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\begin{flashcard}[Hzero-iso-lemma]
\begin{lemma*}
  There is an isomorphism
  \[ \Hgp_0(K) \cong \cloze{\ZZ \{\piz(\polyh|K|)\}} .\]
\end{lemma*}

\begin{proof}
  \cloze{Note we have a homomorphism
  \begin{align*}
    \phi : \Cgp_0(K)
    &\to \ZZ \{\piz(\polyh|K|)\} \\
    [v]
    &\mapsto \text{the path component of $v \in
    \polyh|K|$}
  \end{align*}
  This is onto: any path-connected component of
  $\polyh|K|$ contains a \gls{vert}. If $[v, w]$ is
  an ordered \nsimp{1}, then $d_1[v, w] = [w] -
  [v]$. But $[v]$ and $[w]$ lie in the same
  path-connected component as the \nsimp{1}
  $\simplex<v, w>$ goes between them. So $\Im(d_1)
  \subset \Ker(\phi)$, so we get an induced
  surjective $\phi : \Hgp_0(K) \to
  \ZZ\{\piz(\polyh|K|)\}$. If $\phi([v]) =
  \phi([w])$, choose a path $\gamma : I \to
  \polyh|K|$ from $v$ to $w$. By
  \gls{simpapprox}, $I = \ssimp^1$ can be
  subdivided so that there is a $g :
  (\ssimp^1)\rsubdiv r \to K$ with $\smapc|g| \homotopic
  \gamma$, i.e. there are $1$-simplices $[v,
  v_1], [v_1, v_2], \ldots, [v_k, w]$. Then
  \[ [w] - [v] = d_1([v, v_1] + [v_1, v_2] +
  \cdots + [v_k, w]) \]
  so $[v] = [w] \in \Hgp_0(K)$.}
\end{proof}
\end{flashcard}

\subsection{Mayer-Vietoris Theorem}

\begin{flashcard}[exact-homs-defn]
\begin{definition*}[Exact homomorphisms]
  \glsadjdefn{exact}{exact}{sequence}
  \glsnoundefn{exs}{exact sequence}{exact sequences}
  \glsnoundefn{ses}{short exact sequence}{short
  exact sequences}
  \glsnoundefn{sescc}{short exact sequence of
  chain complexes}{N/A}
  \cloze{Say that a pair of homomorphisms
  \[ A \stackrel{f}{\to} B \stackrel{g}{\to} C \]
  is \emph{exact} at $B$ if $\Im(f) = \Ker(g)$.
  More generally, a collection of homomorphisms
  \[ \cdots \to A_i \to A_{i + 1} \to A_{i + 2} \to A_{i
  + 3} \to \cdots \]
  is \emph{exact} if it is exact at each $A_j$,
  where $A_j$ has homomorphisms in and out.

  A \emph{short exact sequence} is an \gls{exs}
  \[ 0 \to A \stackrel{f}{\to} B \stackrel{g}{\to}
  C \to 0 .\]
  Note that in this case, $f$ is injective, $g$ is
  surjective and $\Im(f) = \Ker(g)$.

  \glsref[cmap]{Chain maps} $i_\bullet : A_\bullet
  \to B_\bullet$ and $j_\bullet : B_\bullet \to
  C_\bullet$ form a \emph{short exact sequence} of
  \glspl{cc} if each $0 \to A_n
  \stackrel{i_n}{\to} B_n \stackrel{j_n}{\to} C_n
  \to 0$ is a \gls{ses}.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[cc-ses-natural-hom-thm]
\begin{theorem*}
  If $0 \to A_\bullet \stackrel{i_\bullet}{\to}
  B_\bullet \stackrel{j_\bullet}{\to} C_\bullet
  \to 0$ is a \gls{sescc}, then there are natural
  homomorphisms $\partial_* : \Hcc_n(C_\bullet)
  \to \Hcc_{n - 1}(A_\bullet)$ such that
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/f0e83b45730e48cd.png}
  \end{center}
  is an \gls{exs}.
\end{theorem*}

\begin{proof}
  \cloze{\textbf{Constructing $\partial_*$ (snake
  lemma)}:
  \[ \begin{tikzcd}[ampersand replacement=\&]
    0 \ar[r]
    \& A_n \ar[r, "i_n"] \ar[d, "d_n"]
    \& B_n \ar[r, "j_n"] \ar[d, "d_n"]
    \& C_n \ar[r] \ar[d, "d_n"]
    \& 0 \\
    0 \ar[r]
    \& A_{n - 1} \ar[r, "i_{n - 1}"]
    \& B_{n - 1} \ar[r, "j_{n - 1}"]
    \& C_{n - 1} \ar[r]
    \& 0
  \end{tikzcd} \]
  Let $[x] \in \Hcc_n(C_\bullet)$. Then $d_n x =
  0$. By surjectivity, there exists $y \in B_n$
  such that $j_n y = x$. Then since the diagram
  commutes, and $d_n j_n y = 0$, we have that
  $j_{n - 1} d_n y = 0$. Then $d_n(y)$ must be in
  the image of $i_{n - 1}$ (since the sequence is
  \gls{exact}). Pick the unique $z$ such that $i_{n - 1} z =
  d_n(y)$ (uniqueness follows since $i_{n - 1}$ is
  injective). Then
  \begin{align*}
    i_{n - 2} d_{n - 1}(z)
    &= d_{n - 1} i_{n - 1}(z) \\
    &= d_{n - 1} d_n(y) \\
    &= 0
  \end{align*}
  Since $i_{n - 2}$ is injective, this implies
  that $d_{n - 1}(z) = 0$. So $z$ is a cycle. We
  now want to define $\partial_* [x] = [z]$, but
  we must check that this is well-defined: we made
  a choice when picking $y$, and also made a
  choice by picking the representative $x$.

  \textbf{$\partial_*$ is well-defined}: Suppose
  $[x] = [x'] \in \Hcc_n(\Ccc_\bullet)$. Then $x -
  x' = d_{n + 1}(a)$, $a \in \Ccc_{n + 1}$. The
  same process for $x'$ gives a $y' \in B_n$. As
  $j_{n + 1}$ is surjective, can write $a = j_{n +
  1}(b)$. Then
  \[ j_n(y - y') = x - x' = j_n d_{n + 1}(b) ,\]
  so by \glsref[exact]{exactness} at $B_n$,
  \[ y - y' = d_{n + 1}(b) + i_n(c) \]
  for some $c \in A_n$. Now $z'$ is such that
  $i_{n - 1}(z') = d_n(y')$, so
  \begin{align*}
    i_{n - 1}(z - z')
    &= d_n(y) - d_n(y') \\
    &= d_n(y - y') \\
    &= d_n(d_{n + 1}() + i_n(c)) \\
    &= d_n i_n(c) \\
    &= i_{n - 1} d_n(c)
  \end{align*}
  injectivity of $i_{n - 1}$ again, shows $z - z'
  = d_n(c)$. So $[z] = [z'] \in \Hcc_{n -
  1}(A_\bullet)$.

  \textbf{$\partial_*$ is a homomorphism}: given
  $[x_1], [x_2] \in \Hcc_n(C_\bullet)$ with
  corresponding $y_1, y_2, z_1, z_2$, choose $y_1
  + y_2$ to be the lift of $x_1 + x_2$. This gives
  $z_1 + z_2$ as the result, so
  \[ \partial_*[x_1 + x_2] = [z_1 + z_2] = [z_1] +
  [z_2] .\]

  \textbf{Exactness at $\Hcc_n(C_\bullet)$}: Let
  $[x] \in \Im(\cmapstar j)$, so there exists $y
  \in B_n$ such that $j_n(y) = x$ and $y$ is a
  cycle. Can use this $y$ to calculate
  $\partial_*[x]$. As $y$ is a cycle, $d_n(y) =
  0$, so $z$ is $0$, so $\partial_*[x] = 0$.

  Suppose now that $\partial_*[x] = 0$. Calculate
  this by choosing $y \in B_n$ and taking the
  corresponding $z$. Then $z = d_n(t)$ (as $[z] =
  0$). Then $j_n(y - i_n(t)) = x$ and $d_n(y -
  i_n(t)) = d_n y - d_n i_n(t) = i_n(z - z) = 0$.
  So $\cmapstar j [y - i_n(t)] = [x]$, so $[x] \in
  \Im(\cmapstar j)$.

  \textbf{Exactness at $\Hcc_n(B_\bullet)$}: As
  $j_n \circ i_n = 0$, $\Im(\cmapstar i) \subseteq
  \Ker(\cmapstar j)$. Suppose $\cmapstar j [y] =
  0$. Then $j_n(y) = d_{n + 1}(a)$ for some $a \in
  \Ccc_{n + 1}$. Let $b \in B_{n + 1}$ be such
  that $j_n(b) = a$. Then
  \[ j_n(y - d_{n + 1} b) = d_{n + 1}(a) - j_n
  d_{n + 1}(b) = d_{n + 1}(a) - d_{n + 1} j_{n +
  1}(b) = 0 .\]
  So let $y - d_{n + 1} (b) = i_n(t)$. Then $d_n(t
  - d_{n + 1}(b)) = 0$, so $[y] = [y - d_{n +
  1}(b)] = [i_n(t)] = \cmapstar i[t]$.

  \textbf{Exactness at $\Hcc_n(A_\bullet)$}: Let
  $[z] \in \partial_*[x]$. Then $i_n(z) = d_{n +
  1}(y)$, so $\cmapstar i[z] = 0$. Let $[z]$ be
  such that $\cmapstar i[z] = 0$. Then $i_n(z) =
  d_{n + 1}(y)$ for some $y$. Thus $[z] =
  \partial_* [j_{n + 1}(y)]$, so $\Ker(\cmapstar
  i) \subseteq \Im(\partial_*)$.}
\end{proof}
\end{flashcard}
