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\input colordvi

\def\cltr{\Red}		  % Red  VERY-Approx PANTONE RED
\def\cltb{\Blue}		  % Blue  Approximate PANTONE BLUE-072
\def\cltg{\PineGreen}	  % ForestGreen  Approximate PANTONE 349
\def\cltp{\DarkOrchid}	  % DarkOrchid  No PANTONE match
\def\clto{\Orange}	  % Orange  Approximate PANTONE ORANGE-021
\def\cltpk{\CarnationPink}	  % CarnationPink  Approximate PANTONE 218
\def\clts{\Salmon}	  % Salmon  Approximate PANTONE 183
\def\cltbb{\TealBlue}	  % TealBlue  Approximate PANTONE 3145
\def\cltrp{\RoyalPurple}	  % RoyalPurple  Approximate PANTONE 267
\def\cltp{\Purple}	  % Purple  Approximate PANTONE PURPLE

\def\cgy{\GreenYellow}     % GreenYellow  Approximate PANTONE 388
\def\cyy{\Yellow}	  % Yellow  Approximate PANTONE YELLOW
\def\cgo{\Goldenrod}	  % Goldenrod  Approximate PANTONE 109
\def\cda{\Dandelion}	  % Dandelion  Approximate PANTONE 123
\def\capr{\Apricot}	  % Apricot  Approximate PANTONE 1565
\def\cpe{\Peach}		  % Peach  Approximate PANTONE 164
\def\cme{\Melon}		  % Melon  Approximate PANTONE 177
\def\cyo{\YellowOrange}	  % YellowOrange  Approximate PANTONE 130
\def\coo{\Orange}	  % Orange  Approximate PANTONE ORANGE-021
\def\cbo{\BurntOrange}	  % BurntOrange  Approximate PANTONE 388
\def\cbs{\Bittersweet}	  % Bittersweet  Approximate PANTONE 167
%\def\creo{\RedOrange}	  % RedOrange  Approximate PANTONE 179
\def\cma{\Mahogany}	  % Mahogany  Approximate PANTONE 484
\def\cmr{\Maroon}	  % Maroon  Approximate PANTONE 201
\def\cbr{\BrickRed}	  % BrickRed  Approximate PANTONE 1805
\def\crr{\Red}		  % Red  VERY-Approx PANTONE RED
\def\cor{\OrangeRed}	  % OrangeRed  No PANTONE match
\def\paru{\RubineRed}	  % RubineRed  Approximate PANTONE RUBINE-RED
\def\cwi{\WildStrawberry}  % WildStrawberry  Approximate PANTONE 206
\def\csa{\Salmon}	  % Salmon  Approximate PANTONE 183
\def\ccp{\CarnationPink}	  % CarnationPink  Approximate PANTONE 218
\def\cmag{\Magenta}	  % Magenta  Approximate PANTONE PROCESS-MAGENTA
\def\cvr{\VioletRed}	  % VioletRed  Approximate PANTONE 219
\def\parh{\Rhodamine}	  % Rhodamine  Approximate PANTONE RHODAMINE-RED
\def\cmu{\Mulberry}	  % Mulberry  Approximate PANTONE 241
\def\parv{\RedViolet}	  % RedViolet  Approximate PANTONE 234
\def\cfu{\Fuchsia}	  % Fuchsia  Approximate PANTONE 248
\def\cla{\Lavender}	  % Lavender  Approximate PANTONE 223
\def\cth{\Thistle}	  % Thistle  Approximate PANTONE 245
\def\corc{\Orchid}	  % Orchid  Approximate PANTONE 252
\def\cdo{\DarkOrchid}	  % DarkOrchid  No PANTONE match
\def\cpu{\Purple}	  % Purple  Approximate PANTONE PURPLE
\def\cpl{\Plum}		  % Plum  VERY-Approx PANTONE 518
\def\cvi{\Violet}	  % Violet  Approximate PANTONE VIOLET
\def\clrp{\RoyalPurple}	  % RoyalPurple  Approximate PANTONE 267
\def\cbv{\BlueViolet}	  % BlueViolet  Approximate PANTONE 2755
\def\cpe{\Periwinkle}	  % Periwinkle  Approximate PANTONE 2715
\def\ccb{\CadetBlue}	  % CadetBlue  Approximate PANTONE (534+535)/2
\def\cco{\CornflowerBlue}  % CornflowerBlue  Approximate PANTONE 292
\def\cmb{\MidnightBlue}	  % MidnightBlue  Approximate PANTONE 302
\def\cnb{\NavyBlue}	  % NavyBlue  Approximate PANTONE 293
\def\crb{\RoyalBlue}	  % RoyalBlue  No PANTONE match
%\def\cbb{\Blue}		  % Blue  Approximate PANTONE BLUE-072
\def\cce{\Cerulean}	  % Cerulean  Approximate PANTONE 3005
\def\ccy{\Cyan}		  % Cyan  Approximate PANTONE PROCESS-CYAN
\def\cpb{\ProcessBlue}	  % ProcessBlue  Approximate PANTONE PROCESS-BLUE
\def\csb{\SkyBlue}	  % SkyBlue  Approximate PANTONE 2985
\def\ctu{\Turquoise}	  % Turquoise  Approximate PANTONE (312+313)/2
\def\ctb{\TealBlue}	  % TealBlue  Approximate PANTONE 3145
\def\caq{\Aquamarine}	  % Aquamarine  Approximate PANTONE 3135
\def\cbg{\BlueGreen}	  % BlueGreen  Approximate PANTONE 320
\def\cem{\Emerald}	  % Emerald  No PANTONE match
%\def\cjg{\JungleGreen}	  % JungleGreen  Approximate PANTONE 328
\def\csg{\SeaGreen}	  % SeaGreen  Approximate PANTONE 3268
\def\cgg{\Green}	  % Green  VERY-Approx PANTONE GREEN
\def\cfg{\ForestGreen}	  % ForestGreen  Approximate PANTONE 349
\def\cpg{\PineGreen}	  % PineGreen  Approximate PANTONE 323
\def\clg{\LimeGreen}	  % LimeGreen  No PANTONE match
\def\cyg{\YellowGreen}	  % YellowGreen  Approximate PANTONE 375
\def\cspg{\SpringGreen}	  % SpringGreen  Approximate PANTONE 381
\def\cog{\OliveGreen}	  % OliveGreen  Approximate PANTONE 582
\def\pars{\RawSienna}	  % RawSienna  Approximate PANTONE 154
\def\cse{\Sepia}		  % Sepia  Approximate PANTONE 161
\def\cbr{\Brown}		  % Brown  Approximate PANTONE 1615
\def\cta{\Tan}		  % Tan  No PANTONE match
\def\cgr{\Gray}		  % Gray  Approximate PANTONE COOL-GRAY-8
\def\cbl{\Black}		  % Black  Approximate PANTONE PROCESS-BLACK
\def\cwh{\White}		  % White  No PANTONE match


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\def\delp{\partial^\prime}
\def\delpp{\partial^{\prime\prime}}
\def\sgn{{\roman{sgn}}}
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\def\bbz{{\Bbb Z}}
\def\bbr{{\Bbb R}}



\ctln{\bf Math 872 \hsk Algebraic Topology}

\ssk

\ctln{Running lecture notes}

\bsk


{\bf Covering spaces:} We can motivate our next topic by looking more
closely at one of our examples above. The projective plane ${\Bbb R}P^2$
has $\pu = {\Bbb Z}_2$ . It is also the quotient of the simply-connected
space $S^2$ by the antipodal map, which, together with the identity map,
forms a group of homeomorphisms of $S^2$ which is isomorphic to ${\Bbb Z}_2$.
The fact that ${\Bbb Z}_2$ has this dual role to play in describing 
${\Bbb R}P^2$ is no accident; codifying this relationship requires the 
notion of a covering space.

\msk

The quotient map $q:S^2\ra {\Bbb R}P^2$ is an example of a {\it covering map}.
A map $p:E\ra B$ is called a covering map if for every point $x\in B$, there
is a neighborhood ${\Cal U}$ of $x$ (an
{\it evenly covered neighborhood}) so that $p^{-1}({\Cal U})$ 
is a disjoint union ${\Cal U}_\alpha$ of open sets in $E$, each mapped
homeomorphically onto ${\Cal U}$ by (the restriction of) $p$ . $B$ is
called the {\it base space} of the covering; $E$ is called the {\it total
space}. The quotient map $q$ is an example; (the image of) the complement
of a great circle in $S^2$ will be an evenly covered neighborhood
of any point it contains. The disjoint union of 43 copies of a space,
each mapping homeomorphically to a single copy, is an example of a 
{\it trivial covering}. As a last example, we have the famous 
exponential map $p:{\Bbb R}\ra S^1$ given by $t\mapsto e^{2\pi it} = 
(\cos (2\pi t),\sin (2\pi t))$. The image of any interval $(a,b)$ of length
less than 1 will have inverse image the disjoint union of the
intervals $(a+n,b+n)$ for $n\in{\Bbb Z}$ .

\msk

OK, maybe not the last. We can build many finite-sheeted (every point
inverse is finite) coverings of a bouquet of two circles, say, by 
assembling $n$ points over the vertex, and then, on either side,
connecting the points by $n$ (oriented) arcs, one each going in and out of
each vertex. By choosing orientations on each 1-cell of the bouquet,
we can build a covering map by sending the vertices above to the
vertex, and the arcs to the one cells, homeomorphically, respecting 
the orientations. We can build infinite-sheeted coverings in much 
the same way.

\msk

\leavevmode


\epsfxsize=3in
\ctln{{\epsfbox{0208f1.ai}}}


\bsk

Covering spaces of a (suitably nice) space $X$ have a very close relationship
to $\pu(X,x_0)$. The basis for this relationship is the

\msk 

{\bf Homotopy Lifting Property:} If $p:\wtl{X}\ra X$ is a covering map, 
$H:Y\times I\ra X$ is a homotopy, $H(y,0)=f(y)$, and
$\wtl{f}:Y\ra \wtl{X}$ is a {\it lift} of $f$ (i.e., $p\circ \wtl{f}=f$),
then there is a unique lift $\wtl{H}$ of $H$ with $\wtl{H}(y,0)=\wtl{f}(y)$ .

\msk

The {\bf proof} of this we will defer to next time, to give us sufficient time to ensure we
finish it!

\msk

 In particular, applying this property in the case $Y=\{*\}$, where a homotopy
$H:\{*\}\times I\ra X$ is really just a a path $\gamma:I\ra X$,
we have the {\bf Path Lifting Property}: ``given 
a covering map $p:\wtl{X}\ra X$, a path 
$\gamma:I\ra X$ with $\gamma(0)=x_0$, and a point 
$\wtl{x}_0\in p^{-1}(x_0)$, there is a unique path $\wtl{\gamma}$
lifting $\gamma$ with $\wtl{\gamma}(0)=\wtl{x}_0$ .'' One of the 
immediate consequences of this is one of the cornerstones of covering space
theory:

\bsk  

If $p:(\wtl{X},\tilde{x}_0)\ra (X,x_0)$ is a covering map, then the 
induced homomorphism $p_*:\pu(\wtl{X},\wtl{x}_0)\ra \pu(X,x_0)$ is
injective.

\msk

{\bf Proof:} Suppose $\gamma:(I,\del I)\ra (\wtl{X},\wtl{x}_0)$ is a 
loop $p_*([\gamma])=1$ in $\pu(X,x_0)$. So there is a homotopy
$H:(I\times I,\del I\times I)\ra (X,x_0)$ between $p\circ\gamma$ and the constant 
path. By homotopy lifting, there is a homotopy $\wtl{H}$ from $\gamma$ to 
the lift of the constant map at $x_0$. The vertical sides 
$s\mapsto \wtl{H}(0,s),\wtl{H}(1,s)$ are also lifts of the 
constant map, beginning from 
$\wtl{H}(0,0),\wtl{H}(1,0)=\gamma(0)=\gamma(1)=\wtl{x}_0$, so
are the constant map at $\wtl{x}_0$. Consequently, the lift at the 
bottom is the constant map at $\wtl{x}_0$. So $\wtl{H}$
represents a null-homotopy of $\gamma$, so $[\gamma]=1$
in $\pu(\wtl{X},\wtl{x}_0)$. So $\pu(\wtl{X},\wtl{x}_0)=\{1\}$ .

\msk

Even more, the image $p_*(\pu(\wtl{X},\wtl{x}_0)))\sset \pu(X,x_0)$
is precisely the elements whose representatives are loops at $x_0$, 
which when
lifted to paths starting at $\wtl{x}_0)$, are loops. 
For if $\gamma$ lifts
to a loop $\wtl{\gamma}$, then $p\circ\wtl{\gamma}=\gamma$, so
$p_*([\wtl{\gamma}])=[\gamma]$ . Conversely, if 
$p_*([\wtl{\gamma}])=[\gamma]$, then $\gamma$ and $p\circ\wtl{\gamma}$
are homotopic rel endpoints, and so the homotopy lifts to 
a homotopy rel endpoints between the lift of $\gamma$ at 
$\wtl{x}_0$, and the lift of $p\circ\wtl{\gamma}$ at $\wtl{x}_0$
(which is $\wtl{\gamma}$, since $\wtl{\gamma}(0)=\wtl{x}_0$ and
lifts are unique). So the lift of $\gamma$ is a loop, as desired.

\bsk


The {\bf proof} of the homotopy lifting property follows a pattern 
that we will become 
very familiar with: we lift maps a little bit at a time. For every $x\in X$
there is an open set ${\Cal U}_x$ evenly covered by $p$ . For each fixed
$y\in Y$, since $I$ is compact and the sets $H^{-1}({\Cal U}_x)$ form an
open cover of $Y\times I$, then since $I$ is compact, 
the Tube Lemma provides an open neighborhood 
${\Cal V}$ of $y$ in $Y$ and finitely many $p^{-1}{\Cal U}_{x}$ whose union
covers ${\Cal V}\times I$ . 

\msk

To define $\wtl{H}(y,t)$, we (using a Lebesgue number argument) cut the
interval $\{y\}\times I$ into finitely many pieces, the $i$th mapping into 
 ${\Cal U}_{x_i}$ under $H$. $\wtl{f}(y)$ is in one of the evenly covered
sets ${\Cal U}_{x_1\alpha_1}$, and the restricted map 
$p^{-1}:{\Cal U}_{x_1}\ra {\Cal U}_{x_1\alpha_1}$ following $H$ restricted
to the first interval lifts $H$ along the first interval to a map 
we will call $\wtl{H}$. We then have 
lifted $H$ at the end of the first interval = the beginning of the second, 
and we continue as before. In this way we can define $\wtl{H}$ for all
$(y,t)$ . To show that this is independent of the choices we have
made along the way, we imagine two ways of cutting up the interval 
$\{y\}\times I$ using evenly covered neighborhoods ${\Cal U}_{x_i}$
and ${\Cal V}_{w_j}$, and take intersections of both sets of intervals
to get a common refinement of both sets, covered by the intersections
${\Cal U}_{x_i}\cap {\Cal V}_{w_j}$, and imagine building $\wtl{H}$ using
the refinement. At the start, at $\wtl{f}(y)$, we are in 
${\Cal U}_{x_1\alpha_1}\cap {\Cal V}_{w_1\beta_1}$. Because at the 
start of the lift $(y,0)$ we lift to the same point, and $p^{-1}$ restricted 
to this intersection agrees with $p^{-1}$ restricted to each of the two
pieces, we get the same lift acroos the first refined subinterval. This
process repeats itself across all of the subintervals, showing that
the lift is independent of the choices made. This also shows that
the lift is unique; once we have decided what $\wtl{H}(y,0)$, the
rest of the values of the $\wtl{H}$ are determined by the requirement
of being a lift. also, once we know the map is well-defined, we can see
that it is continuous, since for any $y$, we can make the same choices
across the entire open set $V$ given by the Tube Lemma, and find
that $\wtl{H}$, restricted to ${\Cal V}\times(a_i-\delta,b_i+\delta)$
(for a small delta; we could wiggle the endpoints in the construction
without changing the resulting function, by its well-definedness)
is $H$ estricted to this set followed by $p^{-1}$ restriced in domain 
and range, so this composition is continuous. So $\wtl{H}$
is locally continuous, hence continuous.

\msk

So, for example, if we build a 5-sheeted cover of the bouquet of 2 circles, 
as before, (after choosing a maximal tree upstairs) 
we can read off the images of the generators of the fundamental group
of the total space; we have labelled each ede by the ereator it
traces out downstairs, and for each ede outside of the maximal tree
chosen, we read from basepoint out the tree to one end, across the edge,
and then back to the basepoint in the tree. In our example, this
gives:


\msk

\ctln{$<ab,aaab^{-1}, baba^{-1},baa,ba^{-1}bab^{-1},bba^{-1}b^{-1} | >$}

\msk

\leavevmode


\epsfxsize=3in
\ctln{{\epsfbox{0208f2.ai}}}


\bsk

This is (from its construction) a copy of the free group on 6 letters,
in the free group $F(a,b)$ . In a similar way, by explicitly building
a covering space, we find that the fundamental group of a closed 
surface of genus 3 is a subgroup of the fundamental group of the 
closed surface of genus 2. 

\msk

The cardinality of a point inverse $p\inv(y)$ is, by the evenly
covered property, constant on (small) open sets, so the set of 
points of $x$ whose point inverses have any given cardinality
is open. Consequently, if $X$ is connected, this number
is constant over all of $X$, and is called the number of {\it sheets}
of the covering $p:\wtl{X}\ra X$ . 

\bsk


The number of sheets of a covering map can also be determined 
from the fundamental groups:

\msk

{\bf Proposition:} If $X$ and $\wtl{X}$ are 
path-connected, then the number of sheets of a covering map equals
the index of the subgroup $H=p_*(\pu(\wtl{X},\wtl{x}_0)$ in 
$G=\pu(X,x_0)$ . 

\msk

To see this, choose loops $\{\gamma_\alpha\}$ representing representatives $\{g_\alpha\}$ of each of the (right) cosets of $H$ in $G$. Then if we lift
each of them to loops based at $\wtl{x}_0$, they will have distinct
endpoints; if $\wtl{\gamma}_1(1)=\wtl{\gamma}_2(1)$, then 
by uniqueness of lifts, $\gamma_1*\overline{\gamma_2}$ lifts to 
$\wtl{\gamma}_1*\overline{\wtl{\gamma}_2}$, so it
lifts to a loop, so $\gamma_1*\overline{\gamma_2}$ represents
an element of $p_*(\pu(\wtl{X},\wtl{x}_0)$, so $g_1=g_2$.
Conversely, every point in $p\inv(x_0)$ is the endpoint of on of these
lifts, since we can construct a path $\wtl{\gamma}$
from $\wtl{x}_0$ to any such point $y$, giving a loop
$\gamma=p\circ \wtl{\gamma}$ representing an element $g\in\pu(X,x_0)$.
But then $g=hg_\alpha$ for some $h\in H$ and $\alpha$, 
so $\gamma$ is homotopic rel endpoints to $\eta*\gamma_\alpha$ for some loop
$\eta$ representing $h$. But then lifting these based at $\wtl{x}_0$, by hmotopy
lifting, $\wtl{\gamma}$ is homotopic rel endpoints to $\wtl{\eta}$, which is a 
loop, followed by the lift $\wtl{\gamma}_\alpha$ of $\gamma_\alpha$
starting at $\wtl{x}_0$. So $\wtl{\gamma}$ and 
$\wtl{\gamma}_\alpha$ have the same value at 1.

\ssk

Therefore, lifts of representatives of coset representatives of $H$ in $G$ give
a 1-to-1 correspondence, given by $\wtl{\gamma}(1)$, with $p\inv{x_0}$.
In particular, they have the same cardinality.

\msk

The path lifting property (because $\pi([0,1],0)=\{1\}$) is actually a special
case of a more general {\bf lifting criterion}: If 
$p:(\wtl{X},\wtl{x}_0)\ra (X,x_0)$ is a covering map, and 
$f:(Y,y_0)\ra (X,x_0)$ is a map, where
$Y$ is path-connected and locally path-connected, then there is a lift 
$\wtl{f}:(Y,y_0)\ra (\wtl{X},\wtl{x}_0)$ of $f$ (i.e., 
$f=p\circ\wtl{f}$) $\Leftrightarrow$ 
$f_*(\pu(Y,y_0))\sset p_*(\pu(\wtl{X},\wtl{x}_0))$ . 
Furthermore, two lifts of $f$ which agree at a single point are equal.

\msk

If the lift exists, then $f=p\circ\wtl{f}$ implies 
$f_*=p_*\circ\wtl{f}_*$, so 
$f_*(\pu(Y,y_0)) = p_*(\wtl{f}_*(\pu(Y,y_0)))\sset p_*(\pu(\wtl{X},\wtl{x}_0))$ , as desired.
Conversely, if $f_*(\pu(Y,y_0))\sset p_*(\pu(\wtl{X},\wtl{x}_0))$,
then we wish to build the lift of $f$. Not wishing to waste our work on the
special case, we will {\it use} path lifting to do it! Given $y\in Y$,
choose a path $\gamma$ in $Y$ from $y_0$ to $y$ and use path
lifting in $X$ to lift the path $f\circ\gamma$ to a path $\wtl{f\circ\gamma}$ with 
$\wtl{f\circ\gamma}(0)=\wtl{x}_0$ . Then define
$\wtl{f}(y)=\wtl{f\circ\gamma}(1)$ . Provided we show that
this is well-defined and continuous, it is our required lift, 
since $(p\circ\wtl{f})(y) = p(\wtl{f}(y))=p(\wtl{f\circ\gamma}(1))
=p\circ\wtl{f\circ\gamma})(1) = (f\circ\gamma)(1) = f(\gamma(1)) = f(y)$. 
To show that it
is well-defined, if $\eta$ is any other path from 
$y_0$ to $y$, then $\gamma*\overline{\eta}$ is a loop in $Y$, 
so $f\circ(\gamma*\overline{\eta})=(f\circ\gamma)*\overline{(f\circ\eta)}$
is a loop in $X$ representing an element of 
$f_*(\pu(Y,y_0))\sset p_*(\pu(\wtl{X},\wtl{x}_0))$, and
so lifts to a loop in $\wtl{X}$ based at $\wtl{x}_0$.
Consequently, as before, $f\circ\gamma$ and $f\circ\eta$ lift to paths
starting at $\wtl{x}_0$ with the same value at 1. So $\wtl{f}$ is
well-defined.  To show that $\wtl{f}$ is continuous, we use the 
evenly covered property of $p$. Given $y\in Y$,
and  a neighborhood $\wtl{\Cal U}$ of 
$\wtl{f}(y)$ in $\wtl{X}$, 
we wish to find a nbhd ${\Cal V}$ of $y$ with 
$\wtl{f}({\Cal V})\sset\wtl{\Cal U}$. Choosing an evenly covered 
neighborhood ${\Cal U}_y$ for $f(y)$, choose the sheet 
$\wtl{\Cal U}_y$ over ${\Cal U}_y$ which contains $\wtl{f}(y)$,
and set ${\Cal W}=\wtl{\Cal U}\cap \wtl{\Cal U}_y$. This is open in 
$\wtl{X}$, and $p$ is a homeomorphism from this 
set to the open set $p({\Cal W})\sset X$. Then if we set ${\Cal V}^\prime= f\inv(p({\Cal W})$
this is an open set containing $y$, and so contains a path-connected open 
set ${\Cal V}$ containing $y$. Then is for every point $z\in {\Cal V}$ we build a path
$\gamma$
from $y_0$ to $z$ by concatenating a path from $y_0$ to $y$ with a path {\it in} ${\Cal V}$
from $y$ to $z$, then by unique path lifting, 
since $f({\Cal V}\sset {\Cal U}_y$ , $f\circ\gamma$ lifts to 
the concatenation of a path from $\wtl{x}_0$ to $\wtl{f}(y)$ and a 
path {\it in} $\wtl{\Cal U}_y$ from $\wtl{f}(y)$ to $\wtl{f}(z)$.
So $\wtl{f}(z)\in\wtl{\Cal U}$.

\ssk

Because $\wtl{f}$ is built by lifting paths, and path
lifting is unique, the last statement of the proposition follows.

\bsk

{\bf Universal covering spaces}: As we shall see, a particularly
important covering space to identify is one which is simply
connected. One thing we can see from the lifting crierion is
that such a covering is essentially unique:

\msk

If $X$ is locally path-connected, and has two connected, simply connected
coverings $p_1:X_1\ra X$ and $p_2:X_2\ra X$, then choosing
basepoints $x_i, i=0,1,2$ , since 
$p_{1*}(\pu(X_1,x_1)) = p_{2*}(\pu(X_2,x_2))=\{1\}\sset \pu(X,x_0)$,
the lifting criterion with each projection playing the role of $f$ in turn
gives us maps $\wtl{p}_1:(X_1,x_1)\ra (X_2,x_2)$ and 
$\wtl{p}_2:(X_2,x_2)\ra (X_1,x_1)$ with $p_2\circ\wtl{p}_1=p_1$
and $p_1\circ\wtl{p}_2=p_2$. Consequently, 
$p_2\circ\wtl{p}_1\circ \wtl{p}_2 = p_1\circ\wtl{p}_2=p_2$
and similarly, 
$p_1\circ\wtl{p}_2\circ \wtl{p}_1 =p_2\circ\wtl{p}_1=p_1$.
So $\wtl{p}_1\circ \wtl{p}_2:(X_2,x_2)\ra (X_2,x_2)$, for example,
is a lift of $p_2$ to the covering map $p_2$. But so is the identity map! By
uniqueness, therefore, $\wtl{p}_1\circ \wtl{p}_2=Id$ . Similarly,
$\wtl{p}_2\circ \wtl{p}_1=Id$. So $(X_1,x_1)$ and $(X_2,x_2)$ 
are homeomorphic. So up to homeomorphism, a space can have
only one connected, simply-connected covering space. It is known
as the {\it universal covering} of the space $X$. 

\msk

Not every (locall path-connected) space $X$ has a universal covering; a 
(further) necessary condition is that $X$ be {\it semi-locally simply connected}.
The idea is that If $p:\wtl{X}\ra X$ is the universal cover, then for every 
point $x\in X$, we have an evenly-covered neighborhood ${\Cal U}$ of $x$.
The inclusion $i:{\Cal U}\ra X$, by definition, lifts to $\wtl{X}$, so
$i_*(\pu({\Cal U},x))\sset p_*(\pu(\wtl{X},\wtl{x}) = \{1\}$, so
$i_*$ is the trivial map. Consequently, every loop in ${\Cal U}$ is 
null-homotopic in $X$. This is semi-local simple connectivity;
every point has a neighborhood whose inclusion-induced homomorphism
is trivial. Not all spaces have this property; the most famous is the 
Hawaiian earrings 
$\displaystyle X=\bigcup_{n}\{x\in {\Bbb R}^2 :  ||x-(1/n,0)||=1/n\}$ .
The point $(0,0)$ has no such neighborhood. 

\bsk


{\bf Building universal coverings:} If a space $X$ is path connected, locally path connected, and
semi-locally simply connected (S-LSC), then it has a universal covering;
we describe a general construction. The idea is that a covering
space should have the path lifting and homotopy
lifting properties, and the universal 
cover can be characterized as the only covering space for 
which {\it only} null-homotopic loops lift to loops. So we build a 
space and a map which \underbar{must} have these properties.
We do this by making a space $\widetilde{X}$ whose
points are (equivalence classes of) $[\gamma]$
based paths $\gamma:(I,0)\ra (X,x_0)$, where two paths are equivalent
if they are homotopic rel endpoints! The projection map is
$p([\gamma])=\gamma(1)$. The S-LSCness is what guarantees that this is a 
covering map; choosing $x\in X$, a path $\gamma_0$ from $x_0$ to $x$,
and a  neighborhood ${\Cal U}$ of $x$ guaranteed by S-LSC, paths from 
$x_0$ to points in ${\Cal U}$ are based equivalent to $\gamma*\gamma_0*\eta$
where $\gamma$ is a based loop at $x_0$ and $\eta$ is a path in ${\Cal U}$.
But by simple connectivity, a path in ${\Cal U}$ is determined up to homotopy
by its endpoints, and so, with $\gamma$ fixed, these paths are in one-to-one
correspondence with ${\Cal U}$. So $p\inv({\Cal U}$ is a disjoint union,
indexed by $\pu(X,x_0)$, of sets in bijective correspondence with ${\Cal U}$.
The appropriate topology on $\widetilde{X}$, essentially given as a basis
by triples $\gamma*,gamma_0,{\Cal U}$ as above, make $p$ a covering map.
Note that the inverse image of 
the basepoint $x_0$ is the equivalence classes of \underbar{loops} at $x_0$,
i.e., $\pu(X,x_0)$. A path $\gamma$ lifts to the path of paths
$[\gamma_t]$, where $\gamma_t(s)=\gamma(ts)$, and so the only 
loop in $X$ which lifts to a loop in $\widetilde{X}$ has endpoint
$[\gamma]=[c_{x_0}]$, i.e., $[\gamma]=1$ in $\pu(X,x_0)$. This
implies that $p_*(\pu(\widetilde{X},[c_{x_0}]))=\{1\}$, so 
$\pu(\widetilde{X},[c_{x_0}])=\{1\}$ . \hhsk However, nobody in their
right minds would go about building $\widetilde{X}$ in this way, in general!
Before describing how to do it ``right'', though, we should perhaps see why
we should want to?

\msk

One reason for the importance of the universal cover is that it gives
us a unified approach to building \underbar{all} connected covering
spaces of $X$. The basis for this is the {\it deck transformation group}
of a covering space $p:\wtl{X}\ra X$; this is the set of all
homeomorphisms $h:\wtl{X}\ra\wtl{X}$ such that $p\circ h = p$.
These homeomorphisms, by definition, permute each of the point inverses
of $p$. In fact, since $h$ can be thought of as a lift of the projection
$p$, by the lifting criterion $h$ is determined by which point in the 
inverse image of the basepoint $x_0$ it takes the basepoint 
$\wtl{x}_0$ of $\wtl{X}$ to. A deck transformation sending
$\wtl{x}_0$ to $\wtl{x}_1$ exists $\Leftrightarrow$
$p_*(\pu(\wtl{X},\wtl{x}_0)=p_*(\pu(\wtl{X},\wtl{x}_1)$
[we need one inclusion to give the map $h$, and the opposite inclusion
to ensure it is a bijection (because its inverse exists)]. These two groups
are in general {\it conjugate}, by the projection of a path from 
$\wtl{x}_0$ to $\wtl{x}_1$; this can be seen by following the change
of basepoint isomorphism down into $G=\pu(X,x_0)$. As we have seen, paths
in $\wtl{X}$ from $\wtl{x}_0$ to $\wtl{x}_1$ are in 1-to-1
correspondence with the cosets of $H=p_*(\pu(\wtl{X},\wtl{x}_0)$ in 
$p_*(\pu({X},{x}_0)$; so deck transformations are in 1-to-1 
correspondence with cosets whose representatives conjugate 
$H$ to itself. The set of such elements in $G$ is called the 
{\it normalizer of $H$ in $G$}, and denoted $N_G(H)$ or simply
$N(H)$. The deck transformation group is therefore
in 1-to-1 correspondence with the group $N(H)/H$ under
$h\mapsto$ the coset represented by the projection of the path from 
$\wtl{x}_0$ to $h(\wtl{x}_0)$. And since $h$ is essentially built
by lifting paths, it follows quickly that this map is a
homomorphism, hence an isomorphism.

\msk

In particular, applying this to the universal covering space
$p:\wtl{X}\ra X$, since in this case $H=\{1\}$, so $N(H)=\pu(X,x_0)$,
its deck transformation group is isomorphic to $\pu(X,x_0)$. 
For example, this gives the quickest possible proof 
that $\pu(S^1)\cong {\Bbb Z}$, since ${\Bbb R}$ is a 
contractible covering space, whose deck transformations
are the translations by integer distances. 
Thus $\pu(X)$ acts on its universal cover as a group of
homeomorphisms. And since this action is {\it simply transitive}
on point inverses [there is exactly one (that's the simple
part) deck transformation carrying any one point in a point 
inverse to any other one (that's the transitive part)], the 
quotient map from $\wtl{X}$ to the orbits of this action \underbar{is}
the projection map $p$. The evenly covered property of $p$ implies
that $X$ does have the quotient topology under this action.

\bsk


Every space it $X$ the quotient of its universal cover (if it has
one!) by its fundamental group $G=\pu(X,x_0)$, realized as the group
of deck transformations. And the quotient map is the covering 
projection. 
So $X\cong \wtl{X}/G$ . In general, a quotient of a 
space $Z$ by a group action $G$ 
need not be 
a covering map; the action must be {\it properly discontinuous}, 
which means that for every point 
 $z\in Z$, there is a neighborhood ${\Cal U}$ of $x$ so that $g\neq 1$ $\Rightarrow$
${\Cal U}\cap g{\Cal U}=\emptyset$ (the group action carries sufficiently
small neighborhoods off of themselves). The evenly covered neighborhoods
provide these for the universal cover. And conversely, the quotient of a space by a 
p.d. group action is a covering space. 

\msk

But! Given $G=\pu(X,x_0)$ and its 
action on a universal cover $\wtl{X}$, we can, instead of quotienting out by $G$,
quotient out by any \underbar{subgroup} $H$ of $G$, to build $X_H=\wtl{X}/H$. 
This is a space with fundamental group $H$, having $\wtl{X}$ as universal covering.
And since the quotient (covering) map $p_G:\wtl{X}\ra X=\wtl{X}/G$ factors through $\wtl{X}/H$,
we get an induced map $p_H:\wtl{X}/H\ra X$, which is a covering map; open sets with
trivial inclusion-induced homomorphism lift homeomorphically to $\wtl{X}$,
hence homeomorphically to $\wtl{X}/H$; taking lifts to each point inverse of $x\in X$
verifies the evenly covering property for $p_H$ . So every subgroup of $G$ is the
fundamental group of a covering of $X$. 

\ssk

We can further refine this to give the {\it Galois correspondence}. Two covering spaces
$p_1:X_1\ra X$ , $p_2:X_2\ra X$ are {\it isomorphic} if there is a homeomorphism
$h:X_1\ra X_2$ with $p_1=p_2\circ h$. Choosing basepoints $x_1,x_2$ mapping to $x_0\in X$,
this implies that, if $h(x_1)=x_2$, then $p_{1*}(\pu(X_1,x_1)) = p_{2*}(h_*(\pu(X_1,x_1))) = 
p_{2*}(\pu(X_2,x_2))$ . On the other hand, our homeomorphism $h$ need not take our
chosen basepoints to one another; if $h(x_1)=x_3$, then $p_{1*}(\pu(X_1,x_1)) = p_{2*}(\pu(X_2,x_3))$.
But  $p_{2*}(\pu(X_2,x_2))$ and $p_{2*}(\pu(X_2,x_3))$ are isomorphic, via a change 
of basepoint isomorphism $\widehat{\eta}$ , where $\eta$ is a path in $X_2$ from $x_2$ to $x_3$.
But such a path projects to $X$ has a loop at $x_0$, and since the change of basepoint isomorphism
is by ``conjugating'' by the path $\eta$, the resulting groups $p_{2*}(\pu(X_2,x_2))$ and $p_{2*}(\pu(X_2,x_3))$
are conjugate, by $p_2\circ \eta$ . So, without reference to basepoints, isomorphic coverings give,
under projection, conjugate subgroups of $\pu(X,x_0)$ . But conversely, given covering spaces
$X_1,X_2$ whose subgroups $p_{1*}(\pu(X_1,x_1))$ and $p_{2*}(\pu(X_2,x_2))$ are conjugate,
lifting a loop $\gamma$ representing the conjugating element to a loop $\wtl{\gamma}$ in
$X_2$ starting at $x_2$ gives, as its terminal endpoint, a point $x_3$ with 
$p_{1*}(\pu(X_1,x_1)) = p_{2*}(\pu(X_2,x_3))$ (since it conjugates back!), and so, by the lifting criterion,
there is an isomorphism $h:(X_1,x_1) \ra (X_2,x_3)$. So conjugate subgroups give isomorphic coverings.
Thus, for a path-connected, locally path-connected, semi-locally simply-connected space $X$,  
the image of the induced homomorphism on \mpu\ 
gives a one-to-one correspondence between 
[isomorphism classes of (connected) coverings of $X$] and 
[conjugacy classes of subgroups of $\pu(X)$].

\msk

So, for example, if you have a group $G$ that you are interested in, you know of a (nice enough) 
space $X$ with $\pu(X)\cong G$, and you know enough about the covering of $X$, then you can
gain information about the subgroup structure of $G$. For example, and in some respects as
motivation for all of this machinery!, a free group $F(\Sigma)$ is \mpu\ of a bouquet of circles $X$. 
Any covering space $\wtl{X}$ of $X$ is a union of vertices and edges, so is a graph.  Collapsing
a maximal tree to a point, $\wtl{X}$ is $\sim$ a bouquet of circles, so has free \mpu . So, every
subgroup of a free group is free. (That is a lot shorter than the original, group-theoretic, proof...) 
A subgroup $H$ of index $n$ in $F(\Sigma)$ corresponds to a $n$-sheeted covering $\wtl{X}$ of $X$. If
$|\Sigma| = m$, then $\wtl{X}$ will have $n$ vertices and $nm$ edges. Collapsing a maximal
tree, having $n-1$ edges to a point, leaves a bouquet of $nm-n+1$ circles, so $H\cong F(nm-n+1)$.
For example, for $m=3$, index $n$ subgroups are free on $2n+1$ generators, so every free subgroup
on 4 generators has infinite index in $F(3)$. Try proving that directly!

\bsk



Every
subgroup of a free group is free, because it is the fundamental group of a
covering of a graph, i.e., of a graph.
A subgroup $H$ of index $n$ in $F(\Sigma)$ corresponds to a $n$-sheeted covering $\wtl{X}$ of $X$. If
$|\Sigma| = m$, then $\wtl{X}$ will have $n$ vertices and $nm$ edges. Collapsing a maximal
tree, having $n-1$ edges to a point, leaves a bouquet of $nm-n+1$ circles, so $H\cong F(nm-n+1)$.
For example, for $m=3$, index $n$ subgroups are free on $2n+1$ generators, so every free subgroup
on 4 generators has infinite index in $F(3)$. Try proving that directly!

\msk

Note that for a graph $\Gamma$ to be a covering of another graph, with $k$ sheets, say,
the number of vertices and edges of $\Gamma$ must both be a mulitple of $k$. This
little observation can be very useful when trying to decide what graphs $\Gamma$ might
cover!

\msk

{\it Kurosh Subgroup Theorem}: If $H < G_1*G_2$ is a subgroup of
a free product, then $H$ is (isomorphic to) a free product of a
collection of conjugates of subgroups of $G_1$ and $G_2$ and a 
free froup. The proof is to build a space by taking 2-complexes
$X_1$ and $X_2$ with $\pu$'s isomorphic to $G_1,G_2$ and join
their basepoints by an arc. The covering space of this space $X$
corresponding to $H$ consists of spaces that cover $X_1,X_2$
(giving, after basepoint considerations, the conjugates)
connected by a collection of arcs (which, suitably interpreted,
gives the free group).

\msk

{\it Residually finite groups}: $G$ is said to be residually finite if for every $g\neq 1$ there is a 
finite group $F$ and a homomorphism $\varphi: G\ra F$ with $\varphi(g)\neq 1$ in $F$. This 
amounts to saying that $g\notin$ the (normal) subgroup $\ker(\varphi)$, which amounts to
saying that a loop corresponding to $g$ does \underbar{not} lift to a loop in the finite-sheeted
covering space corresponding to $\ker(\varphi)$. So residual finiteness of a group can be
verified by building coverings of a space $X$ with $\pu(X)=G$. For example, free groups can be
shown to be residually finite in this way. 

\msk

{\it Ranks of free (sub)groups:} A free group on $n$
generators is isomorphic to a free group on $m$ generators
\lra\ $n=m$; this is because the abelianizations of the two 
groups are ${\Bbb Z}^n,{\Bbb Z}^m$. The (minimal) number of 
generators for a free group is called its {\it rank}.
Given a free group
$G=F(a_1,\ldots a_n)$ and a collection of words $w_1,\ldots w_m\in G$,
we can determine the rank and ndex of the subgroup it $H$ they
generate by building the corresponding cover. The idea is
to start with a bouquet of $m$ circles, each subdivided 
and labelled to spell
out the words $w_i$. Then we repeatedly identify edges sharing
on common vertex if they are labelled precisely the same (same
letter {\it and} same orientation). This process is known
as {\it folding}. One can inductively show that the (obvious)
maps from these graphs to the bouquet of $n$ circles $X_n$ both
have image $H$ under the induced maps on \mpu ; since the map 
for the unfolded graph
factors through the one for the folded graph, the image from the
folded graph can only get smaller, but we can still spell out
the same words as loops in the folded graph, so the image can
also only have gotten bigger! We continue this folding process until there
is no more folding to be done; the resulting graph $X$ is what is 
known (in combinatorics) as a {\it graph covering}; the map to $X_n$
is locally injective. If this map is a covering map, then our subgroup
$H$ has finite index (equal to the degree of the
covering) and we can compute the rank of $H$ (and a basis!) from the 
folded graph. If it is not a covering map, then the map is not locally surjective at
some vertices; if we graft trees onto these vertices, we can extend the map
to an (infinite-sheeted) covering map without changing the homotopy
type of the graph. $H$ therefore has infinite index in $G$, and its
rank can be computed from $H\cong \pu(X)$. 

\bsk


{\bf Postscript: why care about covering spaces?} The preceding discussion
probably makes it clear that covering places play a central role in
(combinatorial) group theory. It also plays a role in embedding 
problems; a common scenario is to have a map $f:Y\ra X$ which is 
injective on \mpu , and we wish to know if we can lift $f$ to a 
finite-sheeted covering so that the lifted map $\widetilde{f}$ is 
homotopic to an embedding. Information that is easier to obtain 
in the case of an embedding can then be passed down to gain information
abut the original map $f$. And covering spaces underlie the 
theory of analytic continuation in complex analysis; starting
with a domain $D\subseteq {\Bbb C}$, what analytic continuation really
builds is an (analytic) function from a covering space of $D$ to ${\Bbb C}$.
For example, the logarithm is really defined as a map from 
the universal cover of ${\Bbb C}\setminus\{0\}$ to ${\Bbb C}$. 
The various ``branches'' of the logarithm refer to which sheet
in this cover you are in.


\vfill
\end







