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My question is really simple, I would like to prove a local diffeomorphism $f:U\to \mathbb R^m$ is a global diffeomorphism over its image $V=f(U)$ if and only if it's an injective function.

The $\Rightarrow$ part is easy, I've already proved a local diffeomorphism is an open function, but I don't know how to use this fact to prove the converse.

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    It's bijective so it has an inverse. The inverse is smooth locally near any $p$ by the inverse function theorem, and thus globally. – 2017-01-10
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    @juanarroyo could you give me what your version of the inverse function theorem is? the version I know is http://www.math.ucsd.edu/~nwallach/inverse[1].pdf – 2017-01-10
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    That's the one. Using that you can define a smooth inverse in a neighborhood of each point in $f(U)$. A map that is smooth in a nbhd of each point is smooth. – 2017-01-10

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Given $x\in V$, let $W\subseteq U$ be an open neighborhood of $f^{-1}(x)$ on which $f$ is a diffeomorphism onto its image. That is, $f$ restricts to a diffeomorphism $f|_W:W\to f(W)$. Then the restriction of $f^{-1}$ to $f(W)$ is smooth, since it is the inverse of the diffeomorphism $f|_W$. But $f(W)$ is an open neighborhood of $x$ and smoothness is a local property, so $f^{-1}$ is smooth at $x$. Since $x\in V$ was arbitrary, this proves $f^{-1}$ is smooth.

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    Thank you for your answer. Did you use the fact $f$ is an open function? – 2017-01-10
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    I used the fact that $f(W)$ is open...this follows from the fact that $f$ is open, but also is probably just directly part of your definition of "local diffeomorphism". And you need to know that $V$ is open for it to even make sense to talk about $f:U\to V$ being a diffeomorphism. – 2017-01-10