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Let $g(x): \mathbb{R}^n \to \mathbb{R}^n$ be a function. Suppose it's locally diffeomorphic $\forall x \in \mathbb{R}^n$. I want to know under what constraints is it also bijective (globally). E.g. if I know that $g(x)$ is not periodic, and is composed of algebraic-like terms, e.g. $x, x/|x|$ etc, can I somehow prove that if it's locally diffeomorphic everywhere then it's also globally bijective?

For example the well-known counterexample $F(x,y) = (e^x \cos(y),e^x \sin(y))^T$ is locally diffeomorphic $\forall x,y \in \mathbb{R}$, but $F(x,y+2\pi) = F(x,y)$.

Any guidance is appreciated.

BadAtMath
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  • A diffeomorphism is by definition a $C^{\infty}$ bijective map with a $C^{\infty}$ inverse.Hence, a diffeomorphism is always bijective... Diffeomorphisms are used to identity when two smooth manifolds are topologically the same.

    Maybe you could tell us what you think a diffeomorphism is so we can better address your question.

     – ADA Apr 22 '17 at 18:33
  • I guess you mean $g:\Bbb R^n\to\Bbb R^n$ a function which is locally diffeomorphism around each $x\in\Bbb R^n$ (i.e. there are open $U,V\subseteq\Bbb R^n$ such that $x\in U$ and $g|_U$ is a diffeomorphism to $V$. – Berci Apr 22 '17 at 18:39
  • Yes Berci, that's exactly what I mean. I will update accordingly. – BadAtMath Apr 22 '17 at 18:42

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I think this only happens if $g$ is (globally) diffeomorphic.

Suppose $g : M \to N$ is bijective and locally diffeomorphic. If $x \in M$ then there is an open set $U$ such that $g\vert_U : U \to V$ is a diffeomorphism. Thus $(Dg\vert_U)_x$ and $(D(g\vert_U)^{-1})_{g(x)}$ exist. But

$$ (Dg\vert_U)_x = (Dg)_x \text{ and } (D(g\vert_U)^{-1})_{g(x)} = (D(g^{-1}))_{g(x)}. $$

Thus $g$ is differentiable on $M$ and $g^{-1}$ is differentiable on $N$. Moreover, $g, g^{-1}$ are smooth since their derivatives agree with those of a smooth function. Therefore $g$ is a diffeomorphism.

I don't think there is any easier criterion for this since being injective and being surjective are fundamentally not local properties.

Trevor Gunn
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    I found what I was really looking for. My question was really how to show that a function which is locally diffeomorphic everywhere is also globally diffeomorphic. The answer is given here: https://math.stackexchange.com/questions/41551/global-invertibility-of-a-map-mathbbrn-to-mathbbrn-from-everywhere-loc – BadAtMath Apr 23 '17 at 16:40
  • For the record, this problem is apparently called the Jacobain Conjecture. That is given a polynomial function $F: \mathbb{R}^n \to \mathbb{R}^n$, under what conditions is $F$ invertible. The answer is (I believe from what I've read thus far) is iff $det(F)$ is constant. Thus it's a constraint on the coefficients of all nonlinear terms. – BadAtMath Apr 23 '17 at 18:40