Let $M$ be a smooth manifold and let $N$ be a manifold with boundary. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}M $.
Solution 1:
By passing to charts near $p$ and $F(p)$, it suffices to prove this in the case $M = \mathbb R^n,N = H^n$. Since $dF_p$ is an isomorphism, by the inverse function theorem (for $\mathbb R^n$, thinking of $H^n$ as a subset of $\mathbb R^n$) we can find an open set $U \ni p$ such that $F(U)$ is open (in $\mathbb R^n$!) and $F : U \to F(U)$ is a diffeomorphism. Since $F(p)\in F(U)$, this rules out $F(p) \in \partial H^n$, because no $\mathbb R^n$-neighbourhood of a boundary point is contained in $H^n$.