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I wonder that whether there exists a version of the inverse function theorem for smooth maps from a smooth manifolds with boundary to a smooth manifold without bounary? More precisely, whether the following assertion is true?

Let $M$ be a smooth manifold with boundary $\partial M$ and $N$ be a smooth manifold without boundary whose dimension $d$ is equal to the dimension of $M$. Let $f$ be a smooth map from $M$ to $N$. Assume that there exists a point $x\in \partial M$ such that the rank of $f_M$ at $x$ is $d$ and the rank of $f_{\partial M}$ on $x$ is $d-1$. Then there exists a open neignborhood $U$ of $x$ on $M$ such that $f_{U}$ is a diffeomorphism from $U$ onto $f(U)$.

Any references or comments are well appreciated. Thanks a lot!

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I think the right approach would be to study the double of the manifold $M$, or better of the neighbourhood $U$ (you take the product of $U$ with the set of two points {$0,1$} and then identify the points $(x,0)$ and $)x,1)$ iff $x\in\partial M\cap U$), the double of $f(U)$, and the obvious extension of the function $f$. Impose the standard conditions on the extended $f$ at $x$ and you should get the correct conditions on the original function $f$ –  Piero D'Ancona Jun 13 '11 at 13:07
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2 Answers

up vote 3 down vote accepted

Note that, the question being local you can work in local charts. Also, recall that, by definition of manifold with boundary, and by definition of smooth maps between manifolds with boundary, you can assume w.l.o.g. that $f$ is the restriction to $U:=V\cap H$ of a $C^1$ map $\tilde f$ defined on a nbd $V$ of $x:=0\in\mathbb{R}^d$, where $H$ is a closed half-space. So $\tilde f$ is locally invertible by the usual inverse mapping theorem on open sets of $\mathbb{R}^d$, and such is f by restriction. Note that you don't have to assume anything on the invertibility of $f_{|\partial M}$. The same argument works for Banach manifolds.

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This seems the correct point of view, the problem is reduced to extending a smooth function across a hyperplane, which can be done easily. The assumption on f should be that Df extends continuously to the boundary and its determinant does not vanish at x. –  Piero D'Ancona Jun 13 '11 at 17:34
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actually, the extension is for free by definition of smooth map between $\partial$-manifolds (I mean, the definition I am using, stating that in local charts they are restrictions of smooth maps defined on full nbd's of the model space). –  Pietro Majer Jun 13 '11 at 18:35
    
Thank you all very much for telling me the key of the problem, it is very helpful to me. –  ProbLe Jun 15 '11 at 12:05
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It seems to me that what you ask is a straightforward consequence of the usual local inversion theorem. More precisely: up to taking a suitable chart at $x$, we can assume $M$ is a half space and $\partial M$ a hyperplane, in $\mathbb{R}^d$. You then just have to apply the inversion theorem in a neighborhood of $X$ in $\mathbb{R}^n$ and restrict to a neighborhood of $x$ in $M$. In other words, manifold with boundary are dealt with just the same as manifold without boundary, but with an embedded hypersurface.

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Thank you very much for your clear explanation! I totally agree with you. –  ProbLe Jun 15 '11 at 11:58
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