Universal Cover

A covering map is a universal cover if is Simply connected.

Theorem

Let be path connected (1), Locally Path Connected (2), and Semi-locally Simply Connected (3). Then there exists a covering with simply connected.

Proof (NONEXAMINABLE)

Fix . Define aas a set:

with Need topology on st

1. cts
2. covering map
3. simply connected

Step 0

Topology on , revisited.

Claim 0

basis for rthe topology on

Proof

Assume open and . Need st

• is semilocally simply connected so there is some open with st trivial
• locally path-connected there is some open, path-connected st
• Have inducing

Second map has image 0, so composition has to be trivial. Still need: trivial for all . Given a pick a path in . Then the following commutes:

\usepackage{tikz-cd}
 
\begin{document}
\begin{tikzcd}
\pi_1(U,x) \arrow[r,"0"] \arrow[d,"Y_{\#}"] & \pi_1(X,x)\arrow[d,"Y_{\#}"]\\
\pi_1(U,y)\arrow[r] & \pi_1(X,y)
\end{tikzcd}
\end{document}

Step 1

Topology on . For , and , define:

Let be the sellection of all these.

Claim 1

is a basis for a topology on (which we’ll then use)

Proof

Assume where Need: st Fix st and Now choosing and for above works.

Step 2

Check this topology has the desired properties.

Claim 2.1

with is cts

Proof

Suppose , . Then and so open. 2. is a covering map?

Claim 2.2.1

homeomorphism

Proof

cts, so is open map.

• surjective? is path-connected so for all there is a path in Now and
• injective? Assume st (ie same image under ) Have paths in starting at and ending at st and Now hence

Claim 2.2.2

Suppose . Then these are equal or disjoint (so is partitioned into such sets)

Proof

Say for paths in Suppose Then for path in Now path in , so Upshot: Similarly swapping and

Claim 2.3

is simply connected

Proof

is path connected. Suppose is a loop based at , st its lift based at , say is a loop. Observe: Suppose any path starting at , then lift to starting at is:

So

So As is a covering map, is injective, so