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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!

Community
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EQJ
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    There is no need to construct an atlas. You have given two manifolds with two differentiable structures. And more! $\phi$ is a diffeomorphism, that's enaugh. You know that $M_2$ is orientable. That means the tangent-space at every point for $M_2$ has on oriented basis of say n-linear independent elements. The tangent map $T\phi$ has an inverse, because its an isomorphism, because $\phi$ is a diffeomorphism. $T\phi$ plays the same role like iso's in linear algebra. $T\phi$ therefore has an inverse, which maps one basis to another. That's the key. – Frieder Mar 13 '15 at 02:52
  • What is your definition of orientable? – Travis Willse Mar 13 '15 at 02:53
  • A manifold $M$ is orientable iff it has ${(x_\alpha, U_\alpha)}$ differentiable structure, such that the differential of the change of coordinates $x_\alpha^{-1}\circ x_{\beta}$ has positive determinant. – EQJ Mar 13 '15 at 02:59
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    @Travis but $\phi$ is not an diffeomorphis is just a local diffeomorphism. – EQJ Mar 13 '15 at 03:03
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    @Travis: The assumption that $\phi$ is a local diffeomorphism is given. To show is that the determinant for inverse of $T_p\phi $ does not change sign. – Frieder Mar 13 '15 at 03:28
  • @YTS I happened to have a question similar to yours, and I finished it. You can see my post: https://math.stackexchange.com/questions/4390675/the-pullback-orientation-on-m-induced-by-a-local-diffeomorphism-fm-to-n – Boar Feb 27 '22 at 10:42

2 Answers2

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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?

Fer Nando
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Hint: Show that you can pick an atlas $(V_{\beta})$ of $M_2$ so that for each $\beta$ the restriction $\phi|_U \to V_{\beta}$ is a diffeomorphism for each component $U$ of $\phi^{-1}(V_{\beta})$.

Travis Willse
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  • I define $W_{\alpha p}=y(W_\alpha)\cap V_p$ then the restriction $y_\alpha \upharpoonright y_\alpha^{-1}W_{\alpha p}=y_{\alpha p}$ gives a new atlas ${(y_{\alpha p}, y_\alpha^{-1}(W_{\alpha p}))}$. Is this atlas you mean? (I am using the notation of my post) – EQJ Mar 13 '15 at 03:30
  • Sorry, there was a typo: The last symbol in the answer should be $\phi^{-1}(V_{\beta})$, not $V_{\beta}$ itself (otherwise, the restriction $\phi|_U$ doesn't even make sense). – Travis Willse Mar 13 '15 at 03:45
  • Anyway, the atlas I have in mind is just the one comprised of the charts $(U, \psi_{\beta} \circ \phi|_U)$. – Travis Willse Mar 13 '15 at 03:45
  • @TravisWillse I understand that you can pull back an orientation locally on $M_1$ but the part I miss is how do you expand it to the whole $M_1$. I would appreciate if you explain this part a bit. – Arvin Rasoulzadeh Jun 12 '20 at 08:04