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\begin{lemma*}
  \glssymboldefn{smapc}{$|f|$}{$|f|$}
  A \gls{simpmap} $f : K \to L$ of
  \glspl{simpcomp} induces a continuous map $|f| :
  \polyh|K| \to \polyh|L|$, and $|f \circ g| = |f|
  \circ |g|$.
\end{lemma*}

\begin{proof}
  For a $\sigma \in K$, $\sigma = \simplex<a_0,
  a_1, \ldots, a_p>$, define
  \begin{align*}
    f_\sigma : \sigma
    &\to \polyh|L| \\
    \sum_{i = 0}^p t_i a_i
    &\mapsto \sum_{i = 0}^p t_i f(a_i)
  \end{align*}
  which is linear in the $t_i$, hence continuous.
  If $\tau \face \sigma$, then $f_\tau =
  f_\sigma|_\tau$, so $f_\sigma|_{\sigma \cap
  \sigma'} = f_{\sigma'}|_{\sigma \cap \sigma'}$,
  so the $f_\sigma$ glue to a continuous $|f| :
  \polyh|K| = \bigcup_{\sigma \in K} \sigma \to
  \polyh|L|$.
  The formula for $|f|$ shows that it behaves as
  claimed under composition.
\end{proof}

So we can recover $f$ from $\smapc|f|$ and the discrete
sets $\vset K \subset \polyh|K|$, $\vset L \subset
\polyh|L|$, i.e. a \gls{simpmap} is the same as a
continuous map $\polyh|K| \to \polyh|L|$ which
sends \glspl{vert} to \glspl{vert}, and is affine
on each \gls{simp}.

\begin{flashcard}[star-link-defn]
\begin{definition*}[Star and link]
  \glssymboldefn{St}{St}{St}
  \glssymboldefn{Lk}{Lk}{Lk}
  \glsnoundefn{star}{star}{stars}
  \glsnoundefn{link}{link}{links}
  For a $x \in \polyh|K|$,
  \begin{enumerate}[(i)]
    \item The (open) \emph{star} of $x$ is the
      union of the interiors of the \glspl{simp}
      which contain $x$
      \[ \St_K(x) = \bigcup_{\sigma \in K}
      \interior \sigma \subset \RR^m \]
      The complement of $\St_K(x)$ is the union of
      \glspl{simp} which do not contain $x$, a
      \gls{polyh}, so closed. Thus $\St_K(x)$ is
      open.
    \item The \emph{link} of $x$, $\Lk_K(x)$ is
      the union of those \glspl{simp} which do not
      contain $x$, but are \glspl{face} of a
      \gls{simp} which does not contain $x$.
  \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/1e163d8bd14c432e.png}
  \end{center}
\end{example*}

\subsection{Simplicial approximation}

\begin{flashcard}[simplicial-approx-defn]
\begin{definition*}[Simplicial approximation]
  \glsnoundefn{simpapprox}{simplicial
  approximation}{simplicial approximations}
  \cloze{Let $f : \polyh|K| \to \polyh|L|$ be a
  continuous \gls{map}. A \emph{simplicial
  approximation} to $f$ is a function $g : \vset K
  \to \vset L$ such that
  \[ f(\St_K(v)) \subset \St_L(g(v)) \]
  for all $v \in \vset K$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[simplicial-approximation-homotopy-lemma]
\begin{lemma*}
  \refsteplabel[simplicial approximations are
  simplicial maps]{simpapprox_are_simpmap}
  If $g$ is a \gls{simpapprox} to a continuous
  \gls{map} $f$, then $g$ is a \gls{simpmap}, and
  $f$ is is \gls{homotopic} to $\smapc|g|$.
  Furthermore, this \gls{homotopy} may be taken
  \glsref[homotopy]{relative} to
  \[ \{x \in \polyh|K| \st f(x) = \smapc|g|(x)\} .\]
\end{lemma*}

\begin{proof}
  \cloze{To show that $g$ defines a \gls{simpmap}, for
  $\sigma \in K$ we must show that the images
  under $g$ of the \glspl{vert} of $\sigma$ span a
  \gls{simp} of $L$.

  For $x \in \interior \sigma$, we have $x \in
  \bigcap_{v \in \vset \sigma} \St_K(v)$, so
  \[ f(x) = \bigcap_{v \in \vset \sigma}
  f(\St_K(v)) \subset \bigcap_{v \in \vset
  \sigma} \St_L(g(v)) .\]
  If $\tau$ is the unique \gls{simp} of $L$ with
  $f(x) \in \interior \tau$, then each $g(v)$ is a
  \gls{vert} of $\sigma$. So the $\{g(v)\}$ span a
  \gls{face} of $\tau$, which is a \gls{simp} of
  $L$.

  Want to show $f \homotopic \smapc|g|$. If
  $\polyh|L| \subseteq \RR^m$, then let
  \begin{align*}
    H : \polyh|K| \times I
    &\to \polyh|L| \subset \RR^m \\
    (x, t)
    &\mapsto t \cdot f(x) + (1 - t) \smapc|g|(x)
  \end{align*}
  This is continuous, so need to show it indeed
  lies in $\polyh|L|$. Let $x \in \interior\sigma
  \subset \polyh|K|$ and suppose $f(x) \in
  \interior \tau \subset \polyh|L|$. If $\sigma =
  \simplex<a_0, \ldots, a_p>$ then by the above
  each $g(a_i)$ is a \gls{vert} of $\tau$. Then
  \[ \smapc|g|(x) = \sum_{i = 0}^p t_i g(a_i) \in
  \tau \]
  as it is a convex linear combination of
  \glspl{vert} of $\tau$. As $f(x) \in \tau$, each
  of $tf(x) + (1 - t)\smapc |g| (x)$ lies in
  $\tau$ too.}
\end{proof}
\end{flashcard}

\begin{flashcard}[barycentre-defn]
\begin{definition*}[Barycentre]
  \glsnoundefn{bcentre}{barycentre}{barycentres}
  \glssymboldefn{bcentre}{hat $\sigma$}{hat $\sigma$}
  \cloze{The \emph{barycentre} of a \gls{simp} $\sigma =
  \simplex<a_0, \ldots, a_p>$ is the point
  \[ \hat{\sigma} = \frac{1}{p + 1} (a_0 + a_1 +
  \cdots + a_p) .\]}
\end{definition*}
\end{flashcard}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/c95f96283b024c5b.png}
\end{center}

\begin{flashcard}[barycentre-subdivision-defn]
\begin{definition*}[Barycentre subdivision]
  \glssymboldefn{rsubdiv}{$K^{(r)}$}{$K^{(r)}$}
  \glssymboldefn{subdiv}{$K'$}{$K'$}
  \glsnoundefn{bsubdiv}{barycentric
  subdivision}{N/A}
  \cloze{The \emph{barycentre subdivision} of a
  \gls{simpcomp} $K$ is
  \[ K' = \{\simplex<\bcentre{\sigma_0}, \ldots,
  \bcentre{\sigma_p}> \st \sigma_i \in K, \text{
  and } \sigma_0 \propface \sigma_1 \propface
  \cdots \propface \sigma_p\} \]
  We will use the notation $K^{(r)} = (K^{(r -
  1)})'$ (where $K^{(0)} = K$).}
\end{definition*}
\end{flashcard}

\begin{note*}
  It is not obvious that this is a \gls{simpcomp}.
\end{note*}

\begin{flashcard}[subdiv-simpcomp-prop]
\begin{proposition*}
  $K\subdiv$ is a \gls{simpcomp}, and
  $\polyh|K\subdiv| = \polyh|K|$.
\end{proposition*}

\begin{proof}
  \cloze{If $\sigma_0 \propface \sigma_1 \propface \cdots
  \propface \sigma_p$ then the
  $\bcentre{\sigma_i}$ are \gls{affindep}: Suppose
  $\sum_{i = 0}^p t_i \bcentre{\sigma_i} = 0$ and
  $\sum_{i = 0}^p t_i = 1$. Let $j = \max\{i \st
  t_i \neq 0\}$. Then
  \[ \bcentre{\sigma_j} = -\sum_{i = 0}^p
  \frac{t_i}{t_j} \bcentre{\sigma_i} \in \sigma_{j
  - 1}, \]
  so $\bcentre{\sigma_j}$ lies in a proper
  \gls{face} of $\sigma_j$, which is not possible.
  Thus all $t_i$ must be $0$.

  $K\subdiv$ is a \gls{simpcomp}: Let
  $\simplex<\bcentre{\sigma_0}, \ldots,
  \bcentre{\sigma_p}> \in K\subdiv$. A \gls{face}
  is given by omitting some
  $\bcentre{\sigma_j}$'s. But omitting some
  $\sigma_j$'s from $\sigma_0 \propface \sigma_1
  \propface \cdots \propface \sigma_p$ still gives
  a strictly increasing chain of \glspl{simp} of
  $K$. Let $\sigma' = \simplex<\bcentre{\sigma_0},
  \ldots, \bcentre{\sigma_p}>$, $\tau' =
  \simplex<\bcentre{\tau_0}, \ldots,
  \bcentre{\tau_1}>$ and consider $\sigma' \cap
  \tau'$. This is inside $\sigma_p \cap \tau_p$,
  which is a \gls{simp} $\delta$ of $K$. So there
  are \glspl{simp}
  \[ \sigma'' = \simplex<\bcentre{\sigma_0} \cap
  \delta, \ldots, \bcentre{\sigma_p} \cap \delta>,
  \qquad \tau'' = \simplex<\bcentre{\tau_0} \cap
  \delta, \ldots, \bcentre{\tau_p} \cap \delta> \]
  of $K$. Now $\sigma' \cap \tau' = \sigma'' \cap
  \tau''$ This reduces to the case that $\sigma''$
  and $\tau''$ are contained in a \gls{simp}
  $\delta$ of $K$. Now we split into cases. If
  $\sigma''$ and $\tau''$ contain $\bcentre \delta$, then
  let $\ol{\sigma}$, $\ol{\tau}$ be the
  \glspl{face} of $\sigma'', \tau''$ opposite to
  $\bcentre \delta$. Then $\sigma'' \cap \tau''$ is
  spanned by $\bcentre \delta$ and $\ol{\sigma}''
  \cap \ol{\tau}''$. But $\ol{\sigma}'' \cap
  \ol{\tau}'' \subset \bound \delta$, which has a
  smaller \gls{scdim} than $\sigma$, so can
  suppose it is a \gls{simp} of $\bound \delta$ by
  induction on \gls{scdim}. If $\sigma''$ or
  $\tau''$ does not contain $\bcentre \delta$,
  then $\sigma'' \cap \tau'' \subset \bound
  \delta$, so again finish by induction on
  \gls{scdim}.

  $\polyh|K\subdiv| = \polyh|K|$: Note $\simplex<
  \bcentre{\sigma_0}, \bcentre{\sigma_1}, \ldots,
  \bcentre{\sigma_p}> \face \sigma_p \face
  \polyh|K|$, so $\polyh|K\subdiv| \subseteq
  \polyh|K|$. Conversely, if $x \in \sigma =
  \simplex<a_0, \ldots, a_p> \subset \polyh|H|$ is
  written as
  \[ x = \sum_{i = 0}^p t_i a_i ,\]
  can reorder the $a_i$ so that $t_0 \ge t_1 \ge
  t_2 \ge \cdots \ge t_p$, so
  \begin{align*}
    x
    &= \ub{(t_0 - 1)}_{\ge 0}a_0 + 2\ub{(t_1 -
    t_2)}_{\ge 0} \left( \frac{a_0 + a_1}{2}
    \right) + 3\ub{(t_2 - t_3)}_{\ge 0} \left(
    \frac{a_0 + a_1 + a_2}{3} \right) + \cdots \\
    &= (t_0 - t_1) \bcentre{\simplex<a_0>} + 2(t_1
    - t_1)\bcentre{\simplex<a_0, a_1>} + 3(t_2 -
    t_3) \bcentre{\simplex<a_0, a_1, a_2>} +
    \cdots \\
    &\in \simplex<\bcentre{\simplex<a_0>},
    \bcentre{\simplex<a_0, a_1>}, \ldots,
    \bcentre{\simplex<a_0, \ldots, a_p>}> \\
    &\subseteq \polyh|K\subdiv| \qedhere
  \end{align*}}
\end{proof}
\end{flashcard}