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\ctln{\bf Math 971 Algebraic Topology}

\ssk

\ctln{February 17, 2005}

\msk

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. 

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