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\newtheorem{Tietze}[thm]{Tietze Extension Theorem}
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\begin{document}


\chapter{Fundamental group and covering spaces}

\section{Fundamental group}

\begin{thm}*
Given topological spaces $X$ and $Y$, $\simeq$ is an equivalence relation
on the set of
all continuous functions from $X$ to $Y$.
\end{thm}

\begin{thm}
Given topological spaces $X$ and $Y$ with $S\subset X$, being
homotopic relative to $S$ is an equivalence relation
on the set of
all continuous functions from $X$ to $Y$.
\end{thm}


\begin{thm}*
Let $\alpha$ be a loop into the topological space $X$.  Then $\alpha =\beta \circ \omega$ where $\omega$ is the standard wrapping map and $\beta$ is a continuous function from $\Sph^1$ into $X$.
\end{thm}


\begin{thm}
Let $X$ be a topological space and let $p$ be a point in $X$. Then a loop  $\alpha =\beta \circ \omega$ (where $\omega$ is the standard wrapping map and $\beta$ is a continuous function from $\Sph^1$ into $X$) is homotopically trivial if and only if $\beta$ can be extended to a
continuous function from $\B^2$ into $X$.
\end{thm}

\begin{thm}
If $\alpha\sim \alpha'$ and $\beta\sim\beta'$,
then $\beta\cdot\alpha\sim\beta'\cdot\alpha'$.
\end{thm}

\begin{thm}
Given $\alpha$, $\beta$, and $\gamma$, then $(\alpha\cdot\beta)\cdot
\gamma \sim\alpha\cdot (\beta\cdot \gamma)$ and $([\alpha]\cdot[\beta])\cdot
[\gamma] \sim[\alpha]\cdot ([\beta]\cdot [\gamma])$.
\end{thm}

\begin{thm}
Let $\alpha$ be a path with $\alpha(0)=x_0$, then
$\alpha\cdot
\alpha^{-1} \sim e_{x_0}$, where $e_{x_0}$ is the
constant path $e_{x_0}: [0,1]\rightarrow x_0$. Stated differently, if $\alpha$ is a path, then $\alpha\cdot
\alpha^{-1}$ is homotopically trivial.
\end{thm}

\begin{thm}*
The fundamental group $\pi_1 (X,x_0)$ is a group. The identity element is the class of homotopically trivial loops based at $x_0$.
\end{thm}

\begin{thm}*
If $X$ is path connected, then $\pi_1(X,p)\cong \pi_1 (X,q)$ for
any points $p,q\in X$.
\end{thm}

\begin{exercise}
We will use $1$ to denote the trivial group:
\be
\item[1.] $\pi_1 ([0,1])\cong 1$.
\item[2.] $\pi_1 (\Sph^0, 1)\cong 1$ where $\Sph^0$ is the zero-dimensional sphere
    $\{-1,1\}\subset\R^1$.
\item[3.] $\pi_1({\textrm {convex\ set}})\cong 1$.
\item[4.] $\pi_1({\textrm{cone}})\cong 1$.
\item[5.] $\pi_1$(cone over Hawaiian earring) $\cong 1$.
\item[6.] $\pi_1$($\R^n$) $\cong 1$ for $n$ for $n\geq 1$.
\item[7.] $\pi_1(\Sph^2) \cong 1$.
\ee
\end{exercise}


This is question can be substituted for two other questions. 
\begin{thm}**
The fundamental group of the circle $\Sph^1$ is infinite cyclic,
that is, $\pi_1(\Sph^1)\cong\Z$.
\end{thm}

If you can't get the proof to work out, remember the result. It is the first
non-trivial fundament group we have seen.

\subsection{Cartesian products}

\begin{thm}*
Let $(X, x_0)$, $(Y, y_0)$ be path connected spaces. Then $\pi_1 (X~\times~Y,~(x_0, y_0)) \cong \pi_1(X, x_0)\times \pi_1 (Y, y_0)$.
\end{thm}

\begin{exercise}
Find:
\be
\item[1.] $\pi_1 (\T^2\cong\Sph^1\times\Sph^1)$;
\item[2.] $\pi_1 (\D^2\times\Sph^1) $;
\item[3.] $\pi_1 (\Sph^2\times\Sph^1) $;
\item[4.] $\pi_1 (\Sph^2\times\Sph^2) $;
\item[5.] $\pi_1 (\Sph^2\times\Sph^2\times\Sph^2) $;
\item[6.] $\pi_1 (\Sph^{p_1}\times\ldots\times\Sph^{p_k}) $ where $p_j\geq 2$
for $1\leq j\leq k$.

\ee
\end{exercise}

\subsection{Induced homomorphisms}

\begin{exercise}
Check that for a continuous function $f:X\to Y$, the induced homomorphism on the fundamental group
$f_*$ is well-defined.
\end{exercise}

\begin{thm}*
If $g: (X, x_0)\to (Y, y_0)$, $f:(Y, y_0)\to (Z, z_0)$ are continuous functions, then $(f\circ g)_*
= f_* \circ g_*$.
\end{thm}

\begin{thm}
If $f$, $g: (X, x_0)\to (Y, y_0)$ are continuous functions and $f$ is homotopic to $g$ relative to $x_0$,
then $f_*: = g_*$ .
\end{thm}

\begin{thm}*
If $h:X\to Y$ is a homeomorphism then $h_* :\pi_1 (X,x_0)\to
\pi_1 (Y,f(x_0))$ is a group isomorphism. So homeomorphic spaces have naturally isomorphic
fundamental groups.
\end{thm}


\section{Retractions and fixed points}

\begin{thm}
Let $A$ be a retract of $X$ and let $i:A\hookrightarrow X$ be the inclusion map.
Then $i_*:\pi_1(A)\to \pi_1(X)$ is injective.
\end{thm}

\begin{question}
Give an example to show why the conclusion of the previous theorem does not follow merely from the assumption that $A$
is a subset of $X$.
\end{question}

\begin{thm}
Let $A$ be a retract of $X$, $a_0\in A$ and $r:A\hookrightarrow X$ the retraction.
Then $r_*:\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective.
\end{thm}

\begin{thm}[No Retraction Theorem for $\D^2$]
\index{No Retraction Theorem!$\D^2$}*
There is no retraction from $\D^2$ to its boundary.
\end{thm}

\begin{thm}
The identity map $i:\Sph^1\to \Sph^1$ is not null homotopic.
\end{thm}

\begin{thm}
The inclusion map $j:\Sph^1\to \R^2-{\mathbf{0}}$ is not null homotopic.
\end{thm}

\begin{thm}[Brouwer Fixed-Point Theorem for $\D^2$]*
\index{Brouwer Fixed-Point Theorem!$\D^2$}
Let $f:\D^2\to\D^2$ be a continuous map, then there is some $x\in \D^2$
for which $f(x)=x$.
\end{thm}

This is one of the coolest theorems of the course. In addition, to proving it
explain why if you droped a map of the United States on the ground 
(in Austin), there 
would be one point on the map that was exactly on top of where it actually is. 

\begin{exercise}
Show that $\R^2-\{2$ points$\}$ strong deformation retracts onto the wedge of two circles.
In addition, show that $\R^2-\{2$ points$\}$ strong deformation 
retracts onto a theta curve. Are the wedge of two circles and the theta curve homeomorphic? (See Figure \ref{wedgeandtheta})
\end{exercise}

\begin{figure}[h]
\begin{center}

\includegraphics[width =3.5 in]{wedgeandtheta}
\caption{\label{wedgeandtheta}The wedge of two circles and the theta curve}
\end{center}
\end{figure}

\begin{thm}*
If $r:X\to A$ is a strong deformation retraction and $a\in A$, then
$\pi_1 (X,a)\cong \pi_1(A,a)$.
\end{thm}

\begin{exercise}
Compute the fundamental groups of the following spaces:
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item $\pi_1(\textrm{solid\ torus}\cong \D^2\times\Sph^1)\cong$
\item $\pi_1(\R^2 - {\mathbf{0}}) \cong$
\item $\pi_1(\textrm{house\ with\ 2\ rooms})\cong $ \index{house with $2$ rooms}

\begin{figure}
\begin{center}
\includegraphics[width= 6 cm]{tworooms} 
\caption{\label{tworooms} House with 2 rooms}
\end{center}
\end{figure}

\item $\pi_1(\textrm{dunce's\ hat})\cong$ \index{dunce's hat}

\begin{figure}
\begin{center}
\includegraphics[width= 6 cm]{duncehat} 
\caption{\label{duncehat} Dunce's hat}
\end{center}
\end{figure}

\ee
\end{exercise}



\begin{thm}
A contractible space is simply connected.
\end{thm}

\begin{thm}
A retract of a contractible space is contractible.
\end{thm}

\begin{thm}
The House with Two Rooms is contractible.
\end{thm}

\begin{thm}
The Dunce's Hat is contractible.
\end{thm}

\end{document}