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\begin{flashcard}[nuk-inverse-lemma]
\begin{lemma*}
  There is a \gls{cmap} $s_\bullet : C_\bullet(K)
  \to C_\bullet(K\subdiv)$ given by sending a
  \gls{simp} $\sigma$ to an appropriate linear
  combination of the \glspl{simp} of $K\subdiv$
  which compose $\sigma$. On $H_*$ it induces
  $\nuk_K^{-1} : \Hgp_n(K) \simto
  \Hgp_n(K\subdiv)$.
\end{lemma*}

\begin{proof}
  \cloze{Start by $s_0([v_0]) = [\bcentre{v_0}]$.
  Supposing $s_0, \ldots, s_{n - 1}$ have been
  defined and satisfy $d_i \circ s_i = s_{i - 1}
  \circ d_i$ for all $i < n$. Then define
  $s_n(\sigma) = [\bcentre{\sigma}, s_{n - 1}
  d_n(\sigma)]$,
  interpreted linearly in the second variable.

  Example: $s_1([v_0, v_1]) =
  [\bcentre{\simplex<v_0, v_1>},
  \bcentre{\simplex<v_1>} -
  \bcentre{\simplex<v_0>}] =
  [\bcentre{\simplex<v_0, v_1>},
  \bcentre{\simplex<v_1>}] -
  [\bcentre{\simplex<v_0, v_1>},
  \bcentre{\simplex<v_0>}]$.

  We calculate
  \begin{align*}
    d_n s_n([v_0, \ldots, v_n])
    &= [\bcentre{\simplex<v_0, \ldots, v_n>}, s_{n
    - 1} d_n[v_0, \ldots, v_n]] \\
    &= s_{n - 1} d_n [v_0, \ldots, v_n] -
    [\bcentre{\simplex<v_0, \ldots, v_n>},
    \ub{d_{n - 1} s_{n -
    1}}_{s_{n - 2} d_{n - 1}} d_n [v_0, \ldots,
    v_n]] \\
    &= s_{n - 1} d_n [v_0, \ldots, v_n]
  \end{align*}
  so $d_n s_n = s_{n - 1} d_n$. So this
  $s_\bullet$ is a \gls{cmap}. Let $\prec$ be an
  ordering of $\vset K$ and $a : K \cong
  \vset{K\subdiv} \to \vset K$ send $\bcentre
  \sigma$ to the smallest \gls{vert} of $\sigma$
  with respect to $\prec$. This is some
  \gls{simpapprox} to $\id$. Now $a_0 \circ s_0 =
  \id$. Suppose $a_i \circ s_i = \id$ for $i < n$.
  If $[v_0, \ldots, v_n]$, $v_0 \prec \cdots \prec
  v_n$ then
  \begin{align*}
    a_n s_n([v_0, \ldots, v_n])
    &= a_n ([\bcentre{\simplex<v_0, \ldots, v_n>},
    s_{n - 1} d_n [v_0, \ldots, v_n]]) \\
    &= [v_0, \ub{a_{n - 1} s_{n - 1}}_{\id} d_n [v_0, \ldots,
    v_n]] \\
    &= [v_0, d_n[v_0, \ldots, v_n]] \\
    &= [v_0, \simplex<v_1, \ldots, v_n> - \sum \pm
    \simplex<v_0, \ldots, \skipel{v_i}, \ldots,
    v_n>] \\
    &= [v_0, \ldots, v_n]
  \end{align*}
  So $a_* \circ s_* = \id$, so $\nuk \circ s_* =
  \id$, so $s_* = \nuk_K^{-1}$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[lefschetz-fp-thm]
\begin{theorem*}[Lefschetz Fixed Point Theorem]
  \cloze{Let $f : X \to X$ be a \gls{map} of
  \glspl{polyh}. If $\lef(f) \neq 0$ then $f$ has
  a fixed point.}
\end{theorem*}

\begin{proof}
  \cloze{Let $X \cong \polyh|K| \subset \RR^N$.
  Suppose $f$ does not have a fixed point. Let
  \[ \delta \defeq \inf\{|x - f(x)|, x \in X\} .\]
  As $X$ is compact, $\delta > 0$. Let$K$ be a
  \gls{triang} of $X$ with $\mesh(K) <
  \frac{\delta}{2}$, and choose a \gls{simpapprox}
  $g : K\rsubdiv r \to K$ to $f$. For $v \in \vset
  {K\rsubdiv r}$ we have $f(v) \in
  f(\St_{K\rsubdiv r}(v)) \subset \St_K(g(v))$, so
  $|f(v) - g(v)| < \frac{\delta}{2}$. But $|f(v) -
  v| > \delta$, so $|g(v) - v| >
  \frac{\delta}{2}$. So if $v \in \sigma \in K$,
  then $g(v) \notin \sigma$. The map $f_*$ is
  defined as $g_* \circ \nuk_{K, r}^{-1} = g_*
  \circ s_*^{(r)}$ where $s_*^{(r)}$ is the
  $r$-fold iteration of the map in the previous
  Lemma. So
  \begin{align*}
    \lef(f)
    &\defeq \sum_i (-1)^i \Trace(f_* :
    H_i(X) \to H_i(X)) \\
    &= \sum_i (-1)^i \Trace(g_i \circ s_i^{(r)} :
    C_i(K) \to C_i(K))
  \end{align*}
  by last lecture. If $\sigma \in K$ is an
  \nsimp{i}, then $s_i^{(r)}(\sigma)$ is a sum of
  \glspl{simp} inside $\sigma$, so $g_i
  s_i^{(r)}(\sigma)$ is a sum of \glspl{simp} not
  including $\sigma$. So the matrix for $g_i \circ
  s_i^{(r)}$ has zeroes on the diagonal, so the
  trace is $0$.}
\end{proof}
\end{flashcard}

\begin{example*}
  If $X$ is a \gls{contractible} \gls{polyh}, then
  \[ \Hgp_n(X) = \begin{cases}
    \ZZ & n = 0 \\
    0 & \text{otherwise}
  \end{cases} \]
  so any $f : X \to X$ has $\lef(f) = 1$, so has a
  fixed point.
\end{example*}

\begin{example*}
  $S_1 = \RR\PP^2$ has
  \[ \Hgp_n(\RR\PP^2) = \begin{cases}
    \ZZ & n = 0 \\
    \ZZ / 2\ZZ & n = 1 \\
    0 & \text{otherwise}
  \end{cases} \]
  so
  \[ \Hqq_n(\RR\PP^2; \QQ) = \begin{cases}
    \QQ & n = 0 \\
    0 & \text{otherwise}
  \end{cases} \]
  so any $f : \RR\PP^2 \to \RR\PP^2$ has $\lef(f)
  = 1$, so has a fixed point.
\end{example*}

\begin{example*}
  Let $G$ be a topological group which is
  path-connected and non-trivial. Suppose it is a
  polyhedron. If $g \neq 1 \in G$, then $g \cdot
  \bullet : G \to
  G$ has no fixed points, so
  \[ 0 = L(g \cdot \bullet) = L(1 \cdot \bullet) =
  L(\id) = \eul(G) \]
  as $g \cdot \bullet \homotopic 1 \cdot \bullet$
  as $G$ is path-connected. This can be used to
  show that certain topological spaces cannot be
  given a topological group structure.
\end{example*}

\begin{example*}
  Consider $S_3$, which is the connected sum of
  three copies of $\RR\PP^2$. Let $f : S_3 \to
  S_3$ be such that $f \circ f = \id$. We have
  \[ \Hqq_n(S_3; \QQ) \cong \begin{cases}
    \QQ & n = 0 \\
    \QQ^2 & n = 1 \\
    0 & \text{otherwise}
  \end{cases} \]
  so $\lef(f) = 1 - \Trace(f_* : \Hqq_1(S_3; \QQ)
  \to \Hqq_1(S_3; \QQ))$. As $f \circ f = \id$,
  $f_* \circ f_* = \id$. So $f_* : \Hqq_1(S_3;
  \QQ) \to \Hqq_1(S_3, \QQ)$ squares to $\id$. So
  its minimal polynomial divides $x^2 - 1$, so it
  is diagonalisable with eigenvalues $\pm 1$. So
  its trace is one of $\{2, 0, -2\}$. So $\lef(f)
  \in \{-1, 1, 3\}$, none of which are $0$, so $f$
  has a fixed point.
\end{example*}
