Induced orientation by a local diffeomorphism and Orientable manifold I have the excersise:

Let $M_1$ and $M_2$ be differentiable manifolds. Let $\phi:M_1\rightarrow M_2$ be a local diffeomorphism. Prove that if $M_2$ is orientable, then $M_1$ is orientable.
My attempt: Since $\phi$ is a local diffeomorphism we can choose for each $p\in M$ two open sets $U_p\subset M_1$ and $V_p\subset M_2$ containing $p$ and $\phi(p)$ respectively such that $\phi\upharpoonright U_p=\phi_p:U_p\rightarrow V_p$ is a diffeomorphism. Since $M_2$ is oriented there exist an atlas $\{(y_\alpha,W_\alpha)\}$ which induces an orientation.

I want a little hint to construct the atlas that gives an orientation to $M_1$.
Note: I have seen
Given a local diffeomorphism $f: N \to M$ with $M$ orientable, then $N$ is orientable.
this is the exactly same question but I don't understand the answer because I have no idea about differentiable forms.
Thanks for help!

I think I have a elementary way to solve, making calculus
$x_\beta^{-1}\circ x_\alpha = (\phi^{-1}\circ y_\beta)^{-1}\circ(\phi^{-1}\circ y_\alpha) = ((\phi^{-1}\circ y_\alpha)^{-1}\circ(\phi^{-1}\circ y_\beta))^{-1} = (y_\alpha^{-1}\circ y_\beta)^{-1} = y_\beta^{-1}\circ y_\alpha$
So, getting de determinant of the differential of change of coordinates in M1 is the same as in M2, so is just to make sure that all is well defined.
If I am right then a question come to my mind, the other implication could be true?, is M1 is orientable then M2 is orientable?
