# If $\phi: M_1 \to M_2$ a diffeomorphism between diff. manifolds, prove that if $M_2$ is oriented then so is $M_1$

Let $\phi: M_1 \to M_2$ a local diffeomorphism between two differentiable manifolds $M_1,M_2$. I want to prove that if $M_2$ is orientable so is $M_1$.

Attempt: In order a manifold to be orientable the determinant of the Jacobian matrix must be positive. That means, for $M_2$ say, that is I have two mappings $g_{1}: V \subset \mathbb{R}^n \to V_1$ and $g_{2}: V \subset \mathbb{R}^n \to V_2$ then it must hold that $$\text{det}\, (g_1 \circ g_2^{-1}) >0.$$

Now the pullback of this transition map to $M_1$, with transition mappings $f_1: U \to U_1$ and $f_2: U \to U_2$, should be $$f_1 \circ f_2^{-1} = \phi^{-1}(g_1 \circ g_2^{-1}),$$ right? Then, I do not know how to show that the determinant of $f_1 \circ f_2^{-1}$ is also positive which is required in order to show that $M_1$ is orientable.

Equivalent condition for manifold $M$ to be orentable is that there exists non-vanishing form $\Omega$ of degree equal to the dimension of manifold.
So let $M_2$ be orientable of dimension $n$. Then there exists non-vanishing $\Omega_2\in\Omega^n(M_2).$ Let $\phi:M_1\rightarrow M_2$ be a diffeomorphism. Set $\Omega_1:=\phi^*\Omega_2.$ By definition of $\Omega_1\in\Omega^n(M_1).$ Now it is sufficient to show that $\Omega_1$ vanish nowhere. In fact for any point $x\in M_1$ and any lineary independet vecotrs $X_1,\dots,X_n\in T_x M_1$ we have the following: $$\Omega_1(X_1,\dots,X_n)=\Omega_2(\phi_*X_1,\dots\phi_*X_n).$$ Since $\phi$ is an diffeomorphism we get that $\phi_*$ is an isomorphism of vector spaces $T_x M_1$ and $T_{\phi(x)} M_2.$ Hence $\phi_*X_1,\dots\phi_*X_n$ are lineary independent as well. So $$\Omega_1(X_1,\dots,X_n)=\Omega_2(\phi_*X_1,\dots\phi_*X_n)\neq 0.$$ Existance of $\Omega_1$ proves that $M_1$ is orientable.