Diffeomorphism mapping interior point to boundary point of manifolds with boundary Why can a diffeomorphism $F: M \to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?
Let $(U, \varphi)$ be a smooth chart for $M$, $(V, \psi)$ a smooth chart for $N$.
I believe it is because it would cause $(\psi \circ F \circ \varphi^{-1}): \varphi(U) \to \psi(V)$ to be a diffeomorphism between an open set in $\mathbb{R}^n$ and an open set in $\mathbb{H}^n$ such that its intersection with $\partial \mathbb{H}^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated
 A: A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods. 
A: The usual proof that neighborhoods that intersect $\partial\mathbb{H}$ are distinct from neighborhoods uses the local homology groups $H^\ast(M,M\setminus\{p\})$ for different choices of $p$.  It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard.  So how about this?
Suppose $f:M\rightarrow N$ is a diffeomorphism and that $p\in \operatorname{int} M$ and $f(p)\in \partial N$.  Then $d_p f:T_p M\rightarrow T_{f(p)} N$ must be an isomorphism.  
But I claim that is not.  To see this, pick a $w\in T_{f(p)}$ which points outside of $N$.  I claim there is no $v\in T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $\gamma_v:(-\epsilon,\epsilon)\rightarrow M$ with $\gamma_v(0) = p$ and $\gamma_v'(0) = v$.  Then $f\circ\gamma$ is a smooth curve in $N$.  By picking a chart near $f(p)$, we can view all this as happening in $\mathbb{R}^n$.
So, without loss of generality, we have a curve $\gamma$ for which $\gamma(0) = \vec{0}\in \mathbb{R}^n$, $\gamma\subseteq H =\{(x_1,...,x_n)\in \mathbb{R}^n: x_n\leq 0\}$, but $\gamma'(0) \notin H$.  That is, the $n$th coordinate of $\gamma'(0)$ is positive.
Since $\gamma$ is smooth, $\gamma'$ is continuous, so the $n$-th coordinate of $\gamma'(t)$ is positive in a neighborhood of $0$.  But now the mean value theorem applied to the $n$th coordinate of $\gamma$ implies that $\gamma(t)\notin H$ for small $t$ with $t>0$.
A: That a diffeomorphism cannot take interior points to boundary points follows immediately from Invariance of domain, which is a theorem stating that if $U \subseteq \mathbb{R}^n$ is an open subset of $\mathbb{R}^n$ and $f : U \to \mathbb{R}^n$ is a continuous injection then $f : U \to f(U)$ is a homeomorphism and $f(U)$ is an open subset of $\mathbb{R}^n$.
Suppose $F : M \to N$ is a diffeomorphism, $p \in \operatorname{Int} M$, but $q := F(p) \in \partial N$.
Passing to coordinates around $p$ and $q := F(p)$ results a homeomorphism (diffeomorphism even) $\widehat{F} : U \to V$ where $U$ is an open subset of $\mathbb{R}^n$ containing $p$'s coordinate representation $\hat{p}$, and $V$ is an open subset of $\mathbb{H} := \left\{ \left(x_1, \ldots, x_n\right) : x_n \geq 0 \right\}$ containing $q$'s coordinate representation $\hat{q} := \widehat{F}\left(\hat{p}\right)$, where in addition $\hat{q} \in \partial \mathbb{H} := \left\{ \left(x_1, \ldots, x_n\right) : x_n = 0 \right\} = \mathbb{R}^{n - 1} \times \{ 0 \}$ because $q \in \partial N$.
Because $\mathbb{H} \subseteq \mathbb{R}^n$, the map $\widehat{F}$ is (also) a continuous injection $\widehat{F} : U \to \mathbb{R}^n$ so that invariance of domain implies that its image $V = \widehat{F}(U)$ is an open subset of $\mathbb{R}^n$. There consequently exists a subset $B \subseteq V$ containing $\hat{q}$ that is open in $\mathbb{R}^n$ ($B$ could, for instance, be taken to be some open ball in $\mathbb{R}^n$). Because $\hat{q} \in \mathbb{R}^{n - 1} \times \{ 0 \}$, the open (in $\mathbb{R}^n$) set $B$ is not a subset of $\mathbb{H} := \left\{ \left(x_1, \ldots, x_n\right) : x_n \geq 0 \right\}$ and so neither is $\widehat{F}$'s image $V = \widehat{F}(U)$ a subset of $\mathbb{H}$ (because it contains $B$). This is a contradiction. $\blacksquare$
