&=(-1)^k\varphi(\tau/[e_0,\dots,e_{k}])\delta \psi(\tau/[e_{k},\dots,e_{k+l+1}])=\\
    &=\sum_{j=0}^{l+1}(-1)^{j+k} \varphi(\delta\tau/[e_0,\dots,e_{k}])\cdot \psi(\tau/[e_{k},\dots,\hat{e}_{k+j},\dots,e_{k+l+1}]).
\end{align*}
Now, the last summand of the first part plus the first summand of the second part yields
\begin{align*}
(-1)^{k+1}&\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])+\\
+(-1)^k&\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=0,    
\end{align*}
and we are done, $\mathsf{LHS=RHS}$.

\end{proof}

\ex Prove $a\cup b=\Delta^*(a\times b)$, where $a\in H^k(X),b\in H^l(X),\Delta\colon X\to X\times X,x\mapsto (x,x)$ and $\times$ is cross product defined $\alpha\times \beta=p_X^* \alpha \cup p_Y^*\beta$ ($p_X,p_Y$ are projections from $X\times Y$).

\begin{proof}
By the following diagram
\[
\xymatrix{
X \ar@{->}[rr]^{\Delta_X}  & & X\times X \ar@{->}[ld]_{p_X=p_1} \ar@{->}[d]^{p_Y=p_1} & \\
& X & X}
\]
we have that $p_X\Delta$ is $\id$. So, compute
$$\Delta^*(a\times b)=\Delta^*(p_X^*a\cup p_Y^*b)=\Delta^*(p_X^*a)\cup\Delta^*(p_Y^*b)=(p_X\Delta)^* a\cup (p_Y\Delta)^*b,$$
and we are done. The thing is that cup product is natural.
\end{proof}

\ex Compute the structure of graded algebra $H^*(S^n\times S^n;\Z)$ for $n$ even and $n$ odd. Use the following:\par 
If $H^n(Y;R)$ is free finitely generated group for all $n$ and $(X,A),Y$ are CW-complexes, then
$$\times\colon H^*(X,A;R)\otimes H^*(Y;R)\to H^*(X\times Y,A\times Y;R)$$
is an isomorphism of graded rings.

\begin{proof}
We will omit writing the $\Z$ coefficients.\par 
Now, $H^*(S^n)\otimes H^*(S^n)\to H^*(S^n\times S^n)$ and we know that for spheres $H^0=\Z$ with generator $1$ and $H^n=\Z$, denote generator $a$. Also, $a\cup a\in H^{2n}=0, a\cup a=0$, so we get $\Z[a]/\langle a^2\rangle$ and $\operatorname{deg}(a)=n$. We can write the same for the second, so denote the other generator $b$ and have $\operatorname{deg}(b)=n$ and we have $\Z[b]/\langle b^2\rangle$.\par 
Now we compute tensor product $\Z[a]/\langle a^2\rangle\otimes \Z[b]/\langle b^2\rangle$, we have four generators: $1_a\otimes 1_b,a\otimes 1_b,1_a\otimes b,a\otimes b$, we will denote them $1,c,d,c\cdot d$. Compute
$$(a\otimes 1_b)\cdot (1_a\otimes b)=(-1)^{0\cdot 0}(a\cdot 1_a)\otimes (1_b\cdot b)=a\otimes b,$$
because $0$ is an idempotent element, i.e. $0\cdot 0=0$, and $(-1)^n=1$ for $n$ even, again, as in the first exercise, we use Evenness of Zero. (We refer the reader to "Principia Mathematica" \textit{Whitehead, Russell},(1910,1912,1913).) Continue with computation
$$(1_a\otimes b)\cdot (a\otimes 1_b)=(-1)^{n\cdot n}(1_a\cdot a)\otimes (b\cdot 1_b)=(-1)^na\otimes b,$$
so the algebra we get is $H^*(S^n\times S^n)=\Z[c,d]/\langle c^2,d^2,dc-(-1)^ncd\rangle$. For $n$  even we have $dc=cd$.
\end{proof}


\ex  Prove that there is no multiplication on even dimensional spheres. Multiplication on the sphere $S^n$ is a map $m\colon S^n\times S^n\to S^n$ such that there is an element $1\in S^n$ satisfying $m(x,1)=x,m(1,x)=x$.\par 
Hint: compute $m^*\colon H^*(S^n)\to H^*{(S^n\times S^n)}$, describe two rings.
\begin{proof}
We have $H^*(S^n)=\Z[\gamma]/\langle\gamma^2\rangle$ and $H^*(S^n\times S^n)=\Z[\alpha,\beta]/\langle\alpha^2,\beta^2\rangle=H^*(S^n)\otimes H^*(S^n)$, because we already know, that $\alpha\beta=\beta\alpha$. Our situation can be described with two diagrams:
\[
\xymatrix{
S^n \ar[r]^{i_1} \ar[dr]_{\id} & S^n\times S^n \ar[d]^m\\
& S^n
}\quad
\xymatrix{
H^{\ast}(S^n) & H^*(S^n\times S^n)  \ar[l]_{i_1^*}\\
& H^*(S^n) \ar[u]_{m^*} \ar[lu]^{\id}
}
\]

Take $m^*(\gamma)=a\alpha+b\beta$ with $a,b\in\Z$ and prove first, that $a=b=1$. Use $m\circ i_1=\id$, so $i_1^*(m^*\gamma)=\gamma$. This gives $i_1^*(a\alpha+b\beta)=\gamma$ and since $i_1^*(a\alpha+b\beta)=a\gamma$, we have $a=1$, same for $b$. Final computation yields
\begin{align*}
0&=m^*(0)=m^*(\gamma^2)=m^*(\gamma\cup\gamma)=m^*\gamma\cup m^*\gamma=\\&=(\alpha+\beta)\cup(\alpha+\beta)=\alpha^2+\alpha\beta+\beta\alpha+\beta^2=0+2\alpha\beta+0\neq 0,
\end{align*}
and that, my friends, is a contradiction.
\end{proof}

\ex (unfinished) Use five lemma to prove that taking any two $\mu$'s iso's, the third $\mu$ is iso as well. (Proving five lemma might be a homework.) Also, show commutativity of the diagram.

\[
\xymatrix{
H^*(X,A) \otimes H^*(Y) \ar[rr] \ar[dd]^{\mu} &  & H^*(X)\otimes H^*(Y)\ar[dd]^{\mu} \ar[ld]\\
 & H^*(A)\otimes H^*(Y)\ar[dd]^<<<<{\mu} \ar[ul]_{\delta^*\otimes\id} & \\
H^*(X\times Y,A\times Y)\ar[rr] &  & H^*(X\times Y)\ar[ld]\\
 & H^*(A\times Y) \ar[lu]_{\delta^*}& 
}
\]

\begin{proof}
We will name the parts of the diagram as follows: back-square, upper-triangle, lower-triangle, left-square, right-square. The triangles come from the long exact sequence of of pairs $(X,A)$ and $(X\times Y, A\times Y)$, the right-square commutativity comes from an inclusion, back-square commutes as well (topology knowledge). The only problematic part is the left-square and it is exercise on computation of connecting homomorphism $\delta^*$.
\end{proof}


\ex[Five lemma] 
Let
$$\begin{xy}
\xymatrix{%@L+2mu@+1em{
 A \ar[r]^*{}
     \ar[d]^*{\cong}
&
 B \ar[r]^*{}
     \ar[d]^*{\cong}
&
 C \ar[r]^*{}
     \ar[d]^*{f}
&
 D \ar[r]^*{}
     \ar[d]^*{\cong}
&
 E \ar[d]^*{\cong}
\\
 \ov{A} \ar[r]_*{}
&
 \ov{B} \ar[r]_*{}
&
 \ov{C} \ar[r]_*{}
&
\ov{D} \ar[r]_*{}
&
 \ov{E}
}\end{xy}
$$
 be a commutative diagram of modules with exact rows. Show that the middle homomorphism $f$ is an isomorphism.
\begin{proof}
We will show the surjectivity of $f$, injectivity is dual. Let $\ov{c}\in \ov{C}$ be arbitrary and suppose it maps to $\ov{d}\in \ov{D}$. This $\ov{d}$ corresponds to some $d\in D$, and since $\ov{d}$ maps to $0$ in $\ov{E}$ by exactness, $d$ also has to map to $0\in E$ by commutativity. By exactness, there exists $c\in C$ which maps to $d$. Since $\ov{c}$ and $f(c)$ both map to $\ov{d}$ by commutativity, there is (by exactness) some $\ov{b}\in \ov{B}$ which maps to $f(c)-\ov{c}$. This $\ov{b}$ corresponds to some $b\in B$, which maps to some $c'\in C$. Then by commutativity, $f(c-c')=f(c)-(f(c)-\ov{c})=\ov{c}$, as desired. (Note that instead of the four vertical maps being isomorphisms, we only needed the surjectivity of $B\to\ov{B}$, $D\to\ov{D}$ and the injectivity of $E\to\ov{E}$).
\end{proof}

\ex 
Show that for a finite CW-complex $X$ and $H^*(Y)$ being finitely generated free group in all dimensions, the cross product
$$H^*(X)\otimes H^*(Y)\xrightarrow{\mu} H^*(X\times Y)$$ is an isomorphism. (In fact, the same is true for $X$ being an infinite CW-complex.)

\begin{proof}
First let $X=pt$ be a point. Then $H^*(pt)=\Z$ with $1\in H^0(pt)$ and $pt\times Y$ is homeomorphic to $Y$, hence $$H^*(pt)\otimes H^*(Y)=\Z\otimes H^*(Y)\cong H^*(Y)\cong H^*(pt \times Y).$$
Now let $X=p_1\sqcup p_2\sqcup\dots \sqcup p_k$ be a finite disjoint union of points (i.e, a discrete set). Then $H^*(X)=\bigotimes_{i=1}^k \Z$, hence
\begin{equation*}
\begin{split}
H^*(X)\otimes H^*(Y)&=\left( \bigoplus_{i=1}^k \Z \right)\otimes H^*(Y)\cong \bigoplus_{i=1}^k\left(  \Z \otimes H^*(Y)\right)\cong \bigoplus_{i=1}^k H^*(Y)\\
&\cong H^*(\underbrace{Y\sqcup Y\sqcup\dots\sqcup Y}_{n \text{ times}})\cong H^*(X \times Y)
\end{split}
\end{equation*}
(We should also show that the isomorphism is indeed given by $\mu$, but if $e_1,\dots,e_k\in H^0(X)$ are such that $$e_i(p_j)=\begin{cases} 1 & \text{ if } i=j\\ 0 & \text{ if } i\neq j\end{cases},$$ it's not hard to see that $$\mu(e_i\otimes a)=(0,0,\dots,\underbrace{a}_{i\text{-th place}},0,\dots,0)$$ using projections and the definition of the cup product.)

Now we can proceed inductively:

\begin{enumerate}[i)]
\item We now know that the theorem is true for $X$ of dimension $0$.
\item Since $D^n$ is homotopy equivalent to a point, the theorem is also true for $X=\bigsqcup_{\alpha=1}^k D_{\alpha}^n$.
\item Suppose that the theorem holds for finite CW-complexes of dimension $n-1$. Then it is also true for pairs $(\bigsqcup D_{\alpha}^n,\bigsqcup S_{\alpha}^{n-1}).$
\item Since $H^*(X,A)\cong H^*(X/A)$ for a subcomplex $A\subseteq X$, the theorem also holds for $\bigvee S_{\alpha}^n\cong\bigsqcup D_{\alpha}^n/\bigsqcup S_{\alpha}^{n-1}$.
\item Now let $X=X^{(n)}$ be an $n$-dimensional CW-complex and consider the diagram
\[
\xymatrix{
H^*(X,A) \otimes H^*(Y) \ar[rr] \ar[dd]^{\mu} &  & H^*(X)\otimes H^*(Y)\ar[dd]^{\mu} \ar[ld]\\
 & H^*(A)\otimes H^*(Y)\ar[dd]^<<<<{\mu} \ar[ul]_{\delta^*\otimes\id} & \\
H^*(X\times Y,A\times Y)\ar[rr] &  & H^*(X\times Y)\ar[ld]\\
 & H^*(A\times Y) \ar[lu]_{\delta^*}& 
},
\]
where the lower triangle represents a long exact cohomology sequence by definition, and the same is true for the upper triangle (since free modules are flat). We know that the theorem holds for $X^{(n-1)}$, and also for $X^{(n)}/X^{(n-1)}\cong \bigvee S_{\alpha}^{n-1}$. Therefore we can use the Five lemma after unfolding the diagram in the appropriate dimensions, from which it follows that the theorem also holds for $X^{(n)}$. This completes the induction.
\end{enumerate}
\end{proof}

\ex Compute the cohomology rings of $\C P^2 \times S^6$ and $\C P^2 \vee S^6$.
\begin{proof}
We have $H^*(\C P^2)=\Z[w]/\langle w^3 \rangle$ for $w\in H^2$ and $H^*(S^6)=\Z[a]/\langle a^2 \rangle$ for $a\in H^6$, hence
$$H^*(\C P^2 \times S^6)=\Z[w]/\langle w^3 \rangle\otimes\Z[a]/\langle a^2 \rangle \cong \Z[w,a]/\langle w^3,a^2 \rangle.$$
Next, it is true in general that $\ov{H}^*(X\wedge Y)\cong \ov{H}^*(X)\oplus \ov{H}^*(Y) $ is an isomoprhism of graded rings (this can be proven proven straight from the definitions, but it takes some time). Since $\C P^2 \vee S^6$ is connected, we have $H^*(\C P^2 \vee S^6)\cong \ov{H}^*(\C P^2)\oplus \ov{H}^*(S^6)\oplus\Z$. Now $w\cup a\in H^8=0$ (more generally, we could use that fact that $(w,0)\cup (0,a)=(0,0)$). Therefore $$H^*(\C P^2 \vee S^6)\cong \Z[w,a]/\langle w^3,a^2,wa \rangle.$$
\end{proof}

\ex Show that the  $\C P^2 \vee S^6$ is not homotopy equivalent to $\C P^3$.
\begin{proof}
It suffices to show that the cohomology rings of these spaces are not isomorphic (note that the additive group structure is not enough to distinguish them). We already know that $$H^*(\C P^2 \vee S^6)\cong \Z[w,a]/\langle w^3,a^2,wa \rangle$$ and we have $H^*(\C P^3)=\Z[b]/\langle b^4 \rangle$ for $b\in H^2$. Any isomorphism would have to map $w$ and $b$ to $\pm$ each other (these are the respective generators in dimension 2), but $w^3=0$ while $b^3\neq 0$, so this is not possible.
\end{proof}

Define the $n$-th homotopy group of the space $X$ with the base point $x_0$ as the group of homotopy classes of the cts\footnote{continuous} maps $(I^n,\dd I^n) \rightarrow (X,x_0)$ with the operation given by prescription:
\[(f+g)(t_1, \dots, t_n) = \begin{cases}
                            f(2 t_1, t_2, \dots, t_n) & 0 \leq t_1 \leq \frac{1}{2},\\
                            g(2 t_1 - 1, t_2, \dots, t_n) & \frac{1}{2} \leq t_1 \leq 1.
                           \end{cases}
\]
Denote it $\pi_n(X,x_0)$.

\ex Show the operation on $\pi_n(X,x_0)$ is associative.

\begin{proof}
We want to show $(f+g)+k \sim f+(g+k)$. We will find prescription for the homotopy by the following diagram:

{\begin{center}
\includegraphics[height=3cm]{assoc.pdf}
\end{center}}

In the following notation\footnote{and some other following notations}, understand $f(t_1)$ as $f(t_1,t_2, \dots, t_n)$ for all $t_2, \dots, t_n \in I$.
\[h(s,t_1)=\begin{cases}
            f(\frac{4}{1+s}t_1) & (s,t_1) \in \mathrm{I}\phantom{\mathrm{II}} \quad (t_1 \in [0,\frac{1}{4}+s\frac{1}{4}], s\in[0,1])\\
            g(4 t_1 - (1+s)) & (s,t_1) \in \mathrm{II}\phantom{\mathrm{I}} \quad (t_1 \in [\frac{1}{4}+s\frac{1}{4},\frac{1}{2}+s\frac{1}{4}], s\in[0,1])\\
            k(\frac{4}{2-s} t_1 - \frac{2+s}{2-s}) & (s,t_1) \in \mathrm{III} \quad (t_1 \in [\frac{1}{2}+s\frac{1}{4},1], s\in[0,1])
           \end{cases}
\]
\end{proof}

\ex Show that the element given by prescription
\[(-f)(t_1,\dots,t_n) = f(1-t_1,t_2,\dots,t_n)\]
is really the inverse element of $f$.

\begin{proof}
 We want to show $f + (-f) \sim const$. The constant will be (function given by) the point $f(2 t_1)=f(0)$. Again let us draw a diagram (the square (its boundary) in the middle sign the same value; the wavy line sign one value too).
 
{\begin{minipage}[b]{0.66\textwidth}
 \[h(s,t_1)=\begin{cases}
             f(2 t_1) & (s,t_1) \in \mathrm{I}\\
             f(\frac{1-s}{2})=(-f)(\frac{1+s}{2}) & (s,t_1) \in \mathrm{II}\\
             (-f)(2 t_1 - 1) = f(2 t_1) & (s,t_1) \in \mathrm{III},
            \end{cases}
 \]
\end{minipage}%
\begin{minipage}[c]{0.33\textwidth}
\centering
\includegraphics[height=2.5cm]{inverse.pdf}
\end{minipage}}

\noindent
 where  $\mathrm{II} = \{(s,t_1) \mid s \in [0,1], t_1 \in [\frac{1-s}{2},\frac{1+s}{2}])$.
\end{proof}
\noindent
\begin{rem} One can see, that proving by pictures is much more pleasant.
\end{rem} 


There is a long exact sequence:
\[ \dots \rightarrow \pi_{n+1}(X,A,x_0) \xrightarrow{\ \dd\ } \pi_n(A,x_0) \xrightarrow{\ i_*\ } \pi_n(X,x_0) \xrightarrow{\ j_*\ } \pi_n(X,A,x_0) \xrightarrow{\ \dd\ } \pi_{n-1}(A,x_0) \rightarrow \dots\]

\ex Show the exactness of this sequence in $\pi_n (X,A,x_0)$ and $\pi_n (A,x_0)$.

\begin{proof}
 At first we will show the exactness in $\pi_n (X,A,x_0)$.\par
 Let us show the inclusion ``$\im j_* \subseteq \krnl \dd$\,''. Take an arbitrary $f \in \pi_n(X,x_0)$, thus $f\colon (I^n,\dd I^n) \rightarrow (X,x_0)$. From definition $j_*(f) = j \circ f\colon (I^n, \dd I^n, J^{n-1}) \rightarrow (X,A,x_0)$, where $J^{n-1}=I^n-I^{n-1}$, and $\dd([f])=[f|_{I^{n-1}}]=const$, since $f$ is constant on whole $\dd I^n \supseteq I^{n-1}$.\par
 ``$\im j_* \supseteq \krnl \dd$\,'': Take an arbitrary $g \in \krnl \dd \subseteq \pi_n(X,A,x_0)$, thus $g\colon (I^n, \dd I^n, J^{n-1}) \rightarrow (X,A,x_0)$. Since $g \in \krnl \dd$, there is the homotopy $h\colon (I^{n-1}, \dd I^{n-1}) \times I \longrightarrow (A,x_0)$ such that $h(x,0) = g|_{I^{n-1}}(x)$ and $h(x,1) = const$. Because $h(x,t) \in A$ and $h(x',t)=x_0$ for all $x \in I^{n-1}$, $x \in \dd I^{n-1}$ and $t \in [0,1]$, we can take $f \in \pi_n(X,x_0)$ defined by %\footnote{consider $x \in I^{n-1}$}
 \[
 f(x,t)= \begin{cases}
          g(x,2t) & \text{for } t \in [0,\tfrac{1}{2}]\\
          h(x,2t-1) & \text{for } t \in [\tfrac{1}{2},1] .
         \end{cases}
 \]
 It is not hard to prove that $j_* (f)$ is homotopic to $g$, see picture below.
 \begin{center}
  \includegraphics[height=5cm]{homotopic.pdf}
 \end{center}
 
 Now, let us show the exactness in $\pi_n (A,x_0)$.\par
 ``$\im \dd \supseteq \krnl i_*$\,'' Let $f \in \krnl i_* \subseteq \pi_n (A,x_0)$ be an arbitrary. Because $i_* f \sim const$, we have homotopy $h\colon (I^n,\dd I^n) \times I \rightarrow (X,x_0)$ such that $h(x,0)=f(x)$ and $h(x,1)=x_0$. It holds $h \in \pi_{n+1} (X,A,x_0)$, since $h(x,0) \in A$, $h(x,1)=x_0$ and $h(x',t)=x_0$ for all $x \in I^n$ and $x' \in \dd I^n$.\par
 ``$\im \dd \subseteq \krnl i_*$\,'' Let $h \in \pi_{n+1} (X,A,x_0)$ be an arbitrary. Denote $h|_{I^n} = f$. Then $h$ gives the homotopy $i_* f \sim const$ in $(X,x_0)$, since $h(x,0) = f(x)$, $h(x,1)=x_0$ and $h(x',t)=x_0$ for all $x' \in \dd I^n$ and $x \in I^n$.
\end{proof}

A map $p\colon E \rightarrow B$ is called a fibration if it has the homotopy lifting property for all $(D^n,\emptyset)$:
\[\xymatrix{
D^n \times \{0\} \ar[r] \ar[d] & E \ar[d]^p\\
D^n \times I \ar@{-->}[ur] \ar[r] & B}
\]
If $p$ is a fibration then it has the homotopy lifting property also for all pairs $(X,A)$ of CW-complexes:
\[\xymatrix{
X \times \{0\} \cup A \times I \ar[r] \ar[d] & E \ar[d]^p\\
X \times I \ar@{-->}[ur] \ar[r] & B}
\]\par
Recall that $p\colon E \rightarrow B$ is a fiber bundle with fibre $F$ if there are open subsets $U_\alpha$ such that
$B = \bigcup_\alpha U_\alpha$ and the following diagram commutes for all $U_\alpha$:
\[
 \xymatrix@C=1.5cm{
 U_\alpha \times F \ar[r]^{\phantom{a}\text{ homeo. }\phantom{a}} \ar[rd]_{pr_1} & p^{-1}(U_\alpha) \ar[d]^p\\
 & U_\alpha}
\]

\ex Show that every fibre bundle is a fibration.

\begin{proof}
 At first consider a trivial fibre bundle $E = B \times F$. Take an arbitrary commutative diagram of the form:
 \[\xymatrix{
 D^n \times \{0\} \ar[r]^f \ar[d] & B \times F \ar[d]^{pr}\\
 D^n \times I \ar[r]^h & B}\]
 Then $h(-,0)=f$ and we can define $H\colon D^n \times I \rightarrow B \times F$ by $H(x,t)=\bigl(h(x,t),g(x)\bigr)$. One can see that the diagram commutes with $H$ too.\par
 Now, let $p\colon E \rightarrow B$ be an arbitrary fibre bundle with fiber $F$ and $B = \bigcup_\alpha U_\alpha$. We can take $I^n$ instead of $D^n$ and consider a diagram:
 \[\xymatrix{
 I^n \times \{0\} \ar[r]^{\quad f} \ar[d] & E \ar[d]^{p}\\
 I^n \times I \ar[r]^{\quad h} & B}\]
 Because $I^n$ is compact, we can divide $I^n \times I$ to finitely many subcubes $C_i \times I_k$ where $I_i = [j_k,j_{k+1}]$ such that $h(C_i \times I_k) \subseteq U_\alpha$ for some $\alpha$. Since $U_\alpha \times F \rightarrow B$ makes a trivial bundle, we can use the same approach as above for each subcube. Since we know $H|_{C_i \times \{0\}} = f|_{C_i \times \{0\}}$, we can find the lift $H$ for all cubes in the first ``column'' (see the picture below) in the same way as for the trivial case:
 \[\xymatrix{
 C_i \times \{0\} \ar[r]^f \ar[d] & U_\alpha \times F \ar[d]^{pr}\\
 C_i \times I_0 \ar@{-->}[ur]^H \ar[r]^h & U_\alpha}\]
 Since we know $H|_{C_i \times \{j_1\}}$ now, we can continue with the second ``column'':
 \[\xymatrix{
 C_i \times \{j_1\} \ar[r]^f \ar[d] & U_\alpha \times F \ar[d]^{pr}\\
 C_i \times I_1 \ar@{-->}[ur]^H \ar[r]^h & U_\alpha}\]
 Thus, we can proceed through all columns in this way until we will get $H$ on the whole $I^n \times I$. The illustration of this situation\footnote{it is drawn as planar, but it should be $n$-dimensional}:
 \begin{center}
  \includegraphics[height=5cm]{subcubes.pdf}
 \end{center}%
\end{proof}

\ex Show the structure of the fibre bundle $S^n \xrightarrow{\ p\ } \R \mathrm{P}^n$.
\begin{proof}
 The fibre is $S^0 = \{-1,1\}$, since $x,-x \longmapsto [x]$. Now, we want to find a neighbourhood $U$ of $[x]$ such that $p^{-1}(U) = U \times S^0$. Set $U=\{[x + v] \mid v \in [x]^\perp\}$ then we have homeomorphism $\varphi\colon U \times S^0 \longrightarrow p^{-1}(U) \subseteq S^n$ given by $\varphi([x+v],1) = \frac{x+v}{\|x+v\|}$ and $\varphi([x+v],-1) = \frac{-x-v}{\|x+v\|}$. We can cover the whole $\R \mathrm{P}^n$ by the open subsets $U_i = \{(x_0 : x_1 : \dots : x_n) \mid x_i \neq 0\}$.
\end{proof}

\ex Show the structure of the fibre bundle $S^{2 n + 1} \xrightarrow{\ p\ } \C \mathrm{P}^n$ with the fibre $S^1$.
\begin{proof}
 Let us look on the special case $S^1 \hookrightarrow S^3 \rightarrow \C \mathrm{P}^1 \cong S^2$ called ``Hopf fibration''. Realise that we can consider $S^3 \subseteq \C^2$, so we can (locally) define the projection $S^3 \rightarrow \C \mathrm{P}^1$ by $(z_1,z_2) \mapsto \frac{z_1}{z_2}$.\par
 In the general case, realise that we can consider $S^{2 n + 1} \subseteq \C^{n+1}$. Take $U_0 = \{[z_0 : z_1 : \dots : z_n] \mid z_0 \neq 0\} \subseteq \C \mathrm{P}^n$. We can consider $U_0 = \{[1 : z_1 : \dots : z_n]\}$. Then the map $U_0 \times S^1 \rightarrow S^{2 n + 1}$ is given by
 \[
 [(1:z_1:\dots:z_n),e^{i t}] \longmapsto \frac{(e^{i t}, e^{i t} z_1, \dots, e^{i t} z_n)}{\| (e^{i t}, e^{i t} z_1, \dots, e^{i t} z_n) \|}
 \]
 We can do the same for other $U_i$ from the covering of $\R \mathrm{P}^n$.
\end{proof}


\ex Long exact sequence of the fibration (Hopf) $S^1\to S^3\to \C P^1=S^2$.

\begin{proof}
This is an important example of a fibration, it deserves our attention. First we write long exact sequence
$$\pi_3(S^1)\stackrel{i_*}{\to}\pi_3(S^3)\stackrel{j_*}{\to}\pi_3(S^2)\stackrel{\partial}{\to}\pi_2(S^1)\to\pi_2(S^3)\to\pi_2(S^2)\to$$
$$\to \pi_1(S^ 1)\to\pi_1(S^3)\to\pi_1(S^2)\to\cdots$$
and also for $\Z\to\R\to S^1=\R/\Z$, that is $\pi_n(\Z,0)\to\pi_n(\R)\to\pi_n(S^1)\to\pi_{n-1}(\Z)$ for $n>0$. Since $\Z$ is discrete, $(S^n,s_0)\to(\Z,0), s_0$ goes to a base point $0$, therefore the map is constant . Hence we get $\pi_n(\Z,0)=0$ for $n\ge 1$. Also, $\pi_n(\R)$ is zero as well, because $\R$ is homotopy equivalent to point. We get that this whole sequence are zeroes for $n\geq 2$. What we are left with is $$0\to\pi_1(S^1)\to\pi_0(\Z)\to\pi_0(\R)=0,$$
so $\pi_1(S^1)\cong \Z$. Continue now with the updated long exact sequence:
$$0\to\pi_3(S^3)\to\pi_3(S^2)\to 0\to\pi_2(S^3)\to\pi_2(S^2)\to\pi_1(S^1)=\Z\to\pi_1(S^3)\cdots,$$
and $\pi_1(S^3)=0$, because $\pi_k(S^n)=0$ for $k<n$ considering the map $S^k\to S^n$ that can be deformed into cellular map and so it is not surjective. So we get
$$\pi_3(S^3)\cong \pi_3(S^2), \quad \pi_2(S^2)\cong  \pi_1(S^1)=\Z.$$
It implies that $\pi_2(S^2)\cong \Z$.\par 
Let us remark that if $\pi_3(S^3)=\Z$, then $\pi_3(S^2)=\Z$. Later we will prove that $$\pi_n(S^n)\cong H_n(S^n)\cong \Z.$$

\end{proof}

\rem (Story time) Eduard Čech was the first who defined higher homotopy groups (1932, Höherdimensionale Homotopiegruppen) but the community od these groups didn't support the study as they were considered not interesting. Were they mistaken? The rest of this remark is left as an exercise for the reader.

\rem (For geometers) For $G$ Lie group and $H$ its subgroup we have a fibre bundle $H\hookrightarrow G\to G/H$.\par As an example consider otronormal group $O(n)$, we have inclusions $O(1)\subseteq O(2)\subseteq\cdots O(n).$ Then
$$O(n-k)\to O(n)\to O(n)/O(n-k)=V_{n,k}$$
is a fibre bundle, we call $V_{n,k}$ Stiefel manifold, $k$-tuples of orthonormal vectors in $\R^n$.\par 
Also, we can take $O(k)\to O(n)/O(n-k)\to O(n)/(O(k)\times O(n-k))$, Grassmannian manifold ($k$-dimensional subspaces in $\R^n$):
$$O(k)\to V_{n,k}\to G_{n,k},$$
is also a fibre bundle.

\ex Long exact sequence of the fibration $F\to E\to B$ ends with
$$\pi_1(B)\to\pi_0(F,x_0)\to\pi_0(E,x_0)\to\pi_0(B).$$
Show exactness in $\pi_0(E)$.

\begin{proof}
We showed $\pi_n(B)\cong\pi_n(E,F)$ for $n\ge 0$ which gave us the exactness of the fibration sequence from the exactness of the sequence of the pair $(E,F)$ till $\pi_0(E,x_0)$.

Denote $S^0=\{-1,1\}$ and consider the composition
$$(\{-1,1\},-1)\to (F,x_0)=p^{-1}(b_0)\to (E,x_0)\to (B,p(x_0)=b_0),$$
where $-1$ goes to $x_0$ and $b_0$ and $1$ goes to $F$ but that is $p^{-1}(b_0)$ so we get the constant map.\par 
For the other part of the exactness we will use homotopy lifting property.\par 
Consider $f:(\{-1,1\},-1)\to (E,x_0)$ such that $pf$ is homotopic to the constant map into
$b_0$. It means that $pf(1)$ is connected with $b_0$ by a curve. Have a diagram
\[
\xymatrix{
(1) \ar@{->}[r] \ar@{^{(}->}[d] & E \ar@{->}[d]^p\\
(1)\times I \ar@{-->}[ur] \ar[r] & B
}
\]
and remark that $f(1)$ is connected by a curve with  $x\in E$ such that  $p( x)=b_0$.
So $x\in F$ and $f$ is homotopic to the map $g: (\{-1,1\},-1)\to (F,x_0)$ which maps $-1$ into $ x_0$ and $1$ into $x$.

\end{proof}

\ex Covering: $G\to X\to X/G$, with action of $G$ on $X$ properly discontinuous, where $X$ is path connected.\par 
Take $$\pi_1(G,1)\to\pi_1(X)\to\pi_1(X/G)\to \pi_0(G)\to \pi_0(X),$$ that is
$$0\to \pi_1(X)\to\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)\to 0.$$
Show that $\dd$ is a group homomorphism. 
\begin{proof}
First note that $\pi_0(G)$ is $G$ taken as a set.
We recall that for $F\to E\to B$ we had
\[
\xymatrix{
\pi_n(E)\ar[r]^{j_*} & \pi_n(E,F,x_0) \ar[r]^{\bar{\dd}} \ar[d]^{p_*}_{\cong} & \pi_{n-1}(F) \\
& \pi_n(B,b_0) \ar@{-->}[ur]_{\dd}
}
\]
We showed the $\cong$ finding a map going from $\pi_n(B,b_0)\to \pi_n(E,F,x_0)$. Also, $$f\colon (I^n,\dd I^n, J^{n-1})\to (E,F,x_0),$$ $\bar{\dd}[F]=[f/\dd I^n]$. The prescription for $\dd\colon\pi_n(B,b_0)\to  \pi_{n-1}(F,x_0)$: take $f\colon (I^n,\dd I^n)\to (B,b_0)$ and make a lift $F\colon (I^n,\dd I^n,J^{n-1})\to (E,F,x_0)$, $\dd[f]=F/I^{n-1}\colon (I^{n-1},\dd I^{n-1}\to (F,x_0)$. The diagram is:

\[\xymatrixcolsep{5pc}
\xymatrix{
I^{n-1}\times \{1\}\cup \dd I^{n-1}\times I \ar[r]^<<<<<{const} \ar@{^{(}->}[d]& E\quad x_0 \ar[d]^p \\
I^{n-1}\times I \ar[r] \ar@{-->}[ur]^F & B\quad b_0
}
\]
and $pF(I^{n-1}\times \{0\})=b_0$, so $F(I^{n-1}\times \{0\})\subseteq p^{-1}(b_0)=F.$\par 
Now, $\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)$, $f\colon(I,\dd I)\to (X/G,[x_0])$. Take two closed curves $\omega,\tau$ at $[x_0]$ and lift them to curves $\bar{\omega},\bar{\tau}$ in $X$ starting in $x_0$. We denote the end point of $\bar{\omega}$ by $g_1x_0$ and the end point of $\bar{\tau}$ by $g_2x_0$. Then the operation $\cdot$ (that is not a composition $\circ$) is:
\begin{align*}
    \overline{\omega\cdot\tau}&=\bar{\omega}\cdot g_1\bar{\tau}\\
    \overline{\omega\cdot\tau}&(0)=x_0\\
    \overline{\omega\cdot\tau}&(1/2)=g_1x_0\\
    \overline{\omega\cdot\tau}&(1)=g_2(g_1x_0)
\end{align*}
And it's a homomorphism.


\end{proof}

\ex Van Kampen theorem - Applications.\par 
%$X=\bigcup U_{\alpha}$, $U_{\alpha}\cap U_{\beta}$ path connected containing $x_0$, $U_{\alpha}\cap U_{\beta}\cap U_{\gamma}$ path connected.

Klein bottle K. Model as a square with identified sides as seen in Fig 1,2,3.

%\vspace{-0.314159265cm}
\begin{figure}[ht]
    \centering
    \includegraphics[scale=0.4]{algtopo_10_obrazek.png}
    %\caption{Triangle}
    %\caption{Awesome Image}
    %\label{fig:awesome_image}
\end{figure}
%\vspace{-1.5cm}

We denote open sets $U_1,U_2$ (disc) and point $x_0$ as in figure 1 and 2. (some of the notation in the solution is established in the theorem)

\begin{proof}
Set $U_1$ is homotopy equivalent to the boundary (Fig 3), and this boundary is in fact a wedge of two circles, so $U_1\simeq S^1\vee S^1$.\par 
We can compute: $\pi_1(U_2,x_0)=\{1\}$ by contractibility, $\pi_1(U_1,x_0)=\pi_1(S^1\vee S^1,x_0)=$ free group on two generators $\alpha,\beta$ as was already shown in lecture.\par 
Then $\pi_1(U_1,x_0)\ast \pi_1(U_2,x_0)=$ free group on two generators $\alpha,\beta$, also $\pi_1(U_1\cap U_2,x_0)=\Z$. \par 
Now, the intersection $U_1\cap U_2\simeq (S^1,x_0)$ and we take generator $\omega$, $i_{2,1\ast}(\omega^{-1})=1$ in $\pi_1(U_2,x_0)$, $i_{1,2\ast}(\omega)=\alpha\beta\alpha^{-1}\beta$.\par 
So, kernel of $\varphi$ (the map from the theorem) is generated by element $\alpha\beta\alpha^{-1}\beta$\\
$\pi_1(K)$ is the group with two generators $\alpha,\beta$ and one relation $\alpha\beta\alpha^{-1}\beta=1$
\end{proof}

\ex Recall definitions of n-connectness and n-equivalence. Prove the following lemma: Inclusion $A\hookrightarrow X$ is n-equivalence if and only if $(X,A)$ is n-connected.

\begin{proof}
"$\Leftarrow$" Take long exact sequence:
$$\to\pi_n(A,x_0)\stackrel{f_1}{\to}\pi_n(X,x_0)\to \pi_n(X,A,x_0)\to \pi_{n-1}(A,x_0)\stackrel{f_2}{\to} \pi_{n-1}(X)\to\pi_{n-1}(X,A)\to$$
and use the assumption that $\pi_i(X,A,x_0)=0$ for $i\leq n$. Then we get that $f_1$ is epimorphism and $f_2$ is isomorphism.\par 
"$\Rightarrow$" Reasoning is the same as in the other direction, the only thing we need to realize is $\pi_0(A,x_0)\stackrel{\cong}{\longrightarrow} \pi_0(X,x_0)$.
\end{proof}

\ex Show $\pi_k(S^{\infty})=0$ for all $k$, where $S^{\infty}$ is colim$\,S^n$.

\begin{proof}
We have $S^1\subset S^2 \subset S^3\subset \cdots \subset S^n \subset\cdots S\nainf$. Take element in $\pi_k(S\nainf)$, that is $f\colon S^k\to S\nainf$. We know that $f(S^k)$ is compact in $S\nainf$. Consider other CW-complex structure than $e^0\cup e^k$ for $S^k$ that is $S^k=\bigcup_{i=0}^{k}e_1^i\cup e_2^i$ (two hemispeheres). Then the following holds: $S\nainf=\bigcup_{i=0}^{\infty}e_1^i\cup e_2^i$. So $f(S^k)\subseteq (S\nainf)^{(N)}=S^N$ for some $N$ where $(S\nainf)^{(N)}$ is $N$-skeleton of $S\nainf$. Now, 
$$f\colon S^k\to S^N\to S^{N+1}\hookrightarrow S\nainf,$$
so the composition $S^k\to S^{N+1}$ is a map that is not onto and therefore $f$ is homotopic to constant map. (map into a disc is homotopic to constant map, disc is contractible) Then we have $[f]=0$. Thus we have proved $\pi_k(S\nainf)=0$.
\end{proof}

\ex Compute homotopy groups of $\R P\nainf$.

\begin{proof}
Suprisingly use previous exercise: We can view $\R P\nainf$ as lines going through origin in $S\nainf$, or...just take $S\nainf/_{\Z/2}$, where the action is $x\mapsto -x$. So we work with the following fibration (we don't write the distinguished points as they are not needed) $\Z/2\to S\nainf \to \R P\nainf$ and the long exact sequence
$$\pi_n(\Z/2)\to\pi_n(S\nainf)\to \pi_n(\R P\nainf)\stackrel{\dd}{\to}\pi_{n-1}(\Z/2)\to \pi_{n-1}(S\nainf),$$
where for all $n\geq 2$ we have all zeroes, for $n=1$ consider $0\to\pi_1(\R P\nainf)\stackrel{\dd}{\to}\pi_0(\Z/2)\to\pi_0(S\nainf)$. Since $\pi_0(S\nainf)=0$ and $\pi_0(\Z/2)=\Z/2$, we get that the homomorphism $\dd$ (it is homomorphism, really, we did it in previous tutorial, but it's still a homomorphism independetly on whether we did it or not) is an isomorphism of groups. By connectness we also know the $\pi_0$ group. So the final results are:
$$\pi_n(\R P\nainf)=0 \text{ for }n\geq 2,\,\pi_1(\R P\nainf)=\Z/2,\, \pi_0(\R P\nainf)=0.$$
\end{proof}

\ex Show that the spaces $S^2\times \R P\nainf$ and $\R P^2$ have the same homotopy groups but they are not homotopy equivalent.

\begin{proof}
Here, also use pre$^2$vious exercise. It is known that $\pi_n(X\times Y)=\pi_n(X)\times \pi_n(Y)$. With this we can compute
$$\pi_0(S^2\times \R P\nainf)=0,\,\pi_1(S^2\times\R P\nainf)=\pi_1(S^2)\times\pi_1(\R P\nainf)=\{0\}\times \Z/2\cong \Z/2,$$
$$\text{for }n\geq 2\, \pi_n(S^2\times \R P\nainf)=\pi_n(S^2)\times \{0\}\cong\pi_n(S^2.$$
Now consider $\R P^2$ as $S^2/_{\Z/2}$, work with the fibration (scheme as follows)
\[\xymatrixrowsep{0.25pc}
\xymatrix{
& \Z/2 \ar[r] & S^2 \ar[r]& \R P^2 &\\
& \pi_n(\Z/2)\ar[r]& \pi_n(S^2)\ar[r]& \pi_n(\R P^2) \ar[r] & \pi_{n-1}(\Z/2)\\
n\geq 2& 0 & \pi_n(S^2)& \cong\pi_n(\R P^2) & 0\\
n=1& & 0 & \pi_1(\R P^2)& \cong \Z/2&0
}
\]
and $\pi_0(\R P^2)=0$. Thus we showed that these two spaces have the same homotopy groups. Marvelous. How to show, that they are not homotopy equivalent? Use cohomology group! That's right. It is well known (or we should already know) that
$$H^*(\R P^2;\Z/2)=\Z/2[\alpha]/\langle\alpha^3\rangle, \alpha\in H^1\text{  and  }$$ $$H^*(S^2\times \R P\nainf;\Z/2)=H^*(S^2;\Z/2)\otimes H^*(\R P\nainf;\Z/2)=\Z/2[\beta]/\langle\beta^2\rangle\otimes \Z/2[\gamma],\beta\in H^2.$$
The former space obviously has no non-zero elements of order $4$, while the latter has a non-zero element of order $4$. This is impossible for homotopy equivalent spaces. We are done.

\end{proof}

\ex Extension lemma: Let $(X,A)$ be a pair of $CW$-complexes, $Y$ a space with $\pi_{n-1}(Y,y_0)=0$ whenever there is a cell of dimension $n$ in $X-A$. Then every map $f\colon A \to Y$ can be extended to a map $F\colon X\to Y$.

\begin{proof}
Set $X_{-1}=A, X_0=X^{(0)}\cup A,X_k=X^{(k)}\cup A$, and $f=f_{-1}\colon X_{-1}\to Y, f_0\colon X_0\to Y$ which extends $f_{-1}$, $f_0(x_0)$ to any point in $Y$.\par 
We have $f_{k-1}\colon X_{k-1}\to Y$ and want to define $f_k\colon X_k\to Y$. Also, $\pi_{k-1}(Y,\bullet)=0$.\par 
Consider following diagram:

\[
\xymatrix{
(D^k,S^{k-1})\ar[r]^{\varphi} \ar[dr]_{f_{k-1}\circ\varphi}& (X_k,X_{k-1})\ar[d]^{f_{k-1}}\\
& Y
}
\]

The map $f_{k-1}\circ \varphi$ is in $\pi_{k-1}(Y,\bullet)$ so it is homotopic to constant map, so we define $f_k$ on $D^k$ as a constant map. Now,

\[
\xymatrix{
D^k\times\{1\}\cup S^{k-1}\times I\ar[r] \ar@{^{(}->}[d]& Y\\
D^k\times I\ar@{-->}[ru]_H &
}
\]
and go back to the first diagram:
\[
\xymatrix{
(D^k,S^{k-1})\ar[r]^{\varphi} \ar[dr]_{g}& (X_k,X_{k-1})\ar[d]^{f_k}\\
& Y
}
\]
We extend $f_{k-1}$ to $f_k\colon X_k\to Y$ and proceed to infinity \textit{(and beyond)}, as we always do.
\end{proof}

\ex Compare $\pi_n(X,x_0)$ and $\pi_n(X,x_1)$, when distinguished points (are / are not) path connected. Use proof with pillows.

\begin{proof}
First the case where $X=S^1\sqcup S^2$ and $x_0\in S^1 $ and $x_1\in S^2$. Then $\pi_1(X,x_0)=\pi_1(S^1)=\Z$ but $\pi_1(X,x_1)=\pi_1(S^2)=0$. If distinguished points are not path connected, homotopy groups can be different, so consider now $\omega$ a curve connecting $x_0$ and $x_1$, $\omega\colon I\to X$, $\omega(0)=x_0$ and $\omega(1)=x_1$. We have $\pi_n(X,x_0)\to\pi_n(X,x_1)$
$f\colon I^n\to X$, $\dd I^n\mapsto x_0$, $g\colon I^n\to X$, $\dd I^n\mapsto x_1$, as seen in Figure 1, $x_0 - x_1$ segments. Proofs with pillows!

$$\includegraphics[scale=0.3]{figure1_2}$$
Denote action $f\mapsto \omega\cdot f$, then $f_1\sim f_2 \Rightarrow \omega \cdot f_1 \sim \omega \cdot f_2$. We are not satisfying algebraist at the moment, only geometers. Let us try to do something about that. \par 
Figure 2 shows other pillow and that $\omega_1\sim \omega_2 \Rightarrow \omega_1\cdot f\sim \omega_2 \cdot f$, we can imagine the segment as in Figure 2, two curves.

$$\includegraphics[scale=0.2]{figure3}$$

Map given by $\omega$ is a bijection $\omega\colon x_0\to x_1$, $\omega_2(\omega_1 f)\sim (\omega_1\omega_2)f$, $\omega^{-1}(\omega f)\sim (\omega\omega^{-1})f\simeq f$, so the map $\pi_n(X,x_0)\to \pi_n(X,x_1)$ is bijection. Figure 3 tries to explain homomorphism.\par 
We get that $x_0,x_1$ are in the same path component and if $\omega\colon x_0\to x_1$ is a curve, then $\pi_n(X,x_0)\to \pi_n(X,x_1)$ is an isomorphism. In particular, if $X$ is simply connected, then every curve gives the same isomorphism.
\end{proof}



\iffalse

\ex Long exact sequence of the fibration (Hopf) $S^1\to S^3\to \C P^1=S^2$.

\begin{proof}
This is an important example of a fibration, it deserves our attention. First we write long exact sequence
$$\pi_3(S^1)\stackrel{i_*}{\to}\pi_3(S^3)\stackrel{j_*}{\to}\pi_3(S^2)\stackrel{\partial}{\to}\pi_2(S^1)\to\pi_2(S^3)\to\pi_2(S^2)\to$$
$$\to \pi_1(S^ 1)\to\pi_1(S^3)\to\pi_1(S^2)\to\cdots$$
and also for $\Z\to\R\to S^1=\R/\Z$, that is $\pi_n(\Z,0)\to\pi_n(\R)\to\pi_n(S^1)\to\pi_{n-1}(\Z)$ for $n>0$. Since $\Z$ is discrete, $(S^n,s_0)\to(\Z,0), s_0$ goes to a base point $0$, therefore homotopic to constant map, we get $\pi_n(\Z,0)=0$. Also, $\pi_n(\R)$ is zero as well, because $\R$ is homotopy equivalent to point, $S^n\to\R\to\ast$. We get that this whole sequence is zero for $n\geq 2$. What we are left with is $$0\to\pi_1(S^1)\to\pi_0(\Z)\to\pi_0(\R)=0,$$
so $\pi_1(S^1)\cong \Z$. Continue now with the updated long exact sequence:
$$0\to\pi_3(S^3)\to\pi_3(S^2)\to 0\to\pi_2(S^3)\to\pi_2(S^2)\to\pi_1(S^1)=\Z\to\pi_1(S^3)\cdots,$$
and $\pi_1(S^3)=0$, because $\pi_k(S^n)=0$ for $k<n$ considering the map $S^k\to S^n$ that can be deformed and so it is not surjective. The final step is
$$0\to\pi_2(S^2)\to \pi_1(S^1)=\Z\to \pi_1(S^3)=0),$$
it follows that $\pi_2(S^2)\cong \Z$.\par 
Let us remark that if $\pi_3(S^3)=\Z$, then $\pi_3(S^2)=\Z$, and (non-trivially) $$\pi_n(S^n)\cong H_n(S^n)\cong \Z.$$

\end{proof}

\rem (Story time) Eduard Čech was the first in defining higher homotopy groups (1932, Höherdimensionale Homotopiegruppen) but the community didn't support the study as these were commutative and considered not interesting. Were they mistaken? The rest of this remark is left as an exercise for the reader.

\rem (For geometers) For $G$ Lie group and $H$ its subgroup we have a fibre bundle $H\hookrightarrow G\to G/H$.\par As an example consider otronormal group $O(n)$, we have inclusions $O(1)\subseteq O(2)\subseteq\cdots O(n).$ Then
$$O(n-k)\to O(n)\to O(n)/O(n-k)=V_{n,k}$$
is a fibre bundle, we call $V_{n,k}$ Stiefel manifold, $k$-tuples of orthonormal vectors in $\R^n$.\par 
Also, we can take $O(k)\to O(n)/O(n-k)\to O(n)/(O(k)\times O(n-k))$, Grassmannian manifold ($k$-dimensional subspaces in $\R^n$):
$$O(k)\to V_{n,k}\to G_{n,k},$$
fibre bundle.

\ex Long exact sequence of the fibration $F\to E\to B$ ends with
$$\pi_1(B)\to\pi_0(F,x_0)\to\pi_0(E,x_0)\to\pi_0(B).$$
Show exactness in $\pi_0(E)$.

\begin{proof}
We showed $\pi_1(B)\cong\pi_1(E,F).$ We have $\pi_1(E,F)\to\pi_0(F)\to\pi_0(E)$.\par 
Denote $S^0=\{-1,1\}$, then
$$(\{-1,1\},-1)\to (F,x_0)=p^{-1}(b_0)\to (E,x_0)\to (B,p(x_0)=b_0),$$
where $-1$ goes to $x_0$ and $b_0$ and $1$ goes to $F$ but that is $p^{-1}(b_0)$ so we get that clearly it is homotopic to constant map.\par 
For the other part of the exactness we will use homotopy lifting property.\par 
Consider $(\{-1,1\},-1)\to (E,x_0)\stackrel{p}{\to}(B,p(x_0))$. Have a diagram
\[
\xymatrix{
(1) \ar@{->}[r] \ar@{^{(}->}[d] & E \ar@{->}[d]^p\\
(1)\times I \ar@{-->}[ur] \ar[r] & B
}
\]
and remark that $x_1\in F, x_1\mapsto b_0, x_1$ connected with $x, x\mapsto b$, and $b_0$ connected with $b$. We have $(\{-1,1\}, -1)\to (F,x_0)$, so $-1\mapsto x_0$ and $1\mapsto x_1$, in $E$ it is homotopic to $f\colon (\{-1,1\},-1)\to (E,x_0)$ ($-1\mapsto x_0,1\mapsto x$) and $pf\colon (\{-1,1\},-1)\to B$ ($-1\mapsto b_0,1\mapsto b$).\par 
It is not obvious that this exercise is easy.

\end{proof}

\ex Covering: $G\to X\to X/G$, with action of $G$ on $X$ properly discontinuous, where $X$ is path connected.\par 
Take $$\pi_1(G,1)\to\pi_1(X)\to\pi_1(X/G)\to \pi_0(G)\to \pi_0(X),$$ that is
$$0\to \pi_1(X)\to\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)\to 0.$$
Show that $\dd$ is a group homomorphism. 
\begin{proof}
First note that $\pi_0(G)$ is $G$ taken as a set.
We recall that for $F\to E\to B$ we had
\[
\xymatrix{
\pi_n(E)\ar[r]^{j_*} & \pi_n(E,F,x_0) \ar[r]^{\bar{\dd}} \ar[d]^{p_*}_{\cong} & \pi_{n-1}(F) \\
& \pi_n(B,b_0) \ar@{-->}[ur]_{\dd}
}
\]
We showed the $\cong$ finding a map going from $\pi_n(B,b_0)\to \pi_n(E,F,x_0)$. Also, $$f\colon (I^n,\dd I^n, J^{n-1})\to (E,F,x_0),$$ $\bar{\dd}[F]=[f/\dd I^n]$. The prescription for $\dd\colon\pi_n(B,b_0)\to  \pi_{n-1}(F,x_0)$: take $f\colon (I^n,\dd I^n)\to (B,b_0)$ and make a lift $F\colon (I^n,\dd I^n,J^{n-1})\to (E,F,x_0)$, $\dd[f]=F/I^{n-1}\colon (I^{n-1},\dd I^{n-1}\to (F,x_0)$. The diagram is:

\[\xymatrixcolsep{5pc}
\xymatrix{
I^{n-1}\times \{1\}\cup \dd I^{n-1}\times I \ar[r]^<<<<<{const} \ar@{^{(}->}[d]& E\quad x_0 \ar[d]^p \\
I^{n-1}\times I \ar[r] \ar@{-->}[ur]^F & B\quad b_0
}
\]
and $pF(I^{n-1}\times \{0\})=b_0$, so $F(I^{n-1}\times \{0\})\subseteq p^{-1}(b_0)=F.$\par 
Now, $\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)$, $f\colon(I,\dd I)\to (X/G,[x_0])$. Take two closed curves $\omega,\tau$ at $[x_0]$ and shift them $\bar{\omega},\bar{\tau}$. We denote end point of $\bar{\omega}$ by $g_1x_0$ and end point of $\bar{\tau}$ by $g_2x_0$. Then the operation $\cdot$ (that is not a composition $\circ$) is:
\begin{align*}
    \overline{\omega\cdot\tau}&=\bar{\omega}\cdot g_1\bar{\tau}\\
    \overline{\omega\cdot\tau}&(0)=x_0\\
    \overline{\omega\cdot\tau}&(1/2)=g_1x_0\\
    \overline{\omega\cdot\tau}&(1)=g_2(g_1x_0)
\end{align*}
...it's a homomorphism.


\end{proof}

\ex Van Kampen theorem - Applications.\par 
%$X=\bigcup U_{\alpha}$, $U_{\alpha}\cap U_{\beta}$ path connected containing $x_0$, $U_{\alpha}\cap U_{\beta}\cap U_{\gamma}$ path connected.

Klein bottle K. Model as a square with identified sides as seen in Fig 1,2,3.


We denote open sets $U_1,U_2$ (disc) and point $x_0$ as in figure 1 and 2. (some of the notation in the solution is established in the theorem)

\begin{proof}
Set $U_1$ is homotopy equivalent to the boundary (Fig 3), and this boundary is in fact a wedge of two circles, so $U_1\simeq S^1\vee S^1$.\par 
We can compute: $\pi_1(U_2,x_0)=\{1\}$ by contractibility, $\pi_1(U_1,x_0)=\pi_1(S^1\vee S^1,x_0)=$ free group on two generators $\alpha,\beta$ as was already shown in lecture.\par 
Then $\pi_1(U_1,x_0)\ast \pi_1(U_2,x_0)=$ free group on two generators $\alpha,\beta$, also $\pi_1(U_1\cap U_2,x_0)=\Z$. \par 
Now, the intersection $U_1\cap U_2\simeq (S^1,x_0)$ and we take generator $\omega$, $i_{2,1\ast}(\omega^{-1})=1$ in $\pi_1(U_2,x_0)$, $i_{1,2\ast}(\omega)=\alpha\beta\alpha^{-1}\beta$.\par 
So, kernel of $\varphi$ (the map from the theorem) is generated by element $\alpha\beta\alpha^{-1}\beta$\\
$\pi_1(K)$ is the group with two generators $\alpha,\beta$ and one relation $\alpha\beta\alpha^{-1}\beta=1$
\end{proof}

\fi



\ex Using homotopy groups show that $\R P^k$ is not a retract of $\R P^n$, $n>k\geq 1$.

\begin{proof}
Retract: The composition $\R P^k\stackrel{i}{\hookrightarrow} \R P^n \stackrel{r}{\to}\R P^k$ is the identity, 
so we have the following diagram and want to use it to obtain a contradiction.
\[%\xymatrixrowsep{0.25pc}
\xymatrix{
\pi_*(\R P^k) \ar[r]^{\pi_*(i)} \ar[rd]_{\operatorname{id}} & \pi_*(\R P^n) \ar[d]^{\pi_*(r)}\\
& \pi_*(\R P^k)
}
\]
From the long exact sequence of the fibration, we know that $0=\pi_1(S^k)\to \pi_1(\R P^k)\stackrel{\cong}{\to}\pi_0(S^0)\to 0$ and $\pi_i(S^k)=\pi_i(\R P^k)$ for $k\geq 2$. Then considering diagram 
\[
\xymatrix{
\pi_k(S^k)=\Z \ar[r] \ar[rd] & \pi_k(S^n)=0 \ar[d]\\
& \pi_k(S^k)=\Z
}
\]
we see that we are factoring identity through zero group and that is Mission Impossible: Factor Zero (in theaters never). Continue with $k=1$. We use the knowledge of $\R P^1\cong S^1$. Then our diagram is a triangle with $\Z,\Z/2,\Z$ with identity $\Z\to\Z$, factoring identity through finite group is Mission Impossible: Group protocol (in theaters maybe one day, one can only hope) ...and we are done.
\end{proof}

\ex Consider the map $q\colon S^1\times S^1\times S^1\to S^3$ defined as a map $S^1\times S^1\times S^1\to D^3/S^2$ where $D^3$ is a small disk in the triple torus which is the identity in the interior of $D^3$ and constant on its complement. Further,  consider the Hopf map $p\colon S^3\to S^2=\C P^1$ (described in Hopf fibration $S^1\to S^3\to S^2$). Compute $q_*$ and $(pq)_*$ in homotopy groups. Show that $pq$ is not homotopic to a constant map.

\begin{proof}
Take following diagram and treat it like your own:

\[
\xymatrix{\xymatrixcolsep{2pc}
& S^1 \ar[d] & \text{(fiber)}\\
S^1\times S^1\times S^1 \ar[r]^<<<q & S^3 \ar[d]^p& \subseteq \C^2\\
& S^2 & =\C P^1
}
\]
where $p(z_1,z_2)=\dfrac{z_1}{z_2}$. We know that only nontrivial homotopy group of the triple torus is $\pi_1(S^1\times S^1\times S^1)$, but  $\pi_1 S^3$ is is trivial, so $q_*$  is zero in homotopy groups and the composition as well. By the following diagram we have $H$, thanks to homotopy lifting property, of course. (denote $(S^1)^3$ the triple torus $S^1\times S^1\times S^1$)

\[
\xymatrix{
& S^1=p^{-1}(s_0) \ar[d]\\
(S^1)^3 \ar[r]^q \ar[ur] \ar[d] & S^3 \ar[d] \\
(S^1)^3\times I \ar[ur]^H \ar[r]_h & (S^2,s_0)
}
\]
where $h(0,-)$ is constant map, $h(1,-)=pq$, $p(H(0))=s_0$,$\operatorname{Im} H(0)\subseteq p^{-1}(s_0)=S^1$ and $H(1)=q\sim H(0)$.
However, $H(0)_*$ and $q_*$ differ in the  third homology groups:
\[
\xymatrix{
& H_3(S^1)=0\ar[d]\\
H_3((S^1)^3)\ar[r]_{q_*=id} \ar[ru]^{H(0)_*}& H_3(S^3)=\Z
}
\]
We are trying to factor through zero, a contradiction.
\end{proof}

\ex (detail for the lecture 9. 5. 2017) If $(X,A)$ is relative $CW$-complex such that there are no cells in dimension $\leq n$ in $X\setminus A$, then $(X,A)$ is $n$-connected.

\begin{proof}
Recall the definition of $n$-connectness of a pair. For $[f]\in\pi_i(X,A,x_0)$, $i\le n$,  use cell approximation of $f$: There is a cell map $q\colon (D^i,S^{i-1},s_0)\to (X,A,x_0)$, such that $q\sim f$ relatively $S^{i-1}$ and $q(D^i)\subseteq X^{(i)}=A$ since $X^{(-1)}=X^{(i)}=\cdots=X^{(n)}=A$. Note the following very useful criterion:\par 
$[f]=0$ in $\pi_i(X,A,x_0)$ $\Longleftrightarrow f\sim q$ relatively $S^{i-1}$, $g(D^i)=A$.\par 
Thus $[f]=0$ in our case, and we are done.
\end{proof}

\ex Let $[X,Y]$ denote a set of homotopy classes of maps from $X$ to $Y$. If $(X,x_0)$ is a CW-complex and $Y$ is path connected, then $[X,Y]\cong [(X,x_0),(Y,y_0)]$.

\begin{proof}
Surely, $[(X,x_0),(Y,y_0)]\subseteq [X,Y]$ and denote the class in the left set as $(g)$ and $[g]$ the class in the $[X,Y]$. The map $(g)\mapsto [g]$ is well defined and injective. To prove that it is also surjective take $[f]\in [X,Y]$. Using HEP for $f:X\times 0\to Y$ and a curve $\omega: x_0\times I\to Y$ which connects $f(x_0)$ with $y_0$ we get $g:X\times 1\to Y$ such that $f\sim g$ and  $g(x_0)=y_0$, then $[f]=[g]$ and $(g)\in[(X,x_0),(Y,y_0)]$. We are done.
\end{proof}

\ex (application)  We know that $deg(f)$ is an invariant of $[S^n,S^n]=\pi_n(S^n)$. \\ 
Study $[S^{2n-1},S^n]\cong \pi_{2n-1}(S^n)$ and describe its co called Hopf invariant $H(f)$.

\begin{proof}
Have $f\colon \partial D^{2n}=S^{2n-1}\to S^n$ and $S^n\cup_f D^{2n}$. For $f\sim g$ we have $S^n\cup_f D^{2n}\simeq S^n\cup_g D^{2n}$, moreover $S^n\cup_f D^{2n}=C_f$ (the cylinder of $f$). For $n\geq 2$ we have $C_f=e^0\cup e^n\cup e^{2n}$. Using cohomology: $H^*(C_f)=\Z$ for $\ast\in\{0,n,2n\}$ and $0$ elsewhere. Take $\alpha\in H^{n}(C_f)$ generator, we have cup product. Then $\alpha\cup\alpha\in H^{2n}(C_f)$ and for $\beta\in H^{2n}(C_f)$ we have $\alpha\cup\alpha=H(f)\beta$, where $H(f)$ is the Hopf invariant.
\end{proof}

\ex (continuation of previous exercise) For $n$ odd, what can we say in this case about Hopf inviariant? And for $n$ even? Thanks.

\begin{proof}
Knowing $\alpha\cup \beta=(-1)^{|\alpha||\beta|} \beta\cup\alpha$ we see that $\alpha\cup\alpha = 0$. So for $n$ odd Hopf invariant is zero.\par 
For $n$ even consider the Hopf fibration $S^1\to S^3\to S^2=\C P^1$. For $\C P^2=D^4\cup_f \C P^1$ (recall how $\C P^n$ is built up from $\C P^{n-1}$) we have $C_f=\C P^2$ and $H^*(\C P^2)=\Z[\alpha]/\langle \alpha^3\rangle$, with $\alpha\in H^2$. The generator of $H^4$ is $\alpha^2$. We get that  $H(f)=1$.
\end{proof}


\ex Let $(X,A)$ be a pair and show that the following diagram commutes:
\[
\xymatrix{
\pi_n(X,A,x_0) \ar[r]^{\dd} \ar[d]_h & \pi_n(A,x_0)\ar[d]^h \\
H_n(X,A) \ar[r]_{\dd_*} & H_{n-1}(A)
}
\]

where $\dd$ is boundary homomorphism, $h$ is Hurewicz homomorphism and $\dd_*$ is the connecting homomorphism. 

\begin{proof}
Take $[f]\in \pi_n(X,A,x_0)$, that is $f\colon (D^n,D^{n-1},s_0)\to (X,A,x_0)$. Then $\dd[f]=[f/S^{n-1}]$ and $h\dd[f]=h[f/S^{n-1}]=(f/S^{n-1})_*(b)$, where $b$ is generator in $H_{n-1}(S^{n-1})$, as we recall in the following: $g\colon S^{n-1}\to A, g_*\colon H_{n-1}(S^{n-1})\to H_{n-1}(A)$ and $h[g]=g_*(b)\in H_{n-1}(A)$. Let $a\in H_n(D^n,S^{n-1})$ be a generator, that is $\dd_*a=b$. We proceed with the following diagram:

\[
\xymatrix{
H_n(D^n,S^{n-1}) \ar[r]^{\dd_*}_{\cong} \ar[d]_{f_*} & H_{n-1}(S^{n-1}) \ar[d]^{(f/S^{n-1})_*} \\
H_n(X,A) \ar[r]_{\dd_*} & H_{n-1}(A)
}
\]
So, now $\dd_*h[f]=\dd (f_*a)=(f/S^{n-1})_*(\dd_*a)$. With $\dd_*a=b$ we conclude that the diagram commutes.
\end{proof}

\ex Homology version of Whitehead theorem. (recall both versions) We show that $f_*\colon \pi_n(X)\to \pi_n(Y)$ is an iso for all n and then use the previous version of the theorem.

\begin{proof}
First, for $f\colon X\hookrightarrow Y$ and inclusion. For pair $(X,A)$ we have the following long exact sequence (with Hurewicz):
\[
\xymatrix{
\pi_{n+1}(Y,X) \ar[r] \ar[d]_h & \pi_{n}(Y) \ar[r] \ar[d]_h & \pi_{n}(X) \ar[r] \ar[d]_h & \pi_{n}(Y,X) \ar[d]_h \\
H_{n+1}(Y,X) \ar[r]_0 & H_n(Y) \ar[r]_{\cong} & H_n(X) \ar[r]_0 & H_n(Y,X)
}
\]

The thing is that $h$ is not always $iso$, we have $\stackrel{0}{\to} \pi_1(Y,X)\stackrel{0}{\to}$, for $Y,X$ simply connected. Now, for $\pi_2\stackrel{h}{\to}H_2$ is iso by Hurewicz theorem, then $H_2=0$ so $\pi_2=0$ and with the same argument we can proceed inductively.\par 
Now, for general map $f\colon X\to Y$ we use the argument with mapping cylinder $M_f$. Do you remember this one from the first lecture? This construction you thought would not be interesting? Well, hold your hats, its usage is advantageous here!

\[
\xymatrix{
  & X \ar[ld]_f \ar[rd]^f \ar[d]^{i_X}  & \\
Y \ar[r]_{i_Y} & M_f & Y \ar[l]^r
}
\]
We know that $f_*\colon H_n(X)\to H_n(Y)$ is an iso and (as $i_Y$ is homotopy equivalence) \\ $i_{Y*}\colon H_n(Y)\to H_n(M_f)$ is an iso. Thus, we conclude that $i_{X*}\colon H_n(X)\to H_n(M_f)$ is an iso. That is, $i_*$ is homotopy equivalence and therefore $f$ is also homotopy equivalence. (Let us remark that for this exercise you should revisit Whitehead theorem, at.pdf study text is also re\-commended)
\end{proof}

\ex (application of Whitehead theorem) Show that $(n-1)$-connected compact manifold of dim n is homotopy equivalent to $S^n$ ($n\geq 2$).

\begin{proof}
Take manifold $M$ as two parts: disk and its complement. Map the disk to $S^n=D^n/S^{n-1}$, that is identity on the disk and complement goes to one point. We have $f\colon M\to S^n$ and for $i<n$ by Hurewicz theorem ($M$ is $(n-1)$-connected) following iso $H_i(M)\cong H_i(S^n)=0$. The same iso for $i>n$ (both $H_i$ are zero).\par 
We conclude our application with this spectacular diagram where all arrows are iso, reasoning goes from the bottom arrows by excision, then with definiton of fundamental class we get the top arrow (our main focus) an iso. (we denote interior of $B$ as $B^O$)
\[
\xymatrix{
H_n(M) \ar[r] \ar[d] & H_n(S^n) \ar[d] \\
H_n(M,M-(B^n)^O)\ar[r] \ar[d] & H_n(S^n,S^n-(B^n)^O)\ar[d]\\
H_n(disk, \dd disk)\ar[r]_{id} & H_n(disk, \dd disk)
}
\]

\end{proof}

\begin{rem}
(Story time) We have that manifold 3-dim (compact) which os pmůy simply connected is homotopy equivalent to $S^3$. There is another result: $M$ 3-dim manifold simply connected is homeomorphic to $S^3$. This other result (it might seem we are close to proving it) is actually famous Poincaré conjecture, one of Millenium Prize Problems and it was already solved by Grigori Perelman in 2002. Interesting story and interesting mathematician for sure. It is well known (as we like to say) that Perelman declined Fields medal (among other prizes).
\end{rem}

\ex Find a map $f$ with Hopf invariant $H(f)=2$.

\begin{proof}
We study a space $X$ with a basepoint $e$. Denote construction $J_2(X)=X\times X /\sim $, where $(x,e)\sim(e,x)$. Apply this idea to $S^n$. We get a projection $p\colon S^n\times S^n \to J_2(S^n)$. On the left we have one 0-cell, two n-cells and one 2n-cell, while on the right we have one of each. We get that $J_2(S^n)$ has to be a space of the form $C_f$, so $H^n(J_2)=\Z$ given by $a$ and $H^{2n}(J_2)=\Z$ given by $b$ and $H^n(S^n\times S^n)=\Z \oplus \Z$ (generators $a_1,a_2$) and $H^{2n}(S^n\times S^n)=\Z$ (with $b_0$). Now, $p^*\colon H^i(J_2)\to H^i(S^n\times S^n)$ and $p^*(a)=a_1+a_2, p^*(b)=b_0$. 
\begin{align*}
    a^2 &= H(f)b \\
    p^*(a^2) &= H(f) p^*(b) \\
    (a_1+a_2)^2&=H(f) b_0\\
    (a_1^2+a_1a_2+a_2a_1+a_2^2) &= H(f)a_1a_2\\
    2a_1a_2&=H(f)a_1a_2\\
    H(f)&=2
\end{align*}
because $b_0=a_1a_2$ and by evenness of the dimension $a_1a_2=a_2a_1$.
\end{proof}

% \ex (hopf map stuff) will not be here...





\end{document}