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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.

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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.


Remarks about definition equivalence.

I encourage you to see for yourself that those definitions are equivalent (in any textbook about differential geometry). The Jacobi's determinant definition for global facts is usually cubersome due to the local nature which makes references to maps and atlases.

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  • $\begingroup$ Thanks a lot for your nice answer. I would still like to understand the proof using the Jacobian since I have been mainly working with it :) $\endgroup$ – Marion Jun 14 '15 at 16:09
  • $\begingroup$ Nevertheless I've done a remark. $\endgroup$ – Fallen Apart Jun 14 '15 at 16:11

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