%! TEX root = AlgT.tex % vim: tw=50 % 04/03/2024 11AM \begin{flashcard}[Mayer-Vietoris-thm] \begin{theorem*}[Mayer-Vietoris Theorem] Let $K$ be a \gls{simpcomp}, $M$, $N$ be subcomplexes, and $L = M \cap N$. Suppose also that $M$ and $N$ cover $K$, i.e. every \gls{simp} of $K$ is in $M$ or $N$. Write \[ \begin{tikzcd}[ampersand replacement=\&] L \ar[r, "j", hook] \ar[d, "i", hook] \& N \ar[d, "l", hook] \\ M \ar[r, "k", hook] \& K \end{tikzcd} \] for the inclusion maps. There are natural homomorphisms $\partial_* : \Hcc_n(K) \to \Hcc_{n - 1}(L)$ such that \begin{center} \includegraphics[width=0.6\linewidth]{images/ad35272b1a474ddd.png} \end{center} is a long \gls{exs}. \end{theorem*} \begin{proof} \cloze{Have a \gls{sescc} \[ 0 \to \Cgp_\bullet(L) \stackrel{(\cmapstar i, \cmapstar j)}{\longrightarrow} \Cgp_\bullet(M) \oplus \Cgp_\bullet(N) \stackrel{\cmapstar k - \cmapstar l}{\longrightarrow} \Cgp_\bullet(K) \to 0 \] because: \begin{enumerate}[(i)] \item $i_\bullet$ and $j_\bullet$ are both injective, so \gls{exact} at left-hand term. \item $k_\bullet$ and $l_\bullet$ are jointly surjective, as $M$ and $N$ cover $K$. \item Suppose $(x, y) \in \Cgp_n(M) \oplus \Cgp_n(N)$ is in $\Ker(k_n - l_n)$, i.e. $k_n(x) = l_n(y)$. So both sides are a linear combination of simplices in $M \cap N = L$, i.e. $\exists z \in \Cgp_n(L)$ such that $x = i_n(z)$, $y = j_n(z)$. So $\Ker(k_n - l_n) \subseteq \Im(i_n \oplus j_n)$. Other constraint is clear. \end{enumerate} Now apply the above algebraic theorem.} \end{proof} \end{flashcard} \subsection{Continuous maps and homotopy inverse} \begin{flashcard}[contiguous-simp-maps-defn] \begin{definition*} \glsadjdefn{contig}{contiguous}{\gls{simpmap}} \cloze{\glsref[simpmap]{Simplicial maps} $f, g : K \to L$ are \emph{contiguous} if for each $\sigma \in K$, $f(\sigma)$ and $g(\sigma)$ are \glspl{face} of some \gls{simp} $\tau$ of $L$.} \end{definition*} \end{flashcard} \begin{flashcard}[simp-approx-contig-lemma] \begin{lemma*} If $f$ and $g$ are both \glspl{simpapprox} to $F : \polyh|K| \to \polyh|L|$, then $f$ and $g$ are \gls{contig}. \end{lemma*} \begin{proof} \cloze{If $x \in \interior{\sigma}\subset \polyh|K|$ and $F(x) \in \interior \tau \subset \polyh|L|$ then as in the proof of ``\nameref{simpapprox_are_simpmap}'', $f(\sigma)$ and $g(\sigma)$ are \glspl{face} of $\tau$.} \end{proof} \end{flashcard} \begin{flashcard}[contig-implies-chomic-lemma] \begin{lemma*} If $f, g : K \to L$ are \gls{contig}, then $f_\bullet \chomic g_\bullet : \Cgp_\bullet(K) \to \Cgp_\bullet(L)$, and so $\cmapstar f = \cmapstar g$. \end{lemma*} \begin{proof} Choose an ordering $\prec$ of $\vset K$, and represent a basis of $\Cgp_n(K)$ by $[a_0, \ldots, a_n]$ with $a_0 \prec a_1 \prec \cdots \prec a_n$. Define a homomorphism \begin{align*} h_n : \Cgp_{n + 1}(K) &\to \Cgp_{n + 1}(L) \\ [a_0, \ldots, a_n] &\mapsto \sum_{i = 0}^n (-1)^i [f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_k)] \end{align*} Direct calculation shows that \[ d_{n + 1} \circ h_n + h_{n - 1} \circ d_n = g_n - f_n .\] This is the \gls{chom}. \end{proof} \end{flashcard} \begin{flashcard}[simpapprox-to-id-lemma] \begin{lemma*} Let $K\subdiv$ be the \gls{bsubdiv}, choose $a : K \cong \vset{K\subdiv} \to \vset K$ any function sending $\bcentre{\sigma}$ to some \gls{vert} of $\sigma$. Such an $a : \vset{K\subdiv} \to \vset K$ is a \gls{simpapprox} to the identity map on $\polyh|K| = \polyh|K\subdiv|$. Furthermore, all \gls{simpapprox} to $\id$ are of this form. \end{lemma*} \begin{proof} \cloze{Seen the first part (just before the definition of \gls{mesh}). If $g : \vset{K\subdiv} \to \vset K$ is a \gls{simpapprox} to $\id$, then \[ \interior{\sigma} \subseteq \id(\St_{K\subdiv}(\bcentre{\sigma})) \subseteq \St_K(g(\bcentre{\sigma})) .\] So $g(\bcentre{\sigma})$ must be a \gls{vert} of $\sigma$.} \end{proof} \end{flashcard} \begin{flashcard}[subdiv-simpapprox-id-iso-prop] \begin{proposition*} If $a : K\subdiv \to K$ is a \gls{simpapprox} to $\id$, then $\cmapstar a : \Hgp_n(K\subdiv) \to \Hgp_n(K)$ is an isomorphism for all $n$. Furthermore, this isomorphism is independent of the choice of $a$. \end{proposition*} \begin{proof} \cloze{First suppose $K$ is a \gls{simp} $\sigma = \ssimp^n \subset \RR^{n + 1}$. Then $K\subdiv$ is a \gls{cone} with \gls{conep} $\bcentre{\sigma}$. Also $K$ is a \gls{cone} with any \gls{vert} as \gls{conep}. So $\cmapstar a : \Hgp_n(K\subdiv) \to \Hgp_n(K)$ is an isomorphism for $n > 0$ (both sides are $0$). Also, $\cmapstar \Hgp_0(K\subdiv) \to \Hgp_0(K)$ is an isomorphism too, with $\cmapstar a[\bcentre{\sigma}] = [a(\bcentre{\sigma})]$. So done if $K$ is a \gls{simp}. \textbf{General case}: double induction on: \begin{enumerate}[(i)] \item $\dim K$ \item number of \glspl{simp} of $K$ \end{enumerate} Let $\sigma \in K$ be a \gls{simp} of maximal dimansion. Then $L = K - \{\sigma\}$ is a \gls{simpcomp}. Let \[ S = \{\text{$\sigma$ and all its faces}\}, \qquad T = S \cap L = \{\text{all proper faces of $\sigma$}\} .\] Any \gls{simpapprox} $a : K\subdiv \to K$ to $\id$ sends $L\subdiv$ into $L$, $S\subdiv$ into $S$, $T\subdiv$ into $T$. So we get a map of Mayer-Vietoris sequences: \[ \begin{tikzcd}[ampersand replacement=\&] \Hgp_n(T\subdiv) \ar[r] \ar[d, "\simeq" description, "\cmapstar{(a|_{T\subdiv})}"] \& \Hgp_n(S\subdiv) \oplus \Hgp_n(L\subdiv) \ar[r] \ar[d, "\simeq" description, "\cmapstar{(a_{K\subdiv})} \oplus \cmapstar{(a|_{L\subdiv})}"] \& \Hgp_n(K\subdiv) \ar[r, "\partial_*"] \ar[d, "\cmapstar a"] \& \Hgp_{n - 1}(T\subdiv) \ar[r] \ar[d, "\simeq" description, "\cmapstar{(a_{T\subdiv})}"] \& \Hgp_{n - 1}(S\subdiv) \oplus \Hgp_{n - 1}(L\subdiv) \ar[d, "\simeq" description] \\ \Hgp_n(T) \ar[r] \& \Hgp_n(S) \oplus \Hgp_n(L) \ar[r] \& \Hgp_n(K) \ar[r, "\partial_*"] \& \Hgp_{n - 1}(T) \ar[r] \& \Hgp_{n - 1}(S) \oplus \Hgp_{n - 1}(L) \end{tikzcd} \] $T\subdiv$ has dimension strictly smaller than $K$, so by induction hypothesis, we may assume that $\cmapstar{(a|_{T\subdiv})}$ and $\cmapstar{(a|_{T\subdiv})}$ are isomorphisms. We can also use the induction hypothesis to assume that the second and fifth downward maps are isomorphisms: we have an isomorphism on the left since $S$ is a \gls{simp} (and we did the \gls{simp} case earlier), and on the right factor we have strictly fewer \glspl{simp}, so we use the induction hypothesis. Now by the ``Five lemma'' (see \es[3]{4}), we deduce that $\cmapstar a$ is an isomorphism.} \end{proof} \end{flashcard}