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\begin{flashcard}[path-group-prop]
\begin{proposition*}
  Let $\gamma_0$ be a \gls{path} from $x_0$ to
  $x_1$, $\gamma_1$ a \gls{path} from $x_1$ to
  $x_2$ and $\gamma_2$ a \gls{path} from $x_2$ to
  $x_3$. Then
  \begin{enumerate}[(i)]
    \item \cloze{$\gamma_0 \concat \gamma_1) \concat
      \gamma_2 \hompath \gamma_0 \concat (\gamma_1
      \concat \gamma_2)$ relative to $x_0$ and
      $x_3$.}
    \item \cloze{$\gamma_0 \concat \constpath{x_1} \hompath
      \gamma_0 \hompath \constpath{x_0} \concat
      \gamma_0$ relative to $x_0$ and $x_1$.}
    \item \cloze{$\gamma_0 \concat \invpath{\gamma_0}
      \hompath \constpath{x_0}$ relative to $x_0$
      and $x_0$, $\invpath{\gamma_0} \concat
      \gamma_0 \hompath \constpath{x_1}$ relative
      to $x_1$ and $x_1$.}
  \end{enumerate}
\end{proposition*}

\begin{proof}
  We illustrate some of the cases (other cases are
  similar) using diagrams and the corresponding
  formulae:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/36d8ca00a0164d84.png}
  \end{center}
  \[ H(s, t) = \begin{cases}
    \gamma_0 \left( \frac{4s}{t + 1} \right) & 0
    \le s \le \frac{t + 1}{4} \\
    \gamma_1 (4s - 1 - t) & \frac{t + 1}{4} \le s
    \le \frac{t + 2}{4} \\
    \gamma_2 \left( 1 - \frac{4(1 - s)}{2 - t}
    \right) & \frac{t + 2}{4} \le s \le 1
  \end{cases} \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/d61f4d81450548f4.png}
  \end{center}
  \[ H(s, t) = \begin{cases}
    \gamma_0 \left( \frac{2s}{t + 1} \right) & 0
    \le s \le \frac{t + 1}{2} \\
    x_1 & \frac{t + 1}{2} \le s \le 1
  \end{cases} \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/41e9abbb7da448b0.png}
  \end{center}
  \[ H(s, t) = \begin{cases}
    \gamma_0(2s) & 0 \le s \le \frac{1 - t}{2} \\
    \gamma_0(1 - t) & \frac{1 - t}{2} \le s \le
    \frac{1 + t}{2} \\
    \gamma_0(2 - 2s) & \frac{1 + t}{2} \le s \le 1
  \end{cases} \]
\end{proof}
\end{flashcard}

\subsection{The fundamental group}

\begin{flashcard}[fundamental-group-defn]
\begin{definition*}[Fundamental group]
  \glsnoundefn{fundgp}{fundamental
  group}{fundamental groups}
  \glssymboldefn{pio}{$\pi_1$}{$\pi_1$}
  \cloze{Let $X$ be a space and $x_0 \in X$. Let
  $\pi_1(X, x_0)$ be the set of
  \glsref[hompath]{homotopy}
  classes of loops in $X$ starting and ending at
  $x_0$. Then the rule
  \[ [\gamma] \cdot [\gamma'] = [\gamma \concat
  \gamma'] \]
  makes $(\pi_1(X, x_0), \cdot,
  [\constpath{x_0}])$ into a group.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\textbf{Claim:} This definition does actually make
$\pio(X, x_0)$ into a group.

\begin{proof}
  Lemma from last lecture shows it is
  well-defined. Previous proposition gives the
  group axioms.
\end{proof}

\begin{flashcard}[based-space-defn]
\begin{definition*}[Based space]
  \glssymboldefn{bs}{$(X, x_0)$}{$(X, x_0)$}
  \glsnoundefn{bsmap}{based map}{based maps}
  \cloze{A \emph{based space} is a space $X$ with a
  chosen point $x_0 \in X$ called the base point,
  $(X, x_0)$.

  \glsnoundefn{bshomotopy}{based homotopy}{based
  homotopies}
  A \emph{map of based spaces} $f : (X, x_0) \to
  (Y, y_0)$ is a \gls{map} $X \stackrel{f}{\to} Y$ such
  that $f(x_0) = y_0$. A \emph{based homotopy} is
  a \gls{homotopy} relative to $[x_0] \subset X$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[based-map-associated-func-prop]
\begin{proposition*}
  To a \gls{bsmap} $f : \bs(X, x_0) \to \bs(Y,
  y_0)$ there is associated a function $\pio(f) :
  \pio\bs(X, x_0) \to \pio\bs(Y, y_0)$ given by
  $\pio(f)([\gamma]) = [f \circ \gamma]$. It
  satisfies:
  \begin{enumerate}[(i)]
    \item \cloze{It is a group homomorphism.}
    \item \cloze{If $f$ is \glsref[bshomotopy]{based
      homotopic} to $f'$, $\pio(f) = \pio(f')$.}
    \item \cloze{If $\bs(A, a) \stackrel{h}{\to} \bs(B,
      b) \stackrel{k}{\to} \bs(C, c)$ are
      \glspl{bsmap}, then $\pio(k \circ h) =
      \pio(k) \circ \pio(h)$.}
    \item \cloze{$\pio(\id_X) = \id_{\pio\bs(X,
      x_0)}$.}
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \cloze{
  \begin{enumerate}[(i)]
    \item $f \circ \constpath{x_0} =
      \constpath{y_0}$ so $\pio(f)$ preserves
      identity element. $f \circ (\gamma \concat
      \gamma') = (f \circ \gamma) \cdot (f \circ
      \gamma')$, hence $\pio(f)$ is a
      homomorphism.
    \item[(ii) -- (iv)] Elementary.
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[f-star-notation]
\begin{notation*}
  \glssymboldefn{ps}{$f_*$}{$f_*$}
  We will write $\cloze{\pio(f)} \eqdef f_*$.
\end{notation*}
\end{flashcard}

\begin{flashcard}[change-of-base-hom-prop]
\begin{proposition*}
  \glssymboldefn{pathisom}{$u_\#$}{$u_\#$}
  Let $u$ be a \gls{path} from $x_0$ to $x_1$ in
  $X$. It induces a group isomorphism
  \begin{align*}
    u_\# : \pio\bs(X, x_0)
    &\to \pio(X, x_1) \\
    [\gamma]
    &\mapsto [u^{-1} \concat \gamma \concat u]
  \end{align*}
  satisfying
  \begin{enumerate}[(i)]
    \item \cloze{If $u \hompath u'$ as \glspl{path} then
      $\pathisom u = \pathisom{u'}$.}
    \item \cloze{$\pathisom{(\constpath{x_0})} = \id_{\pio\bs(X,
      x_0)}$.}
    \item \cloze{If $v$ is a \gls{path} from $x_1$ to
      $x_2$ then $\pathisom{(u \concat v)} = \pathisom v \concat
      \pathisom u$.}
    \item \cloze{If $f : X \to Y$ sends $x_0$ to $y_0$
      and $x_1$ to $y_1$, then
      \[
        \begin{tikzcd}[ampersand replacement=\&]
          \pio\bs(X, x_0) \ar[r, "f"] \ar[d,
          "\pathisom u"]
          \&\pio\bs(Y, y_0) \ar[d, "\pathisom{(f
          \circ u)}"] \\
          \pio\bs(X, x_1) \ar[r, "\ps f"]
          \&\pio\bs(Y, y_1)
        \end{tikzcd}
      \]
      ``this diagram commutes'', i.e.
      \[ \pathisom{(f \circ u)} \circ \ps f = \ps f \circ
      \pathisom u \]}
    \item \cloze{If $u$ is a \gls{path} from $x_0$ to
      $x_0$, then $\pathisom u$ is conjugation by $[u]
      \in \pio\bs(X, x_0)$.}
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \cloze{$\pathisom u$ is a group homomorphism via
  \begin{align*}
    \pathisom u([\gamma]) \cdot \pathisom u([\gamma'])
    &= [a^{-1} \concat \gamma \concat a] \cdot
    [a^{-1} \concat \gamma' \concat u] \\
    &= [u^{-1} \concat \gamma \concat u \concat u^{-1}
    \concat \gamma' \concat u] \\
    &= [u^{-1} \concat \gamma \concat \gamma'
    \concat u] \\
    &= \pathisom u([\gamma] \cdot [\gamma'])
  \end{align*}
  It is an isomorphism as $\pathisom{(u^{-1})}$ is an
  inverse. For (iv),
  \begin{align*}
    (\pathisom{(f \circ u)} \circ \ps f)([\gamma])
    &= \pathisom{(f \circ u)} ([f \circ \gamma]) \\
    &= [(f \circ u)^{-1} \concat f \circ \gamma
    \concat f \circ u] \\
    &= [f \circ (u^{-1} \concat \gamma \concat
    u)] \\
    &= \ps f([u^{-1} \concat \gamma \concat u]) \\
    &= (\ps f \circ \pathisom u)([\gamma]) \qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

\begin{lemma*}
  If $H : X \times I \to Y$ is a \gls{homotopy} fomr
  $f$ to $g$ and $x_0 \in X$ is a base point
  \glsref[bsmap]{base point}, then $u = H(x_0,
  \bullet) : I \to Y$ is a \gls{path} from $f(x_0)$
  to $g(x_0)$.

  Then
  \[ \begin{tikzcd}
    \pio\bs(X, x_0) \ar[r, "\ps f"] \ar[d, "\ps g"]
    &\pio\bs(Y, f(x_0)) \ar[ld, "\pathisom u"] \\
    \pio\bs(Y, g(x_0))
  \end{tikzcd} \]
  commutes (i.e. $\pathisom u \circ \ps f = \ps
  g$).
\end{lemma*}

\begin{proof}
  $I \times I \stackrel{\gamma \times \id}{\to} X
  \times I \stackrel{H}{\to} Y$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/2ddaa2621f2b43ce.png}
  \end{center}
  Want $g \circ \gamma \homotopic \invpath u \concat (f
  \circ \gamma) \concat u$ as loops. The top edge
  in the square is \gls{homotopic} to the
  concatenation of the other three edge. Applying
  $H \circ (\gamma \times \id_I)$ gives the
  required \gls{homotopy}.
\end{proof}