# Covering spaces

• Dec 6th 2010, 01:17 PM
Sogan
Covering spaces
Hello,

I want to show, that any covering of the rectangle $\displaystyle [0,1]^n$ is a trivial covering $\displaystyle \forall n \in \mathbb{N}$.
I shall use the Homotopy Lifting thm and induction on n.

I tried to solve this problem but i couldn't proceed very much.

My first idea was the following:

n=1:

if we have a covering map p:Y->[0,1] , and a path in [0,1] for instance the identity path Id: [0,1]->[0,1]. for a given point $\displaystyle y \in p^{-1}(0)$
there is a lifting of our path Id*:[0,1]->Y ,s.t. $\displaystyle p \circ Id*=Id$.

we know that $\displaystyle W:=Id*([0,1]) \subset Y$ is a connected subset and p(W)=[0,1].

But why is p a trivial covering, i.e. why $\displaystyle p^{-1}([0,1])$ is a disjoint union of open sets, each homeomorphic to [0,1] under p??? I don't see it.

I hope you can help me.

Regards
• Dec 6th 2010, 01:38 PM
Tinyboss
Do you know yet that universal covers are unique up to homeomorphism?
• Dec 6th 2010, 01:42 PM
Sogan
No, i don't. We don't discuss universal covers yet.
• Dec 7th 2010, 09:09 AM
Sogan
I could solve the problem for n=1. Does someone know how i can generalize, i.e. make the "induction step" (n-1)->n?

Thanks a lot.