# Let $M$ be a smooth manifold and let $N$ be a manifold with boundary. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}M$.

Let $M$ be a smooth manifold and let $N$ be a manifold with boundary, both with the same dimension $n$. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}N$.

I am trying to prove this theorem to prove a result about smooth embeddings. Here is how I am thinking the problem could be solved. Assume thet $F(p)$ is a boundary point of $N$. Then there exists a chart $(V,\psi)$ at $F(p)$ such that $\psi(V)$ is an open subset of the upper half space $\mathbb{H}^{n}$. I guess we have to use some fact about $M$ having no boundary and the the fact that $dF_p$ is an isomorphism to show that there is a contradiction, but I cannot figure that out.

• You're almost there - choose another chart near $p$ and then apply the inverse function theorem. – Anthony Carapetis May 11 '17 at 9:46
• I knew that the inverse function theorem might have something to do with it, but I don't know if the inverse function theorem applies for manifolds without boundary. – Eigenfield May 11 '17 at 9:48
• After passing to the charts, all you need is the inverse function theorem for $\mathbb R^n$. – Anthony Carapetis May 11 '17 at 9:51
• So we choose a chart $(U,\varphi)$ at $p$ and a chart $(V,\psi)$ at $F(p)$ (here we assume $F(p)$ is a boundary point which so that $V$ is diffeomorphic to an open subset of the half-space), we get a smooth map $\psi\circ F\circ\varphi^{-1}$ from an open subset of $\mathbb{R}^{n}$ to an open subset of $\mathbb{H}^{n}$. How can you apply the IVF for $\mathbb{H}^{n}$ to that? – Eigenfield May 11 '17 at 10:05
• Possible duplicate of If $dF_p$ is nonsingular, then $F(p)\in$ Int$N$ – rych May 13 '17 at 13:32

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 $H^n$.