TITLE: Product of manifolds & orientability
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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 $\phi^{-1}\circ\xi$.