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\begin{flashcard}[based-uniqueness-prop]
\begin{proposition*}[Based uniqueness]
  \label{based_uniqueness}
  \cloze{Let $\bs(X, x_0)$ satisfy the conditions for
  \nameref{unicov_existence}. If $p_1 :
  \bs(\tilde{X}_1, \tilde{x}_1) \to \bs(X, x_0)$, $p_2 :
  \bs(\tilde{X}_2, \tilde{x}_2) \to \bs(X, x_0)$
  are \gls{pathconn} \glspl{covsp}, then the
  following are equivalent:
  \begin{itemize}
    \item there exists homeomorphism $h :
      \bs(\tilde{X}_1, \tilde{x}_1) \to
      \bs(\tilde{X}_2, \tilde{x}_2)$ such that $p_2
      \circ h = p_1$
    \item $\ps{(p_1)}
      \pio\bs(\tilde{X_1}, \tilde{x}_1)
      = \ps{(p_2)} \pio\bs(\tilde{X}_1, \tilde{x}_1)$
  \end{itemize}
  }
\end{proposition*}

\begin{proof}
  \cloze{If $h$ exists, then
  \[ \Im(\ps{(p_1)}) = \Im(\ps{(p_2)} \circ \ps h) \]
  as $\ps h$ is an isomorphism.

  For the other direction, let $H \le \pio\bs(X,
  x_0)$ be the common image. Will show that
  $\tilde{X}_1$ and $\tilde{X}_2$ are homeomorphic
  to $\ol{X}_H$ (over $X$). Consider
  \begin{align*}
    r : \ol{X}
    &\to \tilde{X}_1 \\
    [\gamma]
    &\mapsto \tilde{\gamma}(1)
  \end{align*}
  end point of the lift $\tilde{\gamma}$ of
  $\gamma$ to $\tilde{X}_1$, starting at
  $\tilde{x}_1$. Notice
  \begin{align*}
    r([\gamma]) = r([\gamma'])
    &\iff \text{$\tilde{\gamma}$ and
    $\tilde{\gamma}'$ end at the same point of
    $\tilde{X}_1$} \\
    \iff [\gamma' \concat \invpath \gamma] \in
    \ps{(p_1)} \pio\bs(\tilde{X}_1, \tilde{x}_1) =
    H \\
    \iff [\gamma] \sim_H [\gamma']
  \end{align*}
  So $r$ descends to a map
  \[ q : \bs(\ol{X}_H, [[\constpath{x_0}]]) \to
  \bs(\tilde{X}_1, \tilde{x}_1) ,\]
  a bijection. It is also an open map, as
  $\ol{X}_H$ and $\tilde{X}_1$ are both locally
  homeomorphic to $X$. So $q$ is a homeomorphism.}
\end{proof}
\end{flashcard}

\begin{flashcard}[unbased-uniqueness-coro]
\begin{corollary*}[Unbased uniqueness]
  \label{unbased_uniqueness}
  \cloze{Let $X$ satisfy the conditions from
  \nameref{unicov_existence}. If $p_1 :
  \tilde{X}_1 \to X$ and $p_2 : \tilde{X}_2 \to X$
  are \gls{pathconn} \glspl{covsp}, then the
  following are equivalent:
  \begin{itemize}
    \item there exists homeomorphism $h : \tilde{X}_1
      \to \tilde{X}_2$ such that $p_2 \circ h = p_1$
    \item $\ps{(p_1)} \pio\bs(\tilde{X}_1,
      \tilde{x}_1)$ and $\ps{(p_2)}
      \pio\bs(\tilde{X}_2, \tilde{x}_2)$ are
      conjugate in $\pio\bs(X, x_0)$ for any
      $\tilde{x}_1 \in p_1^{-1}(x_0)$,
      $\tilde{x}_2 \in p_2^{-1}(x_0)$.
  \end{itemize}
  }
\end{corollary*}

\begin{proof}
  \cloze{If $h$ exists, choose $\ol{x}_1 \in
  p_1^{-1}(x_0)$ and $\tilde{x}_2 =
  h(\tilde{x}_1)$. \nameref{based_uniqueness}
  applies to see that the groups are determined.

  Conversely, suppose $[\gamma] \in \pio\bs(X,
  x_0)$ is such that
  \[ [\gamma]^{-1} \concat \ps{(p_1)}
  \pio\bs(\tilde{X}_1, \tilde{x}_1) \concat
  [\gamma] = \ps{(p_2)} \pio\bs(\tilde{X}_2,
  \tilde{x}_2) .\]
  Lif $\gamma$ to $\tilde{X}_1$ starting
  $\tilde{x}_1$, and say it ends at
  $\tilde{x}_1'$. Then
  \[ \LHS = \ps{(p_1)} \pio\bs(\tilde{X}_1,
  \tilde{x}_1') .\]
  Now by \nameref{based_uniqueness}, we can find a
  homeomorphism $h : \bs(\tilde{X}_1,
  \tilde{x}_1') \to \bs(\tilde{X}_2,
  \tilde{x}_2)$, which can be viewed as an unbased
  homeomorphism.}
\end{proof}
\end{flashcard}

So we now have bijections:
\[ \left\{ \substack{
  \text{based \gls{pathconn}} \\
  \text{\glspl{covmap} of $\bs(X, x_0)$}
} \right\} / 
\substack{
  \text{based homeomorphism} \\
  \text{over $X$}
}
\leftrightarrow
\left\{ \substack{
  \text{subgroups of} \\
  \text{$\pio\bs(X, x_0)$}
} \right\}
\]
\[ \left\{ \substack{
  \text{\gls{pathconn}} \\
  \text{\glspl{covmap} of $X$}
} \right\} /
\substack{
  \text{homeomorphism} \\
  \text{over $X$}
}
\leftrightarrow
\left\{ \substack{
  \text{conjugacy classes of} \\
  \text{subgroups of $\pio\bs(X, x_0)$}
} \right\}
\]

\setcounter{section}{4}
\setcounter{subsection}{0}
\subsection{Free groups and presentations}

\begin{flashcard}[alphabet-words-defn]
\begin{definition*}[Alphabet and words]
  \glsnoundefn{alpha}{alphabet}{alphabets}
  \glssymboldefn{alpha}{$S$}{$S$}
  \glsnoundefn{word}{word}{words}
  \glsadjdefn{redword}{reduced}{word}
  \glsnoundefn{elred}{elementary
  reduction}{elementary reductions}
  \cloze{Let $S = \{s_\alpha\}_{\alpha \in I}$ be a set,
  called the \emph{alphabet}, $S^{-1} =
  \{s_\alpha^{-1}\}_{\alpha \in I}$ (suppose $S
  \cap S^{-1} = \emptyset$.

  A \emph{word} in the \gls{alpha} $\aS$ is a
  (possibly empty) finite sequence $(x_1, x_2,
  \ldots, x_n)$ of elements in $\aS \cup
  \aS^{-1}$. A word is \emph{reduced} if it does
  not contain $(s_\alpha, s_\alpha^{-1})$ or
  $(s_\alpha^{-1}, s_\alpha)$ as a subword. An
  \emph{elementary reduction} consists of removing
  such a subword from a word.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[free-group-defn]
\begin{definition*}[Free group]
  \glsnoundefn{freeg}{free group}{free groups}
  \glssymboldefn{F}{$F(S)$}{$F(S)$}
  \cloze{The \emph{free group} on the \gls{alpha}
  $\aS$, denoted $\F(S)$,
  is the set of \gls{redword} (possibly empty)
  \glspl{word} in this \gls{alpha}. The group
  operation is given by concatenation, and doing
  \glspl{elred} until it is \gls{redword}.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
Note that the group operation here is not
obviously well-defined. We will not prove here
that it is well-defined. See the lecturer's
notes for a proof.

Clearly we have that the empty string is the
identity, and for any word $(x_1, \ldots, x_n)$,
it has inverse $(x_n^{-1}, \ldots, x_1^{-1})$.

There is a function $i : S \to \F(S)$, $s \mapsto
(s)$.

\begin{flashcard}[universal-property-of-free-groups-lemma]
\begin{lemma*}[Universal property of free groups]
  \cloze{For any group $H$, the function
  \[ \left\{ \substack{
    \text{homomorphisms} \\
    \phi : \F(S) \to H
  } \right\} \stackrel{- \circ i}{\to}
  \left\{ \substack{functions} \\ \phi : S \to H
  \right\} \]
  is a bijection.}
\end{lemma*}

\begin{proof}
  Given $\phi : S \to H$, want a $\phi : \F(S) \to
  H$ such that $\phi((s)) = \phi(s)$. Let, on a
  not-necessarily-\gls{redword} \gls{word}
  $(s_{\alpha_1}^{\eps_1}, \ldots,
  s_{\alpha_n}^{\eps_n})$,
  \[ \phi((s_{\alpha_1}^{\eps_1}, \ldots,
  s_{\alpha_n}^{\eps_n})) =
  \phi(s_{\alpha_1})^{\eps_1} \cdots
  \phi(s_{\alpha_n}^{\eps_n}) \in H .\]
  If the word contained $(s_\alpha,
  s_\alpha^{-1})$, then the result contains
  $\phi(s_\alpha) \phi(s_\alpha)^{-1} = e \in H$.
  So $\phi$ is well-defined. As the group
  operation on $\F(S)$ is by concatenation, we see
  $\phi$ is a homomorphism.
\end{proof}
\end{flashcard}

\begin{flashcard}[group-with-relations]
\begin{definition*}[Group with relations]
  \glssymboldefn{gpgen}{$\langle S \mid R
  \rangle$}{$\langle S \mid R \rangle$}
  \glssymboldefn{relgen}{$\llangle R
  \rrangle$}{$\llangle R \rrangle$}
  \glsnoundefn{pres}{presentation}{presentations}
  \glsadjdefn{finp}{finite}{presentation}
  \cloze{Let $S$ be a set and $R \subseteq \F(S)$. Then
  \[ \langle S \mid R \rangle = \F(S) / \llangle R
  \rrangle \]
  with
  \[ \llangle R \rrangle = \{(r_1^{\eps_1})^{g_1}
  \cdots (r_n^{\eps_n})^{g_n} \st r_i \in R,
  \eps_i \in \{\pm 1\}, g \in \F(S)\} \normalsub
  \F(S) ,\]
  where $h^g$ denotes $g^{-1} hg$.

  Call this ($S$ and $R$) a \emph{presentation} of
  the group $\langle S \mid R \rangle$. If $S$ and
  $R$ are finite, call it a \emph{finite
  presentation}.}
\end{definition*}
\end{flashcard}