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.
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.
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, $\displaystyle \widetilde{X}\leftrightarrow\widetilde{Y}$, show these are homotopic to deck transformations, and then use this to conclude.
Ok, this is the approach so far. If $\displaystyle p:\bar{X}\to X$ is a covering map, then $\displaystyle p\times id_{I}:\bar{X}\times I\to X\times I$ is also a covering map. Moreover,since $\displaystyle \bar{X}$ is simply connected,$\displaystyle \bar{X}\times I$ is also simply connected and locally path-connected.Similarly with $\displaystyle Y$ and $\displaystyle \bar{Y}$ and I guess I can lift the homotopy but I'm not sure how to continue fron this point.
Drexel28, can you help me?