# If $f:M\rightarrow N$ is $C^{\infty}$, bijective, and everywhere non-singular, then $f$ is a diffeomorphism

I am not able to solve this problem:

Prove that if $f:M\rightarrow N$ is $C^{\infty}$, one-to-one, onto, and everywhere non-singular, then $f$ is a diffeomorphism.

This $f$ is a diffeomorphism $\iff$ $df$ is surjective everywhere, right? Then how to proceed?

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My guess is that he's referring to this particular $f$, in the last line. – Dylan Moreland Jun 28 '12 at 22:46
Ah, good point Dylan. – Zev Chonoles Jun 28 '12 at 22:47
Since $f$ is a bijective smooth map it is a diffeomorphism iff $f^{-1}$ is smooth. This is guaranteed by the non-singularity condition , which means that the differential $T_mf:T_mM\to T_{f(m)}N$ is invertible at all $m\in M$. – Georges Elencwajg Jun 28 '12 at 23:06
Exactly. As Georges points out this is an application of the Inverse Function Theorem and doesn't require you to worry about $df$ being surjective everywhere. – Matt Jun 28 '12 at 23:19
Can I ask whether or not these problems are homework? In the past day you've asked questions ranging from confusion about the definition of differentiable structure on $\mathbb{R}$ to flows on manifolds. Are these review questions for a final or qual or something? Are you self-studying? It would be easier to know how to answer this string of questions if we knew more background. – Matt Jun 28 '12 at 23:31