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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 06, 2017}}}  %
\fancyhead[R]{{\emph{30. 3. 2017}}}                 % Nastavení pro titulní stranu
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%13:45



\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}

\newpage
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}


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