Thread: Covering spaces

Hello,

I want to show, that any covering of the rectangle is a trivial covering .
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
there is a lifting of our path Id*:[0,1]->Y ,s.t. .

we know that is a connected subset and p(W)=[0,1].

But why is p a trivial covering, i.e. why 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" (n-1)->n?

Thanks a lot.