%! TEX root = AlgT.tex
% vim: tw=50
% 12/02/2024 11AM

\begin{example*}
  The function
  \begin{align*}
    \{a, b\}
    &\to \ZZ / 3\ZZ \\
    a
    &\mapsto 1 \\
    b
    &\mapsto 1
  \end{align*}
  gives a homomorphism $\varphi : \gpgen<a, b|> =
  \pio\bs(S^1 \swedge S^2, x_0) \to \ZZ / 3\ZZ$,
  surjective. So
  \[ K = \Ker(\varphi) \le \pio\bs(S^1 \swedge
  S^1, x_0) \]
  is a subgroup of index $3$. This corresponds to
  a \gls{covsp}. What is it? It is a $p :
  \tilde{X} \to X$ with
  \[ p^{-1}(x_0) \cong \frac{\pio\bs(S^1 \swedge
  S^1, x_0)}{K} \cong \ZZ / 3\ZZ .\]
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/5183103afee243af.png}
  \end{center}
\end{example*}

\begin{example*}
  Consider
  \begin{align*}
    \{a, b\}
    &\to \ZZ / 3\ZZ \\
    a
    &\mapsto 1 \\
    b
    &\mapsto 0
  \end{align*}
  again giving a $\varphi : \pio\bs(S^1 \swedge
  S^1, x_0) = \gpgen<a, b|> \surjto \ZZ / 3\ZZ$,
  with kernel $K$. What is the \gls{covsp}
  corresponding to $K$?
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/0dd93c733be04f15.png}
  \end{center}
\end{example*}

\begin{example*}
  The \gls{unicov} of $S^1 \swedge S^1$?
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/399920e58c8046a7.png}
  \end{center}
\end{example*}

\begin{flashcard}[SVK-proof]
\prompt{Proof of \nameref{SVK}? \\}
\begin{proof}[Proof of \nameref{SVK}
  (non-examinable)]
  \cloze{Without loss of generality can assume $A, B$ are
  \gls{pathconn}.
  \begin{enumerate}[(1)]
    \item \textbf{$\phi$ surjective:} Let $\gamma
      : I \to X$ be a gloop. Then
      $\{\gamma^{-1}(A), \gamma^{-1}(B)\}$ is an
      open cover of $I$, so by
      \nameref{lebesgue_number_lemma} there is a
      $n \gg 0$ such that each $\left[
      \frac{i}{n}, \frac{i + 1}{n} \right]$ is
      sent into $A$ or $B$ (or both). By
      concatenating intervls which lie in $A$, or
      in $B$, can write $\gamma = \gamma_1 \concat
      \gamma_2 \cdots \gamma_k$ with each
      $\gamma_i$ being endpoints in $A \swedge B$.
      Chooce paths $u_i$ from $\gamma_i(1)$ to
      $x_0$ in $A \swedge B$.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/a9d841ee262f4187.png}
      \end{center}
      Then $\gamma \homotopic (\gamma_1 \concat
      u_i) \concat (\invpath{u_i} \concat \gamma_2
      u_2) \concat (\invpath{u_2} \concat \gamma_3
      \concat u_3) \cdots (\invpath{u_{k - 1}}
      \concat \gamma_k)$. Each thing in a pair of
      brackets is a loop based at $x_0$, lying in
      $A$ or in $B$. So $[\gamma] \in \Im(\phi)$
      as required.
    \item \textbf{$\phi$ is injective:} The group
      \[ \pio\bs(A, x_0) \freepH{\pio\bs(A \swedge
      B, x_0)} \pio\bs(B, x_0) \]
      by the following description. It is
      generated by
      \begin{enumerate}[(i)]
        \item For $\gamma : I \to A$ a loop in
          $A$, $[\gamma]_A$.
        \item For $\gamma : I \to B$ a loop in
          $B$, $[\gamma]_B$.
      \end{enumerate}
      With relations:
      \begin{enumerate}[(i)]
        \item If $\gamma \homotopic \gamma'$ in
          $A$, then $[\gamma]_A = [\gamma']_A$,
          and similarly for $B$.
        \item If $\gamma : I \to A \cap B$ then
          $[\gamma]_A = [\gamma]_B$.
        \item $[\gamma]_A \concat [\gamma']_A =
          [\gamma \concat \gamma']_A$, and similarly
          for $B$.
      \end{enumerate}
      Suppose
      \[ \phi([\gamma_1]_{A_{i_1}} \concat
      [\gamma_2]_{A_{i_2}} \cdots
      [\gamma_n]_{A_{i_n}}) = [\constpath{x_0}] \]
      so $\gamma_1 \cdots \gamma_k \homotopic_H
      \constpath{x_)}$ in $X$, $H : I \times I \to
      X$. Can subdivide $I \times I$ into squares
      of size $\frac{1}{N}$ with $n \mid N$ such
      that each square is sent into $A$ or $B$ (or
      both).
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/fba11cca6c644141.png}
      \end{center}
      Choose \glspl{path} $i_{ij}$ from $H
      \left( \frac{i}{N}, \frac{j}{N} \right)$ to
      $x_0$ such that:
      \begin{itemize}
        \item If it is a vertex of a box labelled
          $A$, the path is in $A$.
        \item If it is a vertex of a box labelled
          $B$, the path is in $B$.
      \end{itemize}
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/9403c4563aee46e0.png}
      \end{center}
      $G$ has the property that it decomposes into
      rectangles with vertices sent to $x_0$, and
      each rectangle in $A$ or in $B$. This shows
      that $[\gamma_1]_{A_1} \cdots
      [\gamma_n]_{A_n}$ can be transformed using
      the 3 kinds of relations described.
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\subsection{Attaching a cell}

Let $f : \bs(S^{n - 1}, *) \to \bs(X, x_0)$, then
\[ Y = X \cup_f D^n = \frac{X \sqcup D^n}{f(x)
\sim x \in \partial D^n = S^{n - 1}} \]
Then $[x_0] \in Y$ is a basepoint, and the
inclusion $\bs(X, x_0) \injto \bs(Y, [x_0])$ is a
\gls{bsmap}.
