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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 04, 2017}}}  %
\fancyhead[R]{{\emph{16. 3. 2017}}}                 % Nastavení pro titulní stranu
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%13:25

\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}

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