# Homotopy equivalence and covering spaces

• Dec 2nd 2011, 03:30 AM
kierkegaard
Homotopy equivalence and covering spaces
Let Cov X and Cov Y be simply connected covering spaces of the path connected and locally path connected spaces X and Y respectively. Show that if X is homotopy equivalent to Y, then Cov X is homotopy equivalent to Cov Y.
• Dec 2nd 2011, 06:43 PM
Drexel28
Re: Homotopy equivalence and covering spaces
Quote:

Originally Posted by kierkegaard
Let Cov X and Cov Y be simply connected covering spaces of the path connected and locally path connected spaces X and Y respectively. Show that if X is homotopy equivalent to Y, then Cov X is homotopy equivalent to Cov Y.

This problem is extremely ugly, at least the way I did it when I did it out of Hatcher. Do you at least have any leads?
• Dec 3rd 2011, 05:16 AM
kierkegaard
Re: Homotopy equivalence and covering spaces
Not so far, except the functions that you have by definition of homotopy equivalence. For example, I don't know why we need that the spaces be locally path connected.
• Dec 3rd 2011, 03:02 PM
Drexel28
Re: Homotopy equivalence and covering spaces
Quote:

Originally Posted by kierkegaard
Not so far, except the functions that you have by definition of homotopy equivalence. For example, I don't know why we need that the spaces be locally path connected.

I can try to find the solution for you, but roughly you just write out what's going on, use the universal lifting property to get maps, $\widetilde{X}\leftrightarrow\widetilde{Y}$, show these are homotopic to deck transformations, and then use this to conclude.
• Dec 3rd 2011, 05:25 PM
kierkegaard
Re: Homotopy equivalence and covering spaces
Thanks, now I'm going to put together to solve the exercise. If I cannot find the solution, I'm going to reply.
• Dec 4th 2011, 06:10 AM
kierkegaard
Re: Homotopy equivalence and covering spaces
Ok, this is the approach so far. If $p:\bar{X}\to X$ is a covering map, then $p\times id_{I}:\bar{X}\times I\to X\times I$ is also a covering map. Moreover,since $\bar{X}$ is simply connected, $\bar{X}\times I$ is also simply connected and locally path-connected.Similarly with $Y$ and $\bar{Y}$ and I guess I can lift the homotopy but I'm not sure how to continue fron this point.

Drexel28, can you help me?