# Is a map with invertible differential that maps boundary to boundary a local diffeomorphism?

Let $M,N$ be smooth manifolds with boundary (of the same dimension). Let $f:M \to N$ be a smooth map satisfying

$(1) \, \,f(\partial M)=\partial N,f(\operatorname{Int} M)=\operatorname{Int} N$.

$(2) \, \, df_p$ is invertible for every $p \in M$.

Is it true that $f$ is a local homeomorphism?

I suspect $f$ must in fact be a local diffeomorphism**:

It is certainly a local diffeomorphism around each point in $\operatorname{Int} M$ (By the inverse function theorem).

The question is what happens at the boundary $\partial M$:

Clearly $f|_{\partial M}:\partial M \to \partial N$ is a local diffeomorphism (again by the inverse function for manifolds without boundary).

However, this does not immediately imply that $f$ (when considered as a map $M \to N$) is a local diffeomorphism.

** 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 $f(U)$.

I think $f$ indeed must be a local diffeomorphism. Let $p \in \partial M$. By assumption $f(p) \in \partial N$.Since the question is local, we can assume $M=N=\mathbb{H}^n, p=f(p)=0 \in \partial\mathbb{H}^n$.
Since $f:\mathbb{H}^n \to \mathbb{H}^n$ is smooth then by definition it is smooth as a map $\mathbb{H}^n \to \mathbb{R}^n$. Hence (again straight from the definition) there exist $r >0$ and a smooth function $\tilde f:B_r(p) \to \mathbb{R}^n$ (where $B_r(p) \subseteq \mathbb{R}^n$), such that $\tilde f|_{B_r(p) \cap \mathbb{H}^n}=f|_{B_r(p) \cap \mathbb{H}^n}$.
Since $df_p$ is invertible (by assumption), and $d\tilde f_p = df_p$, $d \tilde f_p$ is also invertible. Hence, by the inverse function theorem (for manifolds without boundary) $\tilde f$ is a local diffeomorphism in some neighbourhood of $p$. From here, it is easy to see $f$ is indeed a local diffeomorphism around $p$. ($f$ maps an open subset of $\mathbb{H}^n$ to an open subset of of $\mathbb{H}^n$, since $\tilde f$ does so, and they coincide on the intersection with $\mathbb{H}^n$. Also, when restricting to small enough neighbourhood of $p$, $f$ has a smooth inverse, which coincides with the inverse of $f$ when restricting the domain).