Product of manifolds & orientability

Question

I'm studying orientability of manifolds currently and I'm having trouble to prove the following: $M\times N$ is orientable iff $M$ and $N$ are orientable.

I am able to prove that the product is orientable if components are orientable (chart is $\{(U_\alpha\times V_{\beta},\phi_\alpha\times \psi_\beta):(\alpha,\beta)\in A\times B \}$, and $\det J=\det J_1 \det J_2>0$ by Cauchy-Binet's theorem), but I don't know how to prove the other direction.

So why this holds: if $M\times N$ is orientable, then $M$ and $N$ are orientable?

Thanks in advance.

Answer 1

If $M\times N$ is orientable, any open submanifold is orientable. We can pick an open subset $U\subset N$ diffeomorphic to $\mathbb R^n$, and $M\times U\equiv M\times\mathbb R^n$ is orientable. By induction it is enough to see that if $M\times\mathbb R$ is orientable, then $M$ is orientable. Pick any open cover $\{W_i\}$ of $M$ such that there are diffeomorphisms $\varphi_i:\mathbb R^m\to W_i$. The cover ${\mathcal A}=\{W_i\times\mathbb R\}$ is an atlas with parametrizations $\psi_i=\varphi_i\times Id:\mathbb R^{m+1}\to W_i\times\mathbb R$. Then, if needed we can modify each $\psi_i$ by changing the sign of the first variable in $\mathbb R^{m+1}$ to make it compatible with a fixed orientation in $M\times\mathbb R$. This changes correspondingly the $\varphi_i$. Thus ${\mathcal A}$ is positive and we have $$ J(\psi_j^{-1}\circ\psi_i)=\begin{pmatrix} J(\varphi_j^{-1}\circ\varphi_i)&0\\0&1 \end{pmatrix}, $$ hence $\det J(\varphi_j^{-1}\circ\varphi_i)=\det J(\psi_j^{-1}\circ\psi_i)>0$. Thus the $\varphi_i$'s are a positive atlas of $M$. We are done.

Answer 2

Take an atlas $K$ of $M\times N$. Then there are parametrizations of $K$ in the form $\phi\times\psi$ where $\phi$ is a parametrization of $M$ and $\psi$ of $N$.

Statement: $A=\{\phi \text{ parametrisation of }M\text{ such that }\phi\times\psi\in K\}$ is a coherent atlas of $M$.

Indeed, if $\phi,\xi\in A$ then $\phi\times\psi,\xi\times\psi\in K$. Therefore, $\operatorname{det} J\left((\phi\times\psi)^{-1}\circ(\xi\times\psi)\right)>0$.

But, $(\phi\times\psi)^{-1}\circ(\xi\times\psi)=(\phi^{-1}\circ\xi,\operatorname{id})$.

Then the jacobian is also positive for $\phi^{-1}\circ\xi$.

Answer 3

This is my attempt, please feel free to correct me if anything's wrong.

Suppose $M \times R^n$ orientable and consider the map $\pi: M \times R^n \to R^n, (m, v) \to v$ which is a submersion. As such it has surjective differential and thus we have that $0$ is a regular value for such map $\pi$.

This implies that $\pi^{-1}(0)$ must be a submanifold. Furthermore, because $M \times R^n$ is orientable, $\pi^{-1}(0)$ must be orientable $\color{red}{*}$. Now observe that $\pi^{-1}(0)$ is simply $M \times \{0\} \cong M$. As such we then must have that $M$ is orientable.

--- Edit: ---

$\color{red}{*}$ As suggested by Ted Shifrin, I shall explain that $\pi^{-1}(0)$ is not orientable by the fact that it is a submanifold of $M \times R^n$. It is orientable because it is the preimage of a regular value of a map between orientable manifolds.

For reference look at Guillemin-Pollack, Differential topology at pages 100 and 101