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\begin{document}

\centerline{\bfseries Mathematics 4530}
\smallskip
\centerline{\bfseries Prelim 2, due December 5, 2013}

\bigskip\bigskip

\noindent
What follows is a generalization of the calculation of $\pi_1(S^1)$.
It is essentially a handout, but with all the proofs removed.  Your
task is to fill in the proofs in Sections
\ref{sec:introduction}--\ref{sec:existence-regular-g}.  [You may read
Section~\ref{sec:cayley-graphs-figure} for fun if you want, but there
is nothing in it for you to write up.]  I have given a clear
indication of the places where you are expected to fill in a proof.  I
have also labeled some proofs as ``extra credit'' to keep the exam
from being too long; they are not necessarily more difficult than the
other proofs.

You may use any result proved in class, in the assigned reading, or in
your homework.  You may not use any sources other than your textbook
and class notes.  You may not use results from unassigned homework
problems or from sections of the book that you have not been told to
read.  Andrew and I will be glad to clear up any ambiguities;
\textbf{you may not discuss the exam with anyone else}.

\section{Introduction}
\label{sec:introduction}

A crucial feature of the exponential covering $p\colon \R\to S^1$ is
that two points in the same fiber differ by an integral multiple
of~$2\pi$.  This was used in the definition of the degree homomorphism
$\pi_1(S^1)\to\Z$, which was then proved to be an isomorphism.  We
wish to generalize this to other covering maps, the so-called
\emph{regular} ones.  As a consequence, one can calculate many
fundamental groups with no more effort than was required for~$S^1$.

\section{Group actions}
\label{sec:group-actions}

We have defined a group to be a set $G$ together with a binary
operation $*$ satisfying certain axioms.  This may give you the wrong
impression of group theory, in that it gives you no clue as to how
groups arise in nature.  The way groups arise in nature is that they
\emph{act} on things, thereby exhibiting the symmetry that those
things have.

\begin{definition}
\label{def:1}
Let $G$ be a group and $X$ a topological space.  By an \emph{action}
of $G$ on~$X$ we mean a function $G\times X\to X$, denoted
$(g,x)\mapsto g*x$, satisfying:
\begin{enumerate}[(1)]
\item %1
$g*(h*x)=(g*h)*x$ for all $g,h\in G$ and $x\in X$.
\item %2
$1*x=x$ for all $x\in X$, where $1$ is the identity element of~$G$.
\item %3
For each $g\in G$, the map $x\mapsto g*x$ is continuous.
\end{enumerate}
\end{definition}
\noindent
It follows that the map $x\mapsto g*x$, which we will call the
\emph{action} of~$g$, is actually a homeomorphism.  Its inverse is
given by the action of~$g^{-1}$.  Group actions are studied
extensively in algebra courses, such as Math~4340.  For our purposes,
however, it suffices to have some examples.

\begin{example}
\label{exam:1}
The additive group $\Z$ acts on $\R$ by $(n,x)\mapsto 2\pi n + x$ for
$n\in\Z$ and $x\in\R$.  (We say ``$\Z$ acts on~$\R$ by translation.'')
This is the action that's implicit in our work on the exponential
covering map, which led to the calculation $\pi_1(S^1) \iso\Z$.
\end{example}

\begin{example}
\label{exam:2}
The multiplicative group $\{\pm1\}$ acts on $S^2$ by scalar
multiplication.  This action is related to the double covering $S^2\to
P^2$ mentioned in class, where $P^2$ is the projective plane.
\end{example}

\begin{example}
\label{exam:3}
More generally, whenever we have a 2-fold covering map $p\colon X\to
Y$, there is an associated action of a group of order~2 on~$X$.  This
is essentially what you proved in additional problem~6 on
Assignment~10.
\end{example}

\begin{example}
\label{exam:4}
Let $\Z_n$ be the additive group of integers mod~$n$.  There is an
action of $\Z_n$ on~$S^1$, in which the integer $k$~mod~$n$ acts as
rotation through $2\pi k/n$ radians.  Alternatively, if we replace
$\Z_n$ by the isomorphic group consisting of the $n$th roots of unity
in~$\C$, then we can describe the action using multiplication of
complex numbers.
\end{example}

\begin{example}
\label{exam:5}
The concept of ``Cayley graph,'' which I will explain in
Section~\ref{sec:cayley-graphs-figure}, leads to many examples of
group actions.  For example, the snowflake that covers the figure~8 is
the Cayley graph of the free group~$F_2$; as a result, there is an
action of $F_2$ on the snowflake.
\end{example}

To simplify the notation in what follows, we will always use
multiplicative notation for our groups (i.e., we will write $gh$
instead of $g*h$), and similarly we will write $(g,x) \mapsto gx$ for
group actions.

Every one of the examples above has the property that it is
\emph{fixed-point free}, i.e., the action of any $g\neq 1$ has no
points $x$ such that $gx=x$.  The significance of this for us is that
if two points $x,x'$ are in the same \emph{$G$-orbit}, i.e., if there
is a $g\in G$ such that $gx=x'$, then the element $g$ is unique.  [If
$gx=x'$ and $hx=x'$, then $gx=hx \implies h^{-1}gx=x \implies
h^{-1}g=1 \implies g=h$.]  We can think of $g$ as something like the
``difference'' between $x$ and~$x'$.  For the case of $\Z$ acting on
$\R$ as in Example~\ref{exam:1}, it really is the difference (divided
by~$2\pi$).

\section{Regular $G$-covers}
\label{sec:regular-g-covers}

\begin{definition}
\label{def:2}
Let $G$ be a group.  A \emph{regular $G$-cover} is a covering map
$p\colon X\to Y$ together with a fixed-point-free action of $G$ on~$X$
whose orbits are the fibers of~$p$.
\end{definition}
\noindent
Let's spell this out explicitly:

\begin{lemma}
\label{lem:1}
Let $p\colon X\to Y$ be a covering map with an action of a group $G$
on~$X$.  Then $p$ is a regular $G$-cover if and only if it satisfies
the following two conditions:
\begin{enumerate}[(a)]
\item%a
$p(gx)=p(x)$ for all $g\in G$, $x\in X$.
\item%b
If $p(x)=p(x')$ then there is a (unique) $g\in G$ such that $x'=gx$.
\end{enumerate}
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\begin{remark}
\label{rem:1}
The group action is useful in connection with path liftings.  Namely,
if $\alpha$ is a path in $Y$ and $\atilde$ is the lift starting at
some point in the fiber over $\alpha(0)$, then we can construct all
the other liftings of $\alpha$ by applying the group action.
\begin{center}
*** Your explanation goes here. ***
\end{center}
\end{remark}

We have already seen many examples of regular $G$-covers.  Indeed,
every example in the previous section has an associated regular
$G$-cover, which was mentioned explicitly except in
Example~\ref{exam:4}.  [You can ask me if you can't figure out the
cover that goes with that example.]  On the other hand, there are many
covering maps that are not regular because they do not have enough
symmetry.  Indeed, conditions (a) and~(b) imply, intuitively, that a
regular $G$-cover looks the same no matter where we stand in a given
fiber.  For example:

\begin{lemma}
\label{lem:2}
Let $p\colon X\to Y$ be a regular $G$-cover for some group $G$, and
let $\alpha$ be a closed curve in~$Y$ that admits a closed
lift~$\tilde\alpha$.  Then every lift of~$\alpha$ is closed.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

One can deduce from this that not all covering maps are regular.

\begin{example}
\label{exam:6}
*** Your example goes here. ***
\end{example}

\noindent
[To make it interesting, make sure that $X$ and $Y$ are both
connected.]

\section{The generalized degree homomorphism}
\label{sec:degree-homomorphism}

Recall that we defined the degree of a closed curve $\alpha$ in $S^1$
by lifting it to~$\R$ and measuring the difference between the
starting and ending points of the lift.  We can do something similar
for any regular cover.

Let $p\colon X\to Y$ be a regular $G$-cover.  Pick a basepoint $y\in
Y$ and a basepoint $x\in p^{-1}(y)$.  Given a path homotopy class
$[\alpha] \in \pi_1(Y,y)$, let $\tilde\alpha$ be the lift of~$\alpha$
starting at~$x$.  Then $\tilde\alpha$ ends in the fiber~$p^{-1}(y)$,
so there is a unique $g\in G$ such that $\tilde\alpha(1)=gx$.  We set
\[
\delta([\alpha]):=g.
\]
In case $p$ is the exponential covering of~$S^1$, this is precisely
the degree defined in class.  Imitating the four steps that occurred
in our calculation of $\pi_1(S^1)$, we will prove:

\begin{theorem}
\label{thr:1}
Suppose that $X$ is simply connected.  Then
\[
\delta \colon \pi_1(Y,y)\to G
\]
is an isomorphism.
\end{theorem}

The four steps are given in the following four lemmas, valid for any
regular $G$-covering.

\begin{lemma}
\label{lem:3}
$\delta$ is well defined.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\begin{lemma}
\label{lem:4}
$\delta$ is a group homomorphism.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\begin{lemma}
\label{lem:5}
If $X$ is path connected, then $\delta$ is surjective.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\begin{lemma}
\label{lem:6}
If $X$ is simply connected, then $\delta$ is bijective.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\section{Examples}
\label{sec:examples}

\begin{example}
\label{exam:7}
Combining Theorem~\ref{thr:1} and Example~\ref{exam:2}, we obtain:
\[
\text{ *** Your conclusion goes here, with full justification. ***}
\]
\end{example}

\begin{example}
\label{exam:8}
Applying Theorem~\ref{thr:1} to a suitable covering of the torus, we
obtain:
\[
\text{ *** Your conclusion goes here, with full justification. ***}
\]
\end{example}

\begin{example}
\label{exam:9}
I promised that I would prove that the fundamental group of the
figure~8 is the free group $F_2$ on two generators.  One way to do
this is to apply Theorem~\ref{thr:1} to the snowflake cover of the
figure~8.  I will sketch how this can be done in
Section~\ref{sec:cayley-graphs-figure}.  A different proof will be
given in class.
\end{example}

\section{Existence of regular $G$-covers}
\label{sec:existence-regular-g}

It is natural to ask how widely applicable Theorem~\ref{thr:1} is.  In
this section we will see that, in principle, Theorem~\ref{thr:1} can
be used to calculate the fundamental group of \emph{every} reasonable
space.  Here ``reasonable'' has a somewhat technical definition:

\begin{definition}
\label{def:3}
A space $Y$ is said to be \emph{semilocally simply connected} (SLSC)
if every point $y\in Y$ has a neighborhood~$U$ such that the inclusion
$U\into Y$ induces the trivial homomorphism $\pi_1(U,y)\to\pi_1(Y,y)$.
\end{definition}
\noindent
This condition may seem strange; but in fact, most spaces that one
meets ``in nature'' satisfy the even stronger condition of being
locally contractible.  [Feel free to ask for examples if you have
trouble seeing this.]  Here's more evidence that the SLSC condition is
reasonable:.

\begin{proposition}
\label{prop:1}
If $Y$ admits a simply connected covering space, then $Y$ is SLSC.
\end{proposition}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

Our goal in this section is to prove the following theorem:

\begin{theorem}
\label{thr:2}
Suppose $Y$ is path connected, locally path connected, and SLSC.  Then
there exists a group $G$ and a regular $G$-cover $p\colon X \to Y$
such that $X$ is simply connected.
\end{theorem}

\noindent
[Necessarily, $G$ will be isomorphic to $\pi_1(Y)$.]

\bigskip

The first step is to figure out what $X$ has to look like.

\begin{lemma}
\label{lem:7}
\begin{enumerate}[(a)]
\item%a
If $U$ satisfies the condition in Definition~\ref{def:3}, then every
smaller neighborhood of~$y$ also satisfies the condition.
\item \label{basis}%b
Suppose $Y$ is SLSC and locally path connected.  Let $\CalB$ be the
collection of path-connected open sets $U$ such that
$\pi_1(U)\to\pi_1(Y)$ is the trivial homomorphism for some (or every)
basepoint in~$U$.  Then $\CalB$ is a basis for~$Y$.
\item \label{inj}%c
For any $U\in\CalB$ let $\beta_1,\beta_2$ be paths in $U$ with
$\beta_1(0)=\beta_2(0)$ and $\beta_1(1)=\beta_2(1)$.  Then
$[\beta_1]=[\beta_2]$ in the fundamental groupoid of~$Y$.
\end{enumerate}
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

Let $Y$ be a path-connected space with basepoint $y_0$, let $\CalG$ be
its fundamental groupoid, and let $\CalG_0\subseteq\CalG$ be the
subset consisting of path classes starting at~$y_0$.  Thus $\CalG_0$
is a set whose elements are equivalence classes $a=[\alpha]$, where
$\alpha$ is a path in~$Y$ with $\alpha(0)=y_0$.  Let $p\colon
X\to Y$ be a covering map, and let $x_0\in p^{-1}(y_0)$ be
a basepoint.  Then one can use path lifting to define a function
$h\colon\CalG_0\to X$.  Namely, given $[\alpha] \in \CalG_0$, we
lift $\alpha$ to a path $\atilde$ in $X$ starting at~$x_0$
and set
\[
h([\alpha]) = \atilde(1).
\]

\begin{lemma}
\label{lem:8}
\begin{enumerate}[(a)]
\item %a
$h$ is well defined.
\item %b
$h$ is surjective if $X$ is path connected and bijective if
$X$ is simply connected.
\end{enumerate}
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

From now on we assume that $Y$ is path connected, locally path
connected, and SLSC.  We have just seen that if $Y$ admits a simply
connected covering space~$X$, then the points of~$X$ must
be in 1--1 correspondence with the elements of the set~$\CalG_0$.
With this as motivation, we will prove Theorem~\ref{thr:2} by starting
with~$\CalG_0$ and putting a suitable topology on it.  Let $\CalB$ be
the basis for~$Y$ given in Lemma~\ref{lem:7}(b).

Let $a=[\alpha]$ be an element of~$\CalG_0$.  For any $U\in\CalB$ with
$\alpha(1)\in U$, let $\Utilde_a\subseteq \CalG_0$ be the set of path
classes of the form $a*b$, where $b=[\beta]$ for some path $\beta$
in~$U$ with $\beta(0)=\alpha(1)$.  We will call $\Utilde_a$ a
\emph{basic neighborhood} of~$a$, and we will call a subset
of~$\CalG_0$ \emph{open} if it contains a basic neighborhood of each
of its points.

\begin{lemma}
\label{lem:9}
The open sets just defined form a topology on~$\CalG_0$, with the sets
$\Utilde_a$ as a basis.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

\noindent 
[Hint: It might help to show that if $a_1\in \Utilde_a$, then
$\Utilde_{a_1} = \Utilde_a$.]

\bigskip

As an aid to the intuition and a reminder of what we're trying to do,
we will denote by~$X$ the set~$\CalG_0$ with the topology that
we have just defined.  Let $p\colon X\to Y$ be defined by
$p([\alpha])=\alpha(1)$.

\begin{lemma}
\label{lem:10}
$p$ is continuous and open and maps $\Utilde_a$ homeomorphically onto
$U$.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

\begin{lemma}
\label{lem:11}
Suppose $a_1$ and $a_2$ are classes of paths starting at $y_0$ and
having a common endpoint~$y$.  Let $U$ be a member of~$\CalB$ such
that $y\in U$.  If $a_1\neq a_2$, then $\Utilde_{a_1}$ and
$\Utilde_{a_2}$ are disjoint.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

\begin{lemma}
\label{lem:12}
$p$ is a covering map.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

Given a path $\alpha$ starting at $y_0$ and given $t\in[0,1]$, let
$\alpha_t$ be the path given by $\alpha_t(s)=\alpha(ts)$ for $0\leq
s\leq 1$.  You may find it helpful to describe by means of words
and/or pictures the path $t\mapsto [\alpha_t]$ in~$X$.  [You
don't need to write this down.]

\begin{lemma}
\label{lem:13}
The path $t\mapsto [\alpha_t]$ in~$X$ is indeed a path, i.e. it
is continuous.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

Note that this lemma shows you how to lift any path $\alpha$ in $Y$
starting at $y_0$ to a path $\atilde$ in $X$ starting at
$x_0:=[e_{y_0}]$.

\begin{lemma}
\label{lem:14}
$X$ is path connected.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

\begin{lemma}
\label{lem:15}
$X$ is simply connected.
\end{lemma}

\begin{proof}
*** Your proof goes here. ***
\end{proof}

\noindent
[Hint: If $[\alpha]\in\pi_1(Y,y_0)$ is nontrivial, show that its lift
$\atilde$ is not a closed curve; deduce the result from this by
covering space theory.]

\begin{lemma}
\label{lem:16}
There is an action of $\pi_1(Y,y_0)$ on $X$ that makes $p$ a regular
$\pi_1(Y,y_0)$-cover.
\end{lemma}

\begin{proof}
*** Extra credit. ***
\end{proof}

This completes the proof of Theorem~\ref{thr:2}.

\begin{remark}
\label{rem:2}
It is possible to show that $X$ covers every other
path-connected covering space of~$Y$.  [One can use
Lemma~\ref{lem:8}.]  For this reason $X$ is often called the
\emph{universal cover} of~$Y$.
\end{remark}

\textbf{This is the end of the exam unless you want to read the
  next section for fun.}

\section{Cayley graphs and the figure 8}
\label{sec:cayley-graphs-figure}

This (optional) section assumes a little more knowledge of group
theory than I have assumed previously.

Let $G$ be a group with a generating set~$S$.  The \emph{Cayley graph}
of $(G,S)$ is the graph $\Gamma = \Gamma(G,S)$ with $G$ as its vertex
set and with an edge joining $g$ and~$gs$ for each $g\in G$ and $s\in
S$.  For example, the Cayley graph of the free group $F_2$ with its
standard generating set is the snowflake.

% A second example is shown
% in the following picture; see if you can guess what $G$ and $S$ are.
% [Hint~1: The number of vertices is the order of the group.  Hint~2:
% Each generator $s\in S$ is assigned a color; so the number of colors
% of edges is the number of generators.]
% \begin{center}
% \includegraphics[scale=.25]{permutahedron}
% \end{center}

The left-translation action of $G$ on itself induces an action of $G$
on the graph~$\Gamma$ and hence an action of $G$ on the geometric
realization $X:=\abs{\Gamma}$.  This action is fixed-point free and
makes $X$ a regular $G$-cover of $Y:=X/G$.  If $G=F_2$ with its
standard generating set, $X$ is a tree and hence is contractible.  In
particular, it is simply connected.  And $Y$ is the figure~8 in this
case.  Theorem~\ref{thr:1} therefore implies that the fundamental
group of the figure~8 is free of rank~2, as claimed.


\end{document}

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