Is a map with invertible differential a diffeomorphism onto its image near a boundary point? 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 $f\colon M\to N$ 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: A map $f : M \to N$ is smooth if for each $p \in M$ there exist charts $\varphi : U \to U' \subset \mathbb{H}^n$ around $p$ and $\psi : V \to V' \subset \mathbb{H}^n$ around $f(p)$ such that $f(U) \subset V$ and $\tilde{f} =\psi f \varphi^{-1}$ is smooth. Here $\mathbb{H}^n = \mathbb{R}^{n-1} \times [0,\infty)$. For an interior point $p$ of $M$ we have $U' \subset int \mathbb{H}^n = \mathbb{R}^{n-1} \times (0,\infty)$, similarly for $f(p)$.
Concerning smoothness we may regard $\tilde{f}$ as a map $U' \to \mathbb{R}^n$. If $\varphi$ is a boundary chart, then smoothness of $\tilde{f}$ means that there exists an open $U'' \subset \mathbb{R}^n$ such that $U'' \cap \mathbb{H}^n = U'$ and a smooth extension $F : U'' \to \mathbb{R}^n$ of $\tilde{f}$.
$df_p$ 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 $F(W'') \subset \mathbb{R}^n$. W.l.o.g. assume that $F(W'') \cap  \mathbb{H}^n \subset V'$.
Now define $W' = W'' \cap  \mathbb{H}^n \subset U'$. Then $F(W')$ is manifold with boundary (note that $W'' \cap (\mathbb{R}^{n-1} \times \{ 0 \})$ 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.
