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\begin{document}
\pagestyle{fancy}                      %Pro větší možnosti práce se záhlavími a zápatími
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 03, 2017}}}  %
\fancyhead[R]{{\emph{9.3.2017}}}                 % Nastavení pro titulní stranu
%\fancyfoot[L]{Školní rok 2016/2017}                %
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%3 hodiny

\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}

\newpage
\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.


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