Inverse function theorem for manifolds with boundary as the domain 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 boundary? 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!
 A: 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.
A: 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.
