Orientability of a product of smooth manifolds implies orientability of each factor

Question

I've been learning a bit about orientability on smooth manifolds. I'm having torubles with this exercise:

Given two smooth manifolds $M$ and $N$, show that the product manifold $M \times N$ is orientable if and only if $M$ and $N$ are orientable.

Using the orientability of $M$ and $N$ one can obtain an oriented atlas for each manifold and the construct an oriented atlas for the product formed by the product charts (i.e. charts of the form $(U \times V, \phi \times \psi)$ with $(U, \phi)$ a chart of $M$ and $(V, \psi)$ a chart of $N$).

I'm stuck on proving the converse. Given an oriented atlas for $M \times N$ one can obtain another oriented atlas formed by charts with basic open sets as domains. But from this atlas I don't know how to extract an atlas for $M$.

I was given the following hint: If $M \times N$ is orientable then $M \times \mathbb{R}^n$ is orientable where $n$ is the dimension of $N$. I don't know either how to prove it nor how to use it.

Answer 1

A manifold is orientable iff it admits a volume form (a nowhere vanishing top degree alternating differential form). It follows that an open submanifold of an orientable manifold is orientable. Take an open subset $U\subset N$ diffeomorphic to $\mathbb R^n$. Then $M\times U$ is an open submanfold of $M\times N$, hence orientable. Now an orientation on $U$ and $M\times U$ defines an orientation on $M$.

Answer 2

$(\Rightarrow)$ If $M^m\times N^n$ is orientable, there is a volume form $\eta\in\Omega^{n+m}(M\times N)$. For any fixed point $q\in N$, take a basis $\{w_1,...,w_n\}$ for $T_qN$. Now define $\omega\in\Omega^m(M)$ by: $$\omega(X_1,...,X_m):=\eta_{(\cdot,q)}\left(X_1,...,X_m,w_1,...,w_n\right)$$

We prove that $\omega$ is a volume form. For a fixed $p\in M$, take a basis $\{v_1,...,v_m\}$ for $T_pM$. Using the identification $T_{(p,q)}M\times N\equiv T_pM\oplus T_qN$, then $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_{(p,q)}M\times N$. Since $\eta$ is a volume form, $\omega_p(v_1,...,v_m)=\eta_{(p,q)}(v_1,...,v_m,w_1,...,w_n)\neq 0$, which means that $\omega$ is a volume form, so $M$ is orientable. By a similar argument, $N$ is orientable.$_\blacksquare$

$(\Leftarrow)$ If $M,N$ are orientable, there are volume forms $\omega\in\Omega^m(M), \sigma\in\Omega^n(N)$. For the natural projections $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$, define: $$\eta:=\pi_M^*(\omega)\wedge \pi_N^*(\sigma)\in\Omega^{n+m}(M\times N)$$

We prove that $\eta$ is a volume form. For a fixed $x=(p,q)\in M\times N$, take basis $\{v_1,...,v_n\}$ for $T_pM$ and $\{w_1,...,w_n\}$ for $T_qN$. Because $\omega$ and $\sigma$ are volume forms, we have $\omega_p(v_1,...,v_m)\neq 0$ and $\sigma_q(w_1,...,w_n)\neq 0$. Since $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_x(M\times N)$, it's enough to check that $\eta_x(v_1,...,v_m,w_1,...,w_n)\neq 0$. Indeed, noticing that $(d\pi_M)_x(v_i)=v_i$, $(d\pi_M)_x(w_j)=0$, $(d\pi_N)_x(v_i)=0$ and $(d\pi_N)_x(w_j)=w_j$, we have: $$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$