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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 11, 2017}}}  %
\fancyhead[R]{{\emph{4. 5. 2017}}}                 % Nastavení pro titulní stranu
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\addtocounter{ex}{56}

% 15:49-17:30, 18:30

\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

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