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I have an open subset $A$ of $\mathbb{R}^k$ and a subset $B$ of $\mathbb{R}^n$, $n>k$, that are homeomorphic and $f:A\longrightarrow B$ is a smooth homeomorphism between two sets. I'm wondering if you know any results as to what additional properties of $f$ (other than its inverse being smooth) would ensure that it is a diffeomorphism.

Such result would be in the spirit of "a continuous bijection is a homeomorphism if and only if it is open (closed)" which lets one prove a function is a homeomorphism without directly proving that its inverse is continuous.

My end goal is to prove that my concrete function $f$ has its Jacobi determinant positive everywhere on $A$ or at least that Jacobian is zero only at isolated points. So if you know any results that would let me reason about the set on which the Jacobian vanishes using the facts (smooth homeomorphism) that I stated, I would very much appreciate it.

Tom
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  • So one thing to consider is that any result in this direction would have to deal with examples like $A=\mathbb{R}$, $B = {(x,0)\in\mathbb{R}^2\ |\ x\in\mathbb{R}}$, and $f:x\mapsto (x^3,0)$. – Neal Jun 19 '16 at 19:28

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Assuming that $B$ is a submanifold of $\mathbb R^n$, $f$ is a diffeomorphism if and only if it is an immersion; i.e. if the differential is injective everywhere. (Without this first assumption you need to think about what you mean by diffeomorphism - what is the smooth structure on $B$ meant to be?)

I doubt this is much help to you, though, since this assumption is exactly equivalent to the Jacobian determinant being non-zero everywhere. Perhaps Sard's Theorem is what you're looking for - it tells you that the image of the set of critical points of $f$ has measure zero in $B$. I'm guessing that $f$ being a smooth homeomorphism is enough for this to imply that the critical set has measure zero in $A$, though all I know for sure offhand is that it is closed with empty interior.

Asking for isolated critical points seems somewhat difficult - whenever we have an example with an isolated critical point we can turn it into a higher-dimensional critical submanifold by taking a product, e.g. extending $f(x)=(x^3,0)$ to $f(x,y) = (x^3,y,0)$. Whatever your assumption is would have to somehow rule this kind of thing out.

Anthony Carapetis
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