Orientation of $X \times Y$

Question

Suppose that $X$ is not orientable. How can I show that $X \times Y$ is never orientable, no matter what manifold $Y$ may be?

I've tried supposing that $X \times Y$ is orientable, then using that $\pi\colon X\times Y\to X$ the projection on first coordinate is orientation preserving map, $X$ would be orientable. But something just not sounds right.

Answer 1

Here is a proof in the smooth setting. One of the (many) equivalent definitions of orientability is the following: Let $M$ be a smooth connected manifold, $p:\tilde M\to M$ is its universal cover and $G$ the group of covering transformations. Note that $\tilde M$ is always orientable. Then $M$ is orientable if and only if $G$ preserves orientation on $\tilde M$.

Suppose now that $M$ is not orientable and let $g\in G$ be an orientation-reversing element, i.e., with respect to the orientation atlas on $\tilde M$, the Jacobian determinant $J_g$ is negative. Now, consider another connected smooth manifold $N$ and its universal cover $\tilde N$ and the group of covering transformations $H$. Then the universal cover of $M\times N$ is $\tilde M \times \tilde N$ and the group of covering transformations is $G\times H$ where every element of $G$ acts on $\tilde N$ via the identity map. Computing the Jacobian determinant of $$ (g,Id): \tilde M \times \tilde N\to \tilde M \times \tilde N$$ we obtain that it equals $J_g$, i.e., is negative. Thus, $M\times N$ is non-orientable.

You can now finish the proof in case $M$ or $N$ is not connected.

The same conclusion holds for topological manifolds, but the proof is a bit more involved (you need to know Kunneth formula and homological interpretation of orientability).