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\begin{document} 

\setcounter{section}{4}
\section{The Fundamental Group}
In this chapter we show how to associate a group with a topological space. When the spaces are the same the groups will be the same.

\begin{defn}
A \emph{path} in a topological space $X$ is a map $\alpha : [0,1] \rightarrow X$. The points $\alpha(0)$ and $\alpha(1)$ are said to be \emph{joined} by $\alpha$.  A space is \emph{path-connected} if any two points can be joined by a path.
\end{defn}

\begin{thm}
$X \approx Y \Rightarrow$  ($X$ is path-connected $\Leftrightarrow$  Y is path-connected).
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove.
\end{xbox}
\end{proof}

\begin{thm}
Every path-connected space is connected.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove, but show that the converse is false.
\end{xbox}
\end{proof}

\begin{defn}
Given a space $X$ and a point $x$ in $X$, a \emph{loop based at $x$} is a path $\alpha$ with $\alpha(0)=\alpha(1)=x$.
\end{defn}

\begin{ex}\ \\
\begin{xbox}
In each of the following, choose an $x$ in $X$ and draw several loops in $X$ based at $x$.  Actually what you will draw is the image of the loop but we will use the same name for both.   $\E^1$, $\E^2$, $\S^1$, $\E^2 \setminus \B^2$, and $\S^1 \times \S^1$.
\end{xbox}
\end{ex}

\begin{defn}
If $\alpha$ and $\beta$ are paths in a path-connected space $X$ such that 
$\alpha (1) = \beta(0)$, define
 \[\alpha \star \beta(s) = 
  \left\{ \begin{array}{ll}
  \alpha(2s)     & 0 \leq s \leq \frac{1}{2} \\
  \beta(2s-1)    & \frac{1}{2} \leq s \leq 1.
  \end{array}
  \right.  \]
  \end{defn}
  
\begin{defn}
Given a path-connected space $X$ and a point $x$ in $X$ define $L(X,x)$ = \{ all loops in $X$ based at $x$\} . 
\end{defn}

\begin{thm}
$(L(X,x),\star)$ forms a group.
 \end{thm}
 \begin{proof}\ \\
\begin{xbox}
Disprove.
\end{xbox}
\end{proof}

\begin{defn}
Let $\alpha, \beta$ be paths connecting $x$ and $y$ in a space $X$.  We will say that \emph{$\alpha$ is homotopic to $\beta$} (written $\alpha \sim \beta$) $\Leftrightarrow$ there exists a map $H : [0,1] \times  [0,1] \rightarrow X$ such that $H(s,0) =  \alpha(s)$, 
$H(s,1) = \beta(s)$, $H(0,t) = x$, and $H(1,t) =y$.  Note that when $\alpha$ and $\beta$ are loops based at $x$, 
$H(0,t) = H(1,t) =x$. 
\end{defn}

\begin{ex}\ \\
\begin{xbox}
In the spaces in Example 5.3 find some pairs of loops that are homotopic and some that are not.
\end{xbox}
\end{ex}

\begin{thm}
The relation $\sim$ is an equivalence relation on $L(X,x)$.
\end {thm}
\begin{proof}\ \\
\begin{xbox}
Prove this by showing:
\begin{enumerate}
\item $\alpha \sim \alpha$
\item $\alpha \sim \beta \Rightarrow  \beta \sim \alpha$
\item $\alpha \sim \beta \text{ and } \beta \sim \gamma \Rightarrow  \alpha \sim \gamma$
\end{enumerate}
\end{xbox}
\end{proof}

\begin{defn}Let $\langle \alpha \rangle$ denote the homotopy equivalence class of $\alpha$, that is the collection of all loops homotopic to $\alpha$.  Then let $\Pi_1(X,x) = \{\langle \alpha \rangle | \alpha \in L(X,x)\}$ and $\langle \alpha \rangle \otimes \langle \beta \rangle = \langle \alpha \star \beta \rangle$.
\end{defn}

\begin{thm} $(\Pi_1 (X,x),\otimes)$, is a group.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove.
\begin{enumerate}
\item Show that the system is closed \dots this follows from the definition.
\item To show associativity, we need to show $(\langle \alpha \rangle \otimes \langle \beta \rangle) \otimes \langle \gamma \rangle = \langle \alpha \rangle \otimes\langle \beta \rangle \otimes\langle \gamma \rangle)$.  To do this we need to show $(\alpha \star \beta) \star \gamma \sim \alpha \star (\beta \star \gamma)$ ... here is a picture that may help....
\includegraphics[width=.45\textwidth]{5_7a.pdf}
\item $e : [0,1] \rightarrow X$  by $e(t)=x$ (for all $t$) seems a likely candidate for the identity.  Show $\langle e \rangle \otimes \langle \alpha \rangle =\langle \alpha \rangle \otimes \langle e \rangle=\langle \alpha \rangle$ \dots here is another picture \dots
\includegraphics[width=.45\textwidth]{5_7b.pdf}
\item What should $\alpha^{-1}$ be? Try drawing the frames of a movie that show $\alpha \star \alpha^{-1} \sim e$ \dots
\end{enumerate}
\end{xbox}
\end{proof}

\begin{defn}We will call $\Pi_1(X,x)$ the \emph{fundamental group of $X$ based at $x$}.
\end{defn}

\begin{defn} Recall the following definitions from algebra.  If $G$ and $H$ are groups, a function 
$h: G \rightarrow H$ is a \emph{homomorphism} $\Leftrightarrow  \forall a,b \in  G, h(a \otimes b)  =  h(a) \otimes h(b)$.  A bijective homomorphism is an \emph{isomorphism}. Two groups are \emph{isomorphic} (denoted $G \cong H) \Leftrightarrow$ there exists an isomorphism between them.
\end{defn}

\begin{defn}
A space $X$ is \emph{simply connected} $\Leftrightarrow$  for all $x \in X, \Pi_1(X,x) \cong \{\langle e \rangle\}$.
\end{defn}

\begin{thm}
Any topological space with the trivial topology is simply connected.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove.  To do this, show that if $\alpha$ is any loop in $X$, then $\alpha \sim e$.
\end{xbox}
\end{proof}

\begin{defn}
Let $\mathbf{x}$ and $\mathbf{y}$ be two points in $\E^n$.  We define the line segment $L(\mathbf{x},\mathbf{y})$ between $\mathbf{x}$ and $\mathbf{y}$ as $L(\mathbf{x},\mathbf{y})=\{(1-t)\mathbf{x}+t\mathbf{y} | 0 \leq t \leq 1 \} $.  A subset $A$ of $\E^n$ is said to be \emph{convex} $\Leftrightarrow$ $\forall \mathbf{x},\mathbf{y} \in A, L(\mathbf{x},\mathbf{y}) \subset A$.
\end{defn}

\begin{thm}
Any convex subset of $\E^n$ with the subspace topology is simply connected.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove. 
\end{xbox}
\end{proof}

\begin{thm}
$\Pi_1 (\S^1,(1,0)) \cong \mathbb{Z}$.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Give some intuitive argument as to why the theorem is true or false.
\end{xbox}
\end{proof}

\begin{thm}
If $X$ is path connected and $x,y \in  X$ then $\Pi_1 (X,x) \cong \Pi_1 (X,y)$.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove...let $\gamma$ be a path connecting $x$ and $y$ \dots consider $\widehat{\gamma} : \Pi_1(X,x) \rightarrow \Pi_1 (X,y)$ defined by $\widehat{\gamma}(\langle \alpha \rangle) = \langle \gamma^{-1} \star \alpha \star  \gamma \rangle$ \dots $\widehat{\gamma}$ is an isomorphism \dots
\end{xbox}
\end{proof}

\begin{rem}
Given the above theorem we will refer to $\Pi_1 (X,x)$ as simply $\Pi_1 (X)$ when $X$ is path-connected.
\end{rem}

\begin{defn}\ \\
\begin{xbox}
Make up a definition for the term \emph{star-shaped} so that the following theorem is true. 
\end{xbox}
\end{defn}

\begin{thm}
Any star-shaped subset of $\E^n$ with the subspace topology is simply connected.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove using the same proof as in Theorem 5.9 . 
\end{xbox}
\end{proof}


\begin{defn} Given a map $f : X \rightarrow Y, x \in  X, y \in  Y$, and $f(x) = y$, we define a function 
$\widehat{f}: \Pi_1(X,x) \rightarrow  \Pi_1(Y,y)$ by $\widehat{f}(\langle \alpha\rangle) = \langle f \circ  \alpha \rangle$.  We say that $\widehat{f}$ is \emph{induced} by $f$.
\end{defn}

\begin{thm}
$\widehat{f}$ as defined above is a group homomorphism.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove. 
\end{xbox}
\end{proof}

\begin{thm}
Given maps $f : X \rightarrow Y$ and $g : Y \rightarrow Z, \widehat{(g \circ f)} = \widehat{g} \circ \widehat{f}$.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove. 
\end{xbox}
\end{proof}

\begin{thm}
$(X \approx Y) \Rightarrow (\Pi_1(X) \cong \Pi_1(Y))$.  
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove \dots then by the previous theorem \dots
\end{xbox}
\end{proof}

\begin{thm}
$(\Pi_1(X) \cong \Pi_1(Y)) \Rightarrow (X \approx Y)$.  
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove or disprove.
\end{xbox}
\end{proof}

\begin{thm}
Let $X$ be a space such that $X = A \cup B$ where $A$ and $B$ are open.  Then any path connecting a point in $A \setminus B$ and a point in $B \setminus A$ must pass through $A \cap B$.
\end{thm}
\begin{proof}
This follows from Lebesgue's Lemma which we will not prove here.
\end{proof}

\begin{thm}
Let $X$ be a path connected space such that $X = A \cup B$, where $A$ and $B$ are open and simply connected and $A \cap B$ is path-connected and non-empty,  then $X$ is simply connected.  
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove....choose a base point $x$ in $A \cap B$, let $\alpha$ be a loop in $X$ based 
at  $x$ \dots find a sequence of points $0=t_0<t_1<t_2<t_3< \dots <t_n=1$ such that $\alpha([t_i,t_{i+1}])$ is contained in $A$ or $B$ \dots $\alpha \sim e$ \dots
\end{xbox}
\end{proof}

\begin{thm}
$\S^n$ is simply connected for $n \geq 2$. 
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove.
\end{xbox}
\end{proof}

\begin{thm}
$\E^3 \setminus \{0\}$ is simply connected. 
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove.
\end{xbox}
\end{proof}

\begin{thm}
$\E^2 \setminus \{0\}$ is simply connected. 
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove or disprove.  Does the same proof work here as in the previous theorem?
\end{xbox}
\end{proof}

\begin{thm}
If $X$ and $Y$ are path-connected spaces, $\Pi_1(X \times Y) \cong \Pi_1(X) \times  \Pi_1(Y)$. 
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove \dots define $\Psi :  \Pi_1(X \times Y) \rightarrow  \Pi_1(X) \times  \Pi_1(Y)$ by  
$\Psi(\langle \alpha \rangle) = (\langle p_1 \circ \alpha \rangle ,\langle p_2 \circ \alpha \rangle)$ \dots
\end{xbox}
\end{proof}

\begin{ex}\ \\
\begin{xbox}
With which spaces does the previous theorem help us?
\end{xbox}
\end{ex}

\begin{defn}
A subset $A$ of a space $X$ is a \emph{retract} of $X \Leftrightarrow$ there exists a surjective map $g : X \rightarrow A$ such that  $\forall a \in A, g(a) = a$ and a map $G : X \times [0,1] \rightarrow  X$ such that: 
\[\begin{array}{ll}
G(x,0) = x     & \forall x \in X\\
G(x,1) = g(x) & \forall x \in X\\
G(a,t) = g(a)=a & \forall a \in A, \forall t \in [0,1]
\end{array}\]
The map $g$ is called a \emph{retraction map} and $G$ is called a \emph{homotopy}.
\end{defn}

\begin{ex}\ \\
$\S^n$ is a retract of $\B^{n+1} \setminus \{0\}$.\\
$\S^n$ is a retract of $\E^{n+1} \setminus \{0\}$.\\
\begin{xbox}
Explicitly write down the retraction map and the homotopy in both cases.
\end{xbox}
\end{ex}

\begin{ex}\ \\
$\S^1$ is a retract of $\S^1 \times [0,1]$.\\
$\S^1$ is a retract of the Mobius Band.\\
A figure eight is a retract of $\B^2 \setminus \{ \text { two copies of } \B^2\}$.\ \\
\begin{xbox}
In these cases just show several frames of the movie as $X$ "retracts" onto $A$.
\end{xbox}
\end{ex}

\begin{ex}\ \\
A figure eight is a retract of $\S^1 \times \S^1 \setminus  \{\B^2\}$.\\
A figure eight is a retract of a Klein Bottle $ \setminus \{\B^2\}$. \\
$\S^1$ is a retract of $\P^2 \setminus \{\B^2\}$.\ \\
\begin{xbox}
Do the same as above.  Hint:  Start with a square with sides identified.
\end{xbox}
\end{ex}

\begin{thm}
If $A$ is a retract of $X$ then $\Pi_1(X) \cong  \Pi_1(A)$.
\end{thm}
\begin{proof}\ \\
\begin{xbox}
Prove \dots think about $\widehat{g}$.
\end{xbox}
\end{proof}

\begin{ex}\ \\
List the fundamental groups that this theorem helps us find.  Does it help us distinguish any of our remaining spaces?
\end{ex}



 \end{document}