\classheader{2012-11-30}

Recall that last time, we proved that $\pi_1(\S^1)\cong\Z$, and introduced covering spaces.

\subsection*{Free products of groups}

\begin{definition}
Let $G$ and $H$ be groups. We define $G\free H$ to be the group generated by $G$ and $H$, with no extra relations; thus,
\[G\free  H=\{g_1h_1\cdots g_kh_k\mid g_i\in G,h_i\in H\},\]
and concatenation of words is the group operation. It is a theorem that $G\ast H$ is in fact a group; associativity is the only tricky part. We can also define, for any collection of groups $G_\alpha$, their free product $\bigfree_\alpha G_\alpha$.
\end{definition}
\begin{example}
The free group on $S=\{s_1,\ldots,s_n\}$ is just $\Z\free\cdots\free\Z$, with one copy of $\Z$ for each $s_i$.
\end{example}
\begin{theorem}[Universal property of free products]
For any collection $\{G_\alpha\}$ of groups and homomorphisms $\{\phi_\alpha:G_\alpha\to H\}$, there is a unique homomorphism $\phi:\bigfree_\alpha G_\alpha\to H$ such that $\phi(g)=\phi_\alpha(g)$ for any $g\in G_\alpha$ for all $\alpha$.
\end{theorem}
Recall that we specify a presentation of a group as
\[G=\langle \underbrace{g_1,\ldots,g_k}_{\text{relations}}\mid \underbrace{r_1,\ldots,r_m}_{\mathclap{\substack{\text{relations, i.e.}\\\text{words in the }g_i}}}\rangle.\]
What this really means is
\[G=F(\{g_1,\ldots,g_k\})/\langle\!\langle r_1,\ldots,r_m\rangle\!\rangle,\]
where $\langle\!\langle\,\cdot\,\rangle\!\rangle$ denotes the normal subgroup generated by a set. Note that we need to use this normal closure because if $b$ is a relation we want to kill, we also need to kill $aba^{-1}$.

\subsection*{Free products with amalgamation}

Suppose we have (not necessarily injective) group homomorphisms as follows:
\begin{center}
\begin{tikzcd}[row sep=0.1cm]
& A_1\\
H \ar{ru}{i_1} \ar{rd}[swap]{i_2} & \\
& A_2
\end{tikzcd}
\end{center}
Then $A_1\free_H A_2$, the free product of $A_1$ and $A_2$ along $H$, is defined to be
\[A_1\free_H A_2=A_1\free A_2/\langle\! \langle \{i_1(h)i_2(h)^{-1}\mid h\in H\}\rangle\!\rangle.\]

\subsection*{The van Kampen theorem}

Suppose that $X=\bigcup_\alpha A_\alpha$, where each of the $A_\alpha$ are path connected and open, and where there is some $x_0\in\bigcap_\alpha A_\alpha$. For example, we might have
\begin{center}
\begin{tikzpicture}[scale=1.5]
\draw[thick] (0:0.6) circle (1);
\draw[thick] (120:0.6) circle (1);
\draw[thick] (240:0.6) circle (1);
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt] at (0,0) {};
\node at (0:1) {$A_1$};
\node at (120:1) {$A_2$};
\node at (240:1) {$A_3$};
\node at (0.25,0) {$x_0$};
\end{tikzpicture}
\end{center}
For any $\alpha$ and $\beta$, we have a commutative diagram of topological spaces
\begin{center}
\begin{tikzcd}
& A_\alpha\ar{rd}{j_\alpha} & \\
A_\alpha\cap A_\beta \ar{ru}{i_{\alpha\beta}} \ar{rd}[swap]{i_{\beta\alpha}} & & X\\
& A_\beta \ar{ru}[swap]{j_\beta}
\end{tikzcd}
\end{center}
which induces a commutative diagram of fundamental groups (all with respect to the basepoint $x_0$):
\begin{center}
\begin{tikzcd}
& \pi_1(A_\alpha)\ar{rd}{j_\alpha} & \\
\pi_1(A_\alpha\cap A_\beta) \ar{ru}{i_{\alpha\beta}} \ar{rd}[swap]{i_{\beta\alpha}} & & \pi_1(X)\\
& \pi_1(A_\beta) \ar{ru}[swap]{j_\beta}
\end{tikzcd}
\end{center}
Clearly, we have that ${j_\alpha}_*({i_{\alpha\beta}}_*([\gamma]))={j_\beta}_*({i_{\beta\alpha}}_*([\gamma])$. There is a map $\Phi:\bigfree_\alpha\pi_1(A_\alpha)\to\pi_1(X)$ induced by taking the ${j_\alpha}_*$ together.
\begin{theorem}[van Kampen]
With this notation, if $A_\alpha\cap A_\beta$ is path connected for any $\alpha,\beta$, then $\Phi$ is surjective. If $A_\alpha\cap A_\beta\cap A_\gamma$ is path-connected for any $\alpha,\beta,\gamma$, then $\ker(\Phi)$ is generated by ${i_{\alpha\beta}}_*([\gamma]){i_{\beta\alpha}}_*([\gamma])^{-1}$ for any loop $\gamma\in A_\alpha\cap A_\beta$.
\end{theorem}
\begin{theorem}[Simpler version]
Suppose that $X=A_1\cup A_2$ such that $A_1\cap A_2$ is path-connected. Then
\[\pi_1(X)\cong \pi_1(A_1)\free_{\pi_1(A_1\cap A_2)}\pi_1(A_2).\]
\end{theorem}
\begin{proof}[Sketch of proof]
$\text{}$

\begin{center}
\begin{tikzpicture}[scale=1.5]
\draw[thick] (0:0.6) circle (1);
\draw[thick] (120:0.6) circle (1);
\draw[thick] (240:0.6) circle (1);
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt] (x) at (0,0) {};
\node at (0:0.85) {$A_1$};
\node at (120:0.85) {$A_2$};
\node at (240:0.85) {$A_3$};
\node at (30:0.25) {$g_1$};
\node at (150:0.25) {$g_2$};
\node at (270:0.25) {$g_3$};
\node (a) at (0:1.4) {};
\node (c) at (120:1.4) {};
\node (e) at (240:1.4) {};
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt] (b) at (60:0.6) {};
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt] (d) at (180:0.6) {};
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=4pt] (f) at (300:0.6) {};
\draw[thick,dashed] plot [smooth cycle, tension=1] coordinates { (a) (b) (c) (d) (e) (f) };
\draw[thick,dashed] (b) -- (x);
\draw[thick,dashed] (d) -- (x);
\draw[thick,dashed] (f) -- (x);
\node (a) at (12:1.4) {$\alpha_1$};
\node (c) at (132:1.4) {$\alpha_2$};
\node (e) at (252:1.4) {$\alpha_3$};
\begin{scope}[shift={(0,-2.5)},xscale=0.5]
\draw[thick] (-3,0) to (3,0);
\draw[thick] (-2.93,-0.10) -- (-3,-0.10) -- (-3,0.10) -- (-2.93,0.10);
\draw[thick] (2.93,-0.10) -- (3,-0.10) -- (3,0.10) -- (2.93,0.10);
\draw[thick] (1,-0.10) -- (1,0.10);
\draw[thick] (-1,-0.10) -- (-1,0.10);
\node at (-2,0.15) {$\alpha_1$};
\node at (0,0.15) {$\alpha_2$};
\node at (2,0.15) {$\alpha_3$};
\draw[thick,-angle 60] (2,0.5) to[bend right=40] node [midway,auto,label=right:{$\alpha$}] {} (3,1.7);
\end{scope}
\end{tikzpicture}
\end{center}

First, let's prove that $\Phi$ is surjective. Given $[\alpha]\in \pi_1(X)$, pull back the open cover $A_\alpha$ via $\alpha$ (terrible notation!) and split the loop $\alpha$ into subpaths $\alpha_i$, each of which maps into some $A_\alpha$.

Now rewrite this as
\begin{center}
\begin{tikzpicture}[yscale=1.5]
\draw[thick] (-3,0) to (6,0);
\draw[thick] (-2.93,-0.10) -- (-3,-0.10) -- (-3,0.10) -- (-2.93,0.10);
\draw[thick] (5.93,-0.10) -- (6,-0.10) -- (6,0.10) -- (5.93,0.10);
\draw[thick] (1,-0.10) -- (1,0.10);
\draw[line width=0.7mm] (-1,-0.10) -- (-1,0.10);
\draw[line width=0.7mm] (2,-0.10) -- (2,0.10);
\draw[thick] (-2,-0.10) -- (-2,0.10);
\draw[thick] (0,-0.10) -- (0,0.10);
\draw[thick] (3,-0.10) -- (3,0.10);
\draw[thick] (4,-0.10) -- (4,0.10);
\draw[line width=0.7mm] (5,-0.10) -- (5,0.10);
\node at (-2.5,0.15) {$\alpha_1$};
\node at (0.5,0.15) {$\alpha_2$};
\node at (3.5,0.15) {$\alpha_3$};
\node at (-1.5,0.15) {$g_1$};
\node at (-0.5,0.2) {$g_1^{-1}$};
\node at (1.5,0.15) {$g_2$};
\node at (2.5,0.2) {$g_2^{-1}$};
\node at (4.5,0.15) {$g_3$};
\node at (5.5,0.2) {$g_3^{-1}$};
\node at (6.5,0.15) {$\cdots$};
\node at (-3.5,0.15) {$\cdots$};
\node at (-2,-0.4) {$\in \pi_1(A_1)$};
\node at (0.5,-0.4) {$\in \pi_1(A_2)$};
\node at (3.5,-0.4) {$\in \pi_1(A_3)$};
\end{tikzpicture}
\end{center}

Now, let's try to understand the kernel $\ker(\Phi)$. Given two elements of $\bigfree_\alpha \pi_1(A_\alpha)$ which are sent to the same thing in $\pi_1(X)$, say
\[[f_1]\cdots[f_k]=[f_1']\cdots[f_m']\in\pi_1(X).\]
Then we have
\[[f_1\cdots f_k]=[f_1'\cdots f_m']\in\pi_1(X),\]
and we want to show that we can get from $[f_1\cdots f_k]$ to $[f_1'\cdots f_m']$ by
\begin{enumerate}
\item combining $[f_i][f_{i+1}]=[f_if_{i+1}]$ if $f_i,f_{i+1}\in A_\alpha$, or
\item taking $[f_i]$ to be in either $\pi_1(A_\alpha)$ or $\pi_1(A_\beta)$ if $[f_i]\in \pi_1(A_\alpha\cap A_\beta)$.
\end{enumerate}
We can create a homotopy from $f_1\cdots f_k$ to $f_1'\cdots f_m'$ by dividing up $[0,1]^2$ into a grid such that, as we pass over each line in the grid, we are using either operation 1 or operation 2.
\begin{center}
\begin{tikzpicture}
\foreach \x in {0.2,0.4,...,1.8} {
\draw[gray] (-1+\x,-1) -- (-1+\x,1);
\draw[gray] (-1,-1+\x) -- (1,-1+\x);}
\draw[thick] (-1,-1) rectangle (1,1);
\node at (0,-1.3) {$f_1'\cdots f_m'$};
\node at (0,1.3) {$f_1\cdots f_k$};
\node at (-1.3,0) {$x_0$};
\node at (1.3,0) {$x_0$};
\draw[thick,-angle 60] (1.8,0) -- (2.6,0) node [label=right:$X$] {};
\end{tikzpicture}
\end{center}
\end{proof}
\pagebreak


\begin{example}
$\text{}$

\begin{itemize}
\item Let's compute $\pi_1(\Sigma_2)$, where $\Sigma_2$ is the two-holed torus. Recall that we can write $\Sigma_2$ as the quotient of an octagon. We break up the octagon into two pieces, $A_1$ and $A_2$:
\begin{center}
\begin{tikzpicture}[scale=0.6]
\coordinate[label=above left: ] (tl) at (-1,2.41);
\coordinate[label=above right: ] (tr) at (1,2.41);
\coordinate[label=above right: ] (ru) at (2.41,1);
\coordinate[label=below right: ] (rd) at (2.41,-1);
\coordinate[label=above left: ] (lu) at (-2.41,1);
\coordinate[label=below left: ] (ld) at (-2.41,-1);
\coordinate[label=below left: ] (bl) at (-1,-2.41);
\coordinate[label=below right: ] (br) at (1,-2.41);
\fill[red!40] (tl.center) -- (tr.center) -- (ru.center) -- (rd.center) -- (br.center) -- (bl.center) -- (ld.center) -- (lu.center) -- cycle;
\begin{scope}[thick,decoration={markings,mark=at position 0.55 with {\arrow[>=angle 90,scale=0.8]{>}}}]
\draw[postaction={decorate}] (tr) -- node[auto,swap] {$c$} (tl);
\draw[postaction={decorate}] (ru) -- node[auto] {$a$} (rd);
\draw[postaction={decorate}] (tr) -- node[auto] {$b$} (ru);
\draw[postaction={decorate}] (br) -- node[auto,swap] {$b$} (rd);
\draw[postaction={decorate}] (bl) -- node[auto,swap] {$a$} (br);
\draw[postaction={decorate}] (bl) -- node[auto] {$d$} (ld);
\draw[postaction={decorate}] (ld) -- node[auto] {$c$} (lu);
\draw[postaction={decorate}] (tl) -- node[auto,swap] {$d$} (lu);
\end{scope}
\draw[thick,fill=purple!60,dashed] (0,0) circle (1.5);
\draw[thick,fill=blue!40] (0,0) circle (1.2);
\node at (0,0) {$A_2$};
\node at (0.7,1.9) {$A_1$};
\begin{scope}[shift={(6,0)}]
\node at (0,0) {$A_1\simeq$};
\node[mypoint] (c) at (2.3,0) {};
\draw[thick] (c.center) to[out=0,in=60,loop,distance=2cm] (c.center);
\draw[thick] (c.center) to[out=90,in=150,loop,distance=2cm] (c.center);
\draw[thick] (c.center) to[out=180,in=240,loop,distance=2cm] (c.center);
\draw[thick] (c.center) to[out=270,in=330,loop,distance=2cm] (c.center);
\end{scope}
\end{tikzpicture}
\end{center}
We have
\[\pi_1(\Sigma_2)=(\pi_1(A_1)\ast 1)/\pi_1(A_1\cap A_2)=\langle a,b,c,d\rangle/\Z,\]
but we need to understand which $\Z$. Looking at $A_1\cap A_2$, we see that
\[\pi_1(\Sigma_2)=\langle a,b,c,d\rangle/[a,b][c,d].\]
\end{itemize}
\end{example}








