TITLE: Orientable double covers for non-orientable manifolds
QUESTION [8 upvotes]: If I have two non-orientable connected manifolds such that their orientable double covers are homeomorphic, can anything be said about the manifolds? Are they homeomorphic?

REPLY [6 votes]: Here's a slightly easier example than mland's.
Let $X = S^2\times S^4$  Then $X$ is the universal (orientation) cover of both $\mathbb{R}P^2\times S^4$ and $S^2\times \mathbb{R}P^4$.  By the Kunneth formula, $H_2(\mathbb{R}P^2\times S^4) \cong 0$ while $H_2(S^2\times \mathbb{R}P^4) = \mathbb{Z}$.
Somewhat relatedly, (if you want a lower dimensional example), $T^2\times S^2$ covers both $K\times S^2$ and $T^2\times \mathbb{R}P^2$ where $K$ denotes the Klein bottle.  In this case, the fundamental group of the first is nonabelian while the fundamental group of the second is abelian.