Vtyjkyuk

Let $M,N$ be smooth manifolds of the same dimension, and suppose $M$ has a non-empty boundary. Let $f:M to N$ be a smooth map, and suppose that $df_p$ is invertible for some $p in partial M$.

Is $f$ a local diffeomorphism onto its image around $p$?

That is, does there exist an open neighbourhood $U$ of $p$, such that $f(U)$ is a smooth submanifold with boundary of $N$, and $f|_U:U to f(U)$ a diffeomorphism?

Is it even true that $f|_U$ must be injective for sufficiently small $U$ around $p$?

Note that to say that $fcolon Mto N$ is a local diffeomorphism means that each point of $M$ has an open neighborhood $U$ such that $f(U)$ is open in $N$ and $f|_U$ is a diffeomorphism from $U$ onto $f(U)$.

When $M$ has a non-empty boundary, a map with invertible differential needs not be open: Take e.g. $M = N=[0, infty)$ and $f(x)=x+1$. However, in this case $f$ is a diffeomorphism onto its image.

A map $f : M to N$ is smooth if for each $p in M$ there exist charts $varphi : U to U' subset mathbbH^n$ around $p$ and $psi : V to V' subset mathbbH^n$ around $f(p)$ such that $f(U) subset V$ and $tildef =psi f varphi^-1$ is smooth. Here $mathbbH^n = mathbbR^n-1 times [0,infty)$. For an interior point $p$ of $M$ we have $U' subset int mathbbH^n = mathbbR^n-1 times (0,infty)$, similarly for $f(p)$.

Concerning smoothness we may regard $tildef$ as a map $U' to mathbbR^n$. If $varphi$ is a boundary chart, then smoothness of $tildef$ means that there exists an open $U'' subset mathbbR^n$ such that $U'' cap mathbbH^n = U'$ and a smooth extension $F : U'' to mathbbR^n$ of $tildef$.

$df_p$ is invertible if and only if $dF_p$ is. In this case $F$ is a local diffeomorphism at $varphi(p)$. Let $W'' subset U''$ be open such that $f(p) in W''$ and $F : W'' to F(W'')$ is a diffeomorphism to an open $F(W'') subset mathbbR^n$. W.l.o.g. assume that $F(W'') cap mathbbH^n subset V'$.

Now define $W' = W'' cap mathbbH^n subset U'$. Then $F(W')$ is manifold with boundary (note that $W'' cap (mathbbR^n-1 times 0 )$ is a submanifold of $W''$).

Next define $U^ast = varphi^-1(W')$ which is an open neighborhood of $p$. We conclude that $f(U^ast) = (psi^-1 F varphi)(U^ast) = psi^-1 F(W')$ is a manifold with boundary. By construction $f : U^ast to f(U^ast)$ is a diffeomorphism.