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\begin{flashcard}[uni-prop-of-group-pres]
\begin{lemma*}[Universal property of group presentations]
  \label{uni_prop_gppres}
  For any group $H$, the function
  \begin{align*}
    \left\{ \substack{
      \text{group homomorphisms} \\
      \psi : \gpgen<S|R> \to H
    } \right\}
    &\to \left\{ \substack{
      \text{functions $\phi : S \to H$} \\
      \text{such that $\phi(r) = e ~\forall r \in R$}
    } \right\} \\
    \psi
    &\mapsto [S \stackrel{\mathrm{inc}}{\to} \F(S)
    \stackrel{\mathrm{quot}}{\to} \gpgen<S|R>
    \stackrel{\psi}{\to} H]
  \end{align*}
  is a bijection.
\end{lemma*}

\begin{proof}
  \cloze{Suppose $\psi, \psi'$ determine functions $\phi
  = \phi' : S \to H$. Then
  \[ \begin{tikzcd}[ampersand replacement=\&]
    \F(S) \ar[r, "\mathrm{quot}"]
    \& \gpgen<S|R> \ar[r, bend left, "\psi"] \ar[r,
    bend right, "\psi'"]
    \& H
  \end{tikzcd} \]
  are equal by the universal property of free
  groups. As ``$\mathrm{quot}$'' is onto, $\psi =
  \psi'$.

  Conversely, given a $\phi : S \to H$ such that
  $\phi(r) = r ~\forall r \in R$, consider
  \[ \varphi : \F(S) \to H \]
  Now $R \subset \Ker(\varphi)$, so as $\Ker(\phi)$ is
  normal, $\relgen R \le \Ker(\varphi)$. This
  $\varphi$ extends to a homomorphism
  \[ \gpgen<S|R> = \frac{\F(S)}{\relgen R} \to H \]
  as required.}
\end{proof}
\end{flashcard}

\begin{example*}
  If $G$ is a group, the function $\phi = \id : G
  \to G$ gives a homomorphism $\varphi : \F(G) \to
  G$, which is onto. Let $R = \Ker(\varphi)$, so
  $\gpgen<G|R> = \frac{\F(G)}{\relgen R} \cong G$.
\end{example*}

\begin{example*}
  $G = \gpgen<a, b | a>$, $H = \gpgen <t|>$.
  Consider
  \begin{align*}
    \phi : \{a, b\}
    &\to H \\
    a
    &\mapsto e \\
    b
    &\mapsto t \\
    \phi' : \{t\}
    &\to G \\
    t
    &\mapsto b
  \end{align*}
  Then $\psi(a) = e$, so we get a homomorphism
  \[ \psi : \gpgen<a, b|a> \to \gpgen<t|> ,\]
  by the \nameref{uni_prop_gppres}. We also get
  \[ \psi' : \gpgen<t|> \to \gpgen<a, b|a> .\]
  Note that
  \begin{align*}
    \psi' \circ \psi([a])
    &= [e] = [a] \\
    \psi' \circ \psi([b]) = [b]
  \end{align*}
  so as $[a], [b]$ generate, we have $\psi' \circ
  \psi = \id$. Similarly $\psi \circ \psi' = \id$.
  So
  \[ \gpgen<a, b|a> \cong \gpgen<t|> .\]
\end{example*}

\begin{example*}
  Let $G = \gpgen<a, b|ab^{-3}, ba^{-2}>$. Have
  \[ [a][b]^{-3} = e, \qquad [b][a]^{-2} = e ,\]
  so
  \[ [a] = [b]^3, \qquad [b] = [a]^2 ,\]
  so $[a] = [a]^6$, so $\boxed{e = [a]^5}$. These
  show that every element is equal to one of $e,
  [a], [a]^2, [a]^3, [a]^4$. Consider:
  \begin{align*}
    \phi : \{a, b\}
    &\to \ZZ / 5\ZZ \\
    a
    &\mapsto 1 \\
    b
    &\mapsto z
  \end{align*}
  Then $\varphi(ab^{-3}) = e = \varphi(ba^{-2})$,
  so we get a homomorphism $\psi : \gpgen<a,
  b|ab^{-3}, ba^{-2}> \to \ZZ / 5\ZZ$, which is an
  isomorphism since $[a] \mapsto 1$.
\end{example*}

\subsection{Free products with amalgamations}

\begin{flashcard}[free-product-defn]
\begin{definition*}[Free product]
  \glssymboldefn{freep}{$G_1 * G_2$}{$G_1 * G_2$}
  \cloze{Consider group homomorphisms
  \[ G_1 \stackrel{i_1}{\leftarrow} H
  \stackrel{i_2}{\to} G_2 \]
  and suppose $G_i = \gpgen<S_i|R_i>$. The
  \emph{free product} of $G_1$ and $G_2$ is
  \[ G_1 * G_2 = \gpgen<S_1 \amalg S_2|R_1 \sqcup R_2>
  .\]}
\end{definition*}
\end{flashcard}

\begin{flashcard}[free-product-with-amalg-over-defn]
\begin{definition*}[Free product with amalgamation
  over $H$]
  \glssymboldefn{freepH}{$G_1 *_H G_2$}{$G_1 *_H
  G_2$}
  The functions
  \[ S_i \to S_1 \amalg S_2 \to \F(S_1 \sqcup S_2)
  \to G_1 \freep G_2 \]
  induce homomorphisms
  \[ G_1 \stackrel{j_1}{\to} G_1 \freep G_2
  \stackrel{j_2}{\leftarrow} G_2 \]

  The \emph{free product with amalgamation over
  $H$} is the quotient
  \[ G_1 \freepH H G_2 = G_1 \freep G_2 / \relgen{ j_1
  i_1(h) (j_2 i_2(h))^{-1}, h \in H} .\]
  So the following diagram commutes:
  \[ \begin{tikzcd}[ampersand replacement=\&]
    H \ar[r, "i_1"] \ar[d, "i_2"]
    \& G_1 \ar[d, "j_1"] \\
    G_2 \ar[r, "j_2"]
    \& G_1 \freepH H G_2
  \end{tikzcd} \]
\end{definition*}
\end{flashcard}

\begin{flashcard}[uni-prop-of-free-prod-with-amalg-lemma]
\begin{lemma*}[Universal property of free products
  with amalgamation]
  \label{uni_prop_amalg}
  For any group $K$,
  \begin{align*}
    \left\{ \substack{
      \text{group homomorphisms} \\
      \phi : G_1 \freepH H G_2 \to K
    } \right\}
    &\to \left\{ \substack{
      \text{group homomorphisms $\phi_1 : G_1 \to
      K$,} \\
      \text{$\phi_2 : G_2 \to K$, such that} \\
      \text{$\phi_1 \circ i_1 = \phi_2 \circ i_2$}
    } \right\} \\
    \phi
    &\mapsto [G_i \stackrel{j_i}{\to} G_1 \freepH
    H G_2 \stackrel{\phi}{\to} K]
  \end{align*}
  is a bijection.
  \[ \begin{tikzcd}[ampersand replacement=\&]
    H \ar[r, "i_1"] \ar[d, "i_2"]
    \&G_1 \ar[d, "j_1"] \ar[rdd, bend left, "\phi_1"] \\
    G_2 \ar[r, "j_2"] \ar[rrd, bend right,
    "\phi_2", swap]
    \& G_1 \freepH H G_2 \ar[rd, "\exists! \phi"] \\
    \& \& K
  \end{tikzcd} \]
\end{lemma*}
\end{flashcard}

\begin{proof}
  Similar to other universal properties.
\end{proof}

\setcounter{section}{5}
\setcounter{subsection}{0}
\subsection{The Seifert-Van Kampen Theorem}

Let $X$ be a space, $A, B \subset X$ be subspaces,
$x_0 \in A \cap B$. We get a commutative diagram
\[ \begin{tikzcd}
  \pio\bs(A \cap B, x_0) \ar[r] \ar[d]
  &\pio\bs(A, x_0) \ar[d] \\
  \pio\bs(B, x_0) \ar[r]
  &\pio\bs(X, x_0)
\end{tikzcd} \]
Then the \nameref{uni_prop_amalg} gives us
\[ \pio\bs(A, x_0) \freepH{\pio\bs(A \cap B, x_0)}
\pio\bs(B, x_0) \to \pio\bs(X, x_0) .\]

\begin{flashcard}[seifert-van-kampen-thm]
\begin{theorem*}[Seifert-van Kampen]
  \label{SVK}
  \cloze{Let $X$ be a space, $A, B \subset X$ be open
  subsets, which cover $X$ and such that $A \cap
  B$ is \gls{pathconn}. Then for any $x_0 \in A
  \cap B$, the induced map
  \[ \phi : \pio\bs(A, x_0) \freepH{\pio\bs(A \cap B,
  x_0)} \pio\bs(B, x_0) \to \pio\bs(X, x_0) \]
  is a group isomorphism.}
\end{theorem*}
\end{flashcard}

\begin{example*}
  Consider $S^n$ for $n \ge 2$. Let
  \begin{align*}
    A
    &= S^n - \{\text{north pole}\} \cong \RR^n
    \homoteq \{*\} \\
    B
    &= S^n - \{\text{south pole}\} \cong \RR^n
    \homoteq \{*\}
  \end{align*}
  $A \cap B \cong S^{n - 1} \times (-1, 1)$,
  \gls{pathconn} for $n \ge 2$, and $\homoteq S^{n -
  1}$. So
  \[ \pio\bs(S^n, \bullet) \cong \{e\}
  \freepH{\pio\bs(S^{n - 1}, \bullet)} \{e\} =
  \{e\} .\]
  So $S^n$ is \gls{simpconn}.
\end{example*}

\begin{example*}
  We saw that there is a
  \glsref[nshtd]{$2$-sheeted} \gls{covmap} $p :
  S^n \to \RR \PP^n$. For $n \ge 2$, $S^n$ is
  \gls{simpconn} , so this is a \gls{unicov}. So
  \[ \pio\bs(\RR\PP^n, \bullet)
  \stackrel{\mathrm{bij}}{\to} p^{-1}(\bullet) \]
  has $2$ elements. So $\pio\bs(\RR \PP^n,
  \bullet) \cong \ZZ / 2\ZZ$.
\end{example*}

\vspace{-1em}
\glssymboldefn{swedge}{$\wedge$}{$\wedge$}
Given $(X, x_0), (Y, y_0)$, then
\[ X \wedge Y = (X \sqcup Y) / x_0 \sim y_0 .\]

\begin{example*}
  Let $S^1 \subset \CC$ have basepoint $1 \in S^1
  \subset \CC$. $S^1 \swedge S^1$ is:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/239716b069ed4fd7.png}
  \end{center}
  Which is covered by
  \[ (S^1 - \{1\}) \swedge S^1 = U \]
  \[ S^1 \swedge (S^1 - \{1\}) = V \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/335c405a169948d6.png}
  \end{center}
  So \nameref{SVK} implies
  \begin{align*}
    \pio\bs(S^1 \swedge S^1, x_0)
    &\cong \gpgen<a|> \freepH{\{e\}} \gpgen<b|> \\
    &= \gpgen<a, b|>
  \end{align*}
\end{example*}