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\begin{flashcard}[nth-simplicial-homology-group-defn]
\begin{definition*}[$n$-th simplicial homology group]
  \glssymboldefn{Hgp}{$H_n(K)$}{$H_n(K)$}
  \cloze{The \emph{$n$-th simplicial homology group of
  $K$} is
  \[ H_n(K) \defeq \frac{\Ker(d_n : C_n(K) \to
  C_{n - 1}(K))}{\Im(d_{n - 1} : C_{n + 1}(K) \to
  C_n(K))} .\]}
\end{definition*}
\end{flashcard}

\begin{example*}
  Let $K$ be the union of all the proper
  \glspl{face} of the standard \nsimp{2} $\ssimp^2
  \subset \RR^3$, i.e.
  \[ K = \{\simplex<e_1>, \simplex<e_2>,
  \simplex<e_3>, \simplex<e_1, e_2>, \simplex<e_2,
  e_3>, \simplex<e_3, e_1>\} .\]
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/9cbdc51d9ec245bb.png}
  \end{center}
  Order the \glspl{vert} as $e_1 \prec e_2 \prec
  e_3$. Then
  \begin{align*}
    \Cgp_0(K)
    &= \ZZ\{[e_1], [e_2], [e_3]\} \\
    \Cgp_1(K)
    &= \ZZ\{[e_1, e_2], [e_2, e_3], [e_1, e_3]\}
    \\
    \Cgp_n(K)
    &= 0
    &&(n \ge 2)
  \end{align*}
  and
  \[ \dhom_1 : \Cgp_1(K) \to \Cgp_0(K), \qquad
  \dhom_1[e_i, e_j] = [e_j] - [e_i] \]
  \[ \begin{pmatrix}
    -1 & 0 & -1 \\
    1 & -1 & 0 \\
    0 & 1 & 1
  \end{pmatrix} \]
  Note $\Im(d_1) = \langle [e_i] -
  [e_j]\rangle_{\ZZ}$, so
  \[ \Hgp_0(K) = \frac{\ZZ\{[e_1], [e_2],
  [e_3]\}}{\langle [e_i] - [e_j] \rangle} \cong
  \ZZ .\]
  For $\Hgp_1$, note
  \begin{align*}
    \Ker(d_1)
    &= \ZZ\{[e_1, e_2] \ub{- [e_1,
    e_3]}_{= [e_3, e_1]} +
    [e_2, e_3]\} \cong \ZZ \\
    \Im(d_2)
    &= 0
  \end{align*}
  So $\Hgp_1(K) \cong \ZZ$.
\end{example*}

\begin{example*}
  Now let $L$ be the standard \nsimp{2} $\ssimp^2
  \subset \RR^3$, i.e.
  \[ L = \{\simplex<e_1>, \simplex<e_2>,
  \simplex<e_3>, \simplex<e_1, e_2>, \simplex<e_1,
  e_3>, \simplex<e_2, e_3>, \simplex<e_1, e_2,
  e_3>\} .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/dd2e4a721abf46bf.png}
  \end{center}
  Then
  \begin{align*}
    \Cgp_0(L)
    &= \ZZ\{[e_1], [e_2], [e_3]\} \\
    \Cgp_1(L)
    &= \ZZ\{[e_1, e_2], [e_1, e_3], [e_2, e_3]\}
    \\
    \Cgp_2(L)
    &= \ZZ\{[e_1, e_2, e_3]\}
  \end{align*}
  So
  \[ 0 \stackrel{\dhom_4}{\to} 0
  \stackrel{\dhom_3}{\to} \ub{\Cgp_2(L)}_{= \ZZ}
  \stackrel{\dhom_2}{\to} \ub{\Cgp_1(L)}_{= \ZZ^3}
  \stackrel{\dhom_1}{\to} \ub{\Cgp_0(L)}_{= \ZZ^3}
  \to 0 .\]
  Note
  \[ \dhom_2[e_1, e_2, e_3] = [e_2, e_3] - [e_1,
  e_3] + [e_1, e_2] \neq 0 ,\]
  so $\dhom_2$ is injective, so $\Hgp_2(L) = 0$.
  But also
  \begin{align*}
    \Hgp_1(L)
    &= \frac{\Ker(\dhom_1)}{\Im(\dhom_2)}
    = \frac{\ZZ\{[e_1, e_2] - [e_1, e_3] + [e_2,
    e_3]\}}{\ZZ\{[e_1, e_2] - [e_1, e_3] + [e_2,
    e_3]\}}
    = 0 \\
    \Hgp_0(L)
    &= \ZZ
  \end{align*}
  (where $\Hgp_0(L) = \ZZ$ because the relevant
  groups $\Cgp_1, \Cgp_0$ haven't changed since
  the previous example).
\end{example*}

\subsection{Some homological algebra}

\begin{flashcard}[chain-complex-defn]
\begin{definition*}[Chain complex]
  \glssymboldefn{Ccc}{$C_\bullet$}{$C_\bullet$}
  \glsnoundefn{cc}{chain complex}{chain complexes}
  \glssymboldefn{Hcc}{$H_n(C_\bullet)$}{$H_n(C_\bullet)$}
  \glssymboldefn{dcc}{$d_n$}{$d_n$}
  \cloze{A \emph{chain complex} is a sequence $C_0, C_1,
  C_2, \ldots$ of abelian groups and homomorphisms
  $d_n : C_n \to C_{n - 1}$ such that $d_{n - 1}
  \circ d_n = 0$ for all $n$. Write this data as
  $C_\bullet$, and call the $d_n$'s the
  \emph{differentials} of $C_\bullet$. Then define
  \[ H_n(C_\bullet) \defeq \frac{\Ker(d_n : C_n
  \to C_{n - 1})}{\Im(d_{n + 1} : C_{n + 1} \to
  C_n)} .\]}
\end{definition*}
\end{flashcard}

\begin{flashcard}[Zcc-Bcc-notations]
\begin{notation*}
  \glssymboldefn{Zcc}{$Z_n(C_\bullet)$}{$Z_n(C_\bullet)$}
  \glssymboldefn{Bcc}{$B_n(C_\bullet)$}{$B_n(C_\bullet)$}
  Write
  \begin{align*}
    Z_n(\Ccc_\bullet)
    &\defeq \cloze{\Ker(\dcc_n : \Ccc_n \to
    \Ccc_{n - 1})}
    &&\text{``the \cloze{$n$-cycles of
    $\Ccc_\bullet$''}} \\
    B_n(\Ccc_\bullet)
    &\defeq \cloze{\Im(\dcc_{n + 1} : \Ccc_{n + 1} \to
    \Ccc_n)}
    &&\text{``the \cloze{$n$-boundaries of
    $\Ccc_\bullet$''}}
  \end{align*}
\end{notation*}
\end{flashcard}

\begin{flashcard}[chain-map-defn]
\begin{definition*}[Chain map]
  \glsnoundefn{cmap}{chain map}{chain maps}
  \cloze{A \emph{chain map} $f_\bullet : \Ccc_\bullet \to
  D_\bullet$ is a sequence of homomorphisms $f_n :
  \Ccc_n \to D_n$, such that $f_n \circ \dcc_{n +
  1} = \dcc_{n + 1} \circ f_{n + 1}$, i.e. the
  diagram
  \[ \begin{tikzcd}[ampersand replacement=\&]
    \Ccc_{n + 1} \ar[r, "f_{n + 1}"] \ar[d,
    "\dcc_{n + 1}"]
    \& D_{n + 1} \ar[d, "\dcc_{n + 1}"] \\
    \Ccc_n \ar[r, "f_n"]
    \& D_n
  \end{tikzcd} \]
  commutes.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[chain-homotopy-defn]
\begin{definition*}[Chain homotopy]
  \glsnoundefn{chom}{chain homotopy}{chain
  homotopies}
  \glsadjdefn{chomic}{chain
  homotopic}{\glspl{cmap}}
  \glssymboldefn{chomic}{$\simeq$}{$\simeq$}
  \cloze{A \emph{chain homotopy} between $f_\bullet,
  g_\bullet : \Ccc_\bullet \to D_\bullet$ is a
  sequence of homomorphisms $h_n : \Ccc_n \to D_{n
  + 1}$ such that
  \[ g_n - f_n = d_{n + 1} \circ f_n + h_{n - 1}
  \circ d_n .\]}
\end{definition*}
\end{flashcard}

\vspace{-1em}
The above definition is hard to motivate at the
moment, but one should just accept it as it is for
now.

\begin{flashcard}[chain-map-induced-hom-lemma]
\begin{lemma*}
  \glssymboldefn{cmapstar}{$f_*$}{$f_*$}
  A \gls{cmap} $f_\bullet : \Ccc_\bullet \to
  D_\bullet$ induces a homomorphism
  \begin{align*}
    f_* : \Hcc_n(C_\bullet)
    &\to \Hcc_n(D_\bullet) \\
    [x]
    &\mapsto [f_n(x)]
  \end{align*}
  Furthermore, if $g_\bullet$ is \gls{chomic} to
  $f_\bullet$, then $g_* = f_*$.
\end{lemma*}

\begin{proof}
  \cloze{Need to show that $\cmapstar f$ is well-defined.
  \begin{enumerate}[(i)]
    \item Let $[x] \in \Hcc_n(\Ccc_\bullet)$, i.e.
      $x \in \Ccc_n$ and $\dcc_n(x) = 0$. Then
      \[ d_n f_n(x) = f_{n - 1} \ub{d_n (x)}_{=0}
      = 0 .\]
      so $f_n(x) \in \Zcc_n(D_\bullet)$.
    \item If $[x] = [y] \in \Hcc_n(C_0)$, then $x
      - y \in \Bcc_n(\Ccc_\bullet)$, $x - y =
      \dcc_{n + 1}(z)$. So
      \[ f_n(x) - f_n(y) = f_n \dcc_{n + 1}(z) =
      \dcc_{n + 1} f_{n + 1}(z) \in
      \Bcc_n(D_\bullet) ,\]
      so $[f_n(x)] = [f_n(y)]$. So $\cmapstar f$
      is a well-defined function. It is a
      homomorphism.
  \end{enumerate}
  Now let $g_\bullet$ be \gls{chomic} to
  $f_\bullet$, i.e. $g_n - f_n = \dcc_{n + 1}
  \circ h_n + h_{n - 1} \circ \dcc_n$. Let $x \in
  \Zcc_n(\Ccc_\bullet)$, so
  \[ g_n(x) - f_n(x) = \dcc_{n + 1} \circ h_n (x) +
  h_{n - 1} \circ \dcc_n (x) \in \Bcc_n(D_\bullet)
  .\]
  So $\cmapstar g([x]) = [g_n(x)] = [f_n(x)] =
  \cmapstar f([x])$.}
\end{proof}
\end{flashcard}