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\subsection{Homology of spheres}

\begin{flashcard}[sphere-homology-lemma]
\begin{lemma*}
  The sphere $S^{n - 1}$ is triangulable, and for
  $n - 1 \ge 1$ we have
  \[ H_i(S^{n - 1}) \cong \begin{cases}
    \ZZ & i = 0, n - 1 \\
    0 & \text{otherwise}
  \end{cases} \]
\end{lemma*}

\begin{proof}
  \cloze{$\partial D^n$ has polyhedron homeomorphic to
  $S^{n - 1}$, and we calculated its
  $\Hgp_\bullet$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[brouwer-higher-dimension-thm]
\begin{theorem*}[Brouwer (higher dimensions)]
  \phantom{}
  \cloze{
  \begin{enumerate}[(i)]
    \item There is no retraction of $D^n$ to
      $\partial D^n = S^{n - 1}$.
    \item Any $f : D^n \to D^n$ has a fixed point.
  \end{enumerate}
  }
\end{theorem*}

\begin{proof}
  \cloze{(i) $\implies$ (ii) by the same argument as the
  case $n = 2$. Let $r = D^n \to S^{n - 1}$ be a
  retraction, and consider $\id_{S^{n - 1}} = r
  \circ i$. Then
  \[ \id = (r \circ i)_* : \ub{\Hgp_{n - 1}(S^{n -
  1})}_{\cong \ZZ} \stackrel{i_*}{\to} \ub{\Hgp_{n -
  1}(D^n)}_{\cong \H_{n - 1}(\{*\}) = 0}
  \stackrel{r_*}{\to} \ub{\Hgp_{n - 1}(S^{n -
  1})}_{\cong \ZZ} \]
  for $n - 1 > 0$.}
\end{proof}
\end{flashcard}

We have seen a different triangulation of $S^n$,
via the \gls{simpcomp} $K$ in $\RR^{n + 1}$ with
\glspl{simp} $\simplex<\pm e_1, \ldots, \pm e_{n +
1}>$ and all their faces. We must have
\[ \Hgp_i(K) \cong \begin{cases}
  \ZZ & i = 0, n \\
  0 & \text{otherwise}
\end{cases} \]
by independence of triangulation.

\begin{lemma*}
  The element
  \[ x = \sum_{\bf{a} \in \{1, -1\}^{n + 1}} a_1
  a_2 \cdots a_{n + 1} [a_1 e_1, a_2 e_2, \ldots,
  a_{n + 1} e_{n + 1}] \]
  is a cycle, and generates $\Hgp_n(K) \cong \ZZ$.
\end{lemma*}

\begin{proof}
  When we apply $d_n$ to $x$, the \gls{simp}
  \[ [a_1 e_1, \ldots, \skipel{a_i e_i}, \ldots,
  a_{n + 1} e_{n + 1}] \]
  shows up twice, corresponding to $a_i = +1$,
  $a_i = -1$. So the coefficients cancel out (so
  $x$ is a cycle).

  It generates as it is clearly not divisible in
  $\Cgp_n(K)$ (coefficients are $\pm 1$).
\end{proof}

The reflection $r_i : \RR^{n + 1} \to \RR^{n + 1}$
in the $i$-th coordinate is
\begin{align*}
  (r_i)_*(x)
  &= \sum a_1 \cdots a_{n + 1} [a_1 e_1, \ldots,
  a_{i - 1} e_{i - 1}, -a_i e_i, a_{i + 1} e_{i +
  1}, \ldots, a_{n + 1} e_{n + 1}] \\
  &= -x
\end{align*}
So $(r_i)_* : \Hgp_n(S^n) \to \Hgp_n(S^n)$ is
multiplication by $-1$.

The antipodal map $a : S^n \to S^n$ is $r_1 r_2
\cdots r_{n + 1}$. So
\[ a_* : \Hgp_n(S^n) \to \Hgp_n(S^n) \]
is multiplication by $(-1)^{n + 1}$.

\begin{corollary*}
  If $n$ is even, then $a : S^n \to S^n$ is not
  \gls{homotopic} to $\id$.
\end{corollary*}

\subsection{Homology of surfaces}

\subsubsection*{Example}

\begin{center}
  \includegraphics[width=0.4\linewidth]{images/021361f67ca74c8f.png}
  \includegraphics[width=0.4\linewidth]{images/07e6d7591d9a4591.png}
\end{center}
Homology of $K$: consider
\begin{center}
  \includegraphics[width=0.2\linewidth]{images/053d929af9564e2e.png}
\end{center}
Then $\polyh|W| \hookrightarrow \polyh|K|$ is a
homotopy equivalence. Apply Mayer-Vietoris to 
\begin{center}
  \includegraphics[width=0.2\linewidth]{images/9d6524ac6c7446b2.png}
\end{center}
to obtain
\[ \Hgp_0(K) = \ZZ\{[1]\}, \qquad \Hgp_1(K) =
\ZZ\{a, b\} \]
where $a = [1, 2] + [2, 3] + [3, 1]$ and $b =
[1, 4] + [4, 5] + [5, 1]$.

Note: $r = [5, 6] + [6, 9] + [9, 5] \sim a + b -
a - b = 0$.

Homology of $K_g$: can decompose $K_g = K_{g - 1}
\cup K$, where $K_{g - 1} \cap K = \ssimp^1$.
Mayer-Vietoris gives
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/910fa29cf58943e7.png}
\end{center}
so $\Hgp_0(K_g) = \ZZ\{\text{a vertex}\}$,
$\Hgp_1(K_g) = \ZZ^{2g}$, $H_2(K_g) = 0$. Note $L
= \partial K_g$ is the cycle $r_1 + r_2 + \cdots +
r_g = 0$.

Let $\Sigma_g = \polyh|K_g| \bigcup_{\polyh|L|} \polyh|CL|$.

Homology of $K_g \cup CL$: Apply Mayer-Vietoris,
using $K_g \cap CL = L$, a a triangulation of
$S^1$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/13151fb0c1ab44a5.png}
\end{center}
so $\Hgp_0(K_g \cup CL) \cong \ZZ$, $\Hgp_1(K_g
\cup CL) \cong \ZZ^{2g}$, $\Hgp_2(K_g \cup CL)
\cong \ZZ$. So this is the homology of $\Sigma_g$,
so in particular, given different values of $g$,
the $\Sigma_g$ will be non-homeomorphic, since
their homologies will differ.
