Product of manifolds & orientability

Solution 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 $\psi_i=\varphi_i\times Id:\mathbb R^{n+1}\to W_i\times\mathbb R$. Then, if needed we can modify each $\psi_i$ by changing the sign of the first variable in $\mathbb R^{n+1}$ 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 $$ J(\psi_j^{-1}\circ\psi_i)=\begin{pmatrix} J(\varphi_j^{-1}\circ\varphi_i)&0\\0&I \end{pmatrix}, $$ hence $J(\varphi_j^{-1}\circ\varphi_i)=\det J(\psi_j^{-1}\circ\psi_i)>0$. Thus the $\varphi_i$'s are a positive atlas of $M$. We are done.

Solution 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 $\phi^{-1}\circ\xi$.