
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

Do you know yet that universal covers are unique up to homeomorphism?

No, i don't. We don't discuss universal covers yet.

I could solve the problem for n=1. Does someone know how i can generalize, i.e. make the "induction step" (n1)>n?
Thanks a lot.