TITLE: $M \times N$ orientable if and only if $M, N$ orientable
QUESTION [7 upvotes]: For two manifolds $M$ and $N$ I'm trying to prove that $M \times N$ is orientable if and only if $M$ and $N$ are orientable.
My attempt so far:
$\impliedby)$ Assume $M, N$ are orientable. Then $\widetilde{M}$ and $\widetilde{N}$ the orientation double covers of $M$ and $N$ are disconnected. But $\widetilde{M} \times \widetilde{N}$ is the orientation double cover for $M \times N$. Since $\widetilde{M}, \widetilde{N}$ are disconnected certainly $\widetilde{M} \times \widetilde{N}$ is disconnected. Therefore, $M \times N$ is orientable.
$\implies)$ Proceed by contradiction. Assume $M \times N$ is orientable and WLOG $M$ is nonorientable. Then the orientation double cover $\widetilde{M\times N}$ is disconnected.
Now note that $\widetilde{M}$ is connected since $M$ is nonorientable. So $\widetilde{M} \times \widetilde{N}$ is connected since the product of a connected space and a disconnected space is connected. But $\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}$, so $\widetilde{M\times N}$ is connected. But this is a contradiction, therefore it must be that $M$ is orientable.
This completes our proof. 
The validity of this proof hinges on two assumptions that I'm unsure about:
1. $\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}$
2. The product of a connected space and a disconnected space is connected.
Are these two facts true? Is my proof valid? How do I go about proving these two necessary assumptions if they are indeed true? If I'm totally going the wrong way could someone send me in the proper direction?
Thanks

REPLY [7 votes]: I'll use that a manifold $X$ is orientable if and only its first Stiefel-Whitney class vanishes: $$w_1(M)\stackrel {\text {def}}{=} w_1(TM)=0\in H^1(X,\mathbb Z/2)$$  Now back to our problem:
If $p_M:M\times N\to M$ and $p_N:M\times N\to N$ are the projections we have $T(M\times N)=p_M^*TM \oplus p_N^*TN$ which allows us to write $$w_1(M\times N)=(w_1M ,w_1M)\in H^1(M\times N,\mathbb Z/2)=H^1(M,\mathbb Z/2)\oplus H^1(N,\mathbb Z/2)$$ Thus $$w_1(M\times N)=0\iff w_1(M)=0 \operatorname {and} w_1(N)=0$$ which proves that $$  M\times N \operatorname {is orientable} \iff  M \operatorname  {and} N \operatorname  {both are} .$$