# A local diffeomorphism can map a boundary point to an interior point

I would like to find an example for a local diffeomorphism between smooth manifolds with boundary which maps some boundary point to an interior point.

I am quite sure such an example exists.

• I'm thought boundary points get mapped to boundary points, else why would they be called boundary points? For example, the charts are local diffeomorphisms to the upper half plane of some $\Bbb R^n$, when you're near the boundary that is, and they should map boundary points to boundary points no? – snulty Oct 24 '16 at 9:54
• @Asaf: Given that the term local diffeomorphism has a standard meaning, and the mappings I gave are not local diffeomorphisms according to this definition, perhaps it's best for posterity if either 1. You unselect my answer so I can delete it, or 2. The question (and my answer) be edited to use another term, perhaps "diffeomorphism onto its image" (assuming that's the concept you wanted to capture). – Andrew D. Hwang Oct 25 '16 at 0:27
• The closed half-line $M = [0, \infty)$ is a manifold with boundary. Though not a local diffeomorphism, the mapping $f(x) = x + 1$ is a diffeomorphism onto its image, sending the boundary to the interior. This example generalizes in obvious ways to arbitrary dimension. For compact examples of the same type, scale the unit $n$-ball: $f(\mathbf{x}) = \lambda \mathbf{x}$, $0 < |\lambda| < 1$. – Andrew D. Hwang Oct 25 '16 at 11:14
• @Asaf: Thank you. I've converted my answer to a comment (reworded to accommodate your intent, and standard terminology), which preserves the content for posterity). I think it's fine to leave your question as-is. :) – Andrew D. Hwang Oct 25 '16 at 11:15

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 $f(U)$.

Theorem. If $M$ and $N$ are smooth manifolds with boundary and $f\colon M\to N$ is a local diffeomorphism, then $f(\partial M)\subseteq \partial N$.

Proof. Assume for contradiction that there is a point $p\in \partial M$ such that $f(p)\in \operatorname{Int} M$. Let $U$ and $V$ be open neighborhoods of $p$ and $f(p)$ respectively, such that $f|_U\colon U\to V$ is a diffeomorphism.

Because $p\in\partial M$, there is a vector $v\in T_pM$ that is not tangent to $\partial M$. This means, in particular, that there is no smooth curve $\gamma\colon (-\varepsilon,\varepsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. Let $w = df_p(v) \in T_{f(p)}N$. Because $f(p)$ is an interior point of $N$, there is a smooth curve $\gamma\colon (-\varepsilon,\varepsilon)\to V$ such that $\gamma(0) = f(p)$ and $\gamma'(0) = w$. Then $\tilde\gamma = f^{-1}\circ \gamma$ is a smooth curve in $M$ such that $\tilde\gamma(0)=p$ and $\tilde\gamma'(0) = v$, which is a contradiction. $\square$

When $M$ and $N$ both have empty boundaries (and the same dimension), it's easy to show using the inverse function theorem that a smooth map $f\colon M\to N$ is a local diffeomorphism if and only if it is an immersion (a map whose differential is injective at each point). But in the case of nonempty boundaries, this isn't true. There are plenty of examples (such as those described by Andrew Hwang) of smooth immersions that take boundary points to interior points, but they're not local diffeomorphisms.

Let $M$ a differentiable manifold with boundary. Let $S$ a submanidold with boundary such that $\partial S\neq \partial M$, and let $i:S\rightarrow M$ the immersion of $S$ in $M$. Then some boundary points go to internal points of $M$.

• Actually the inclusion $i$ won't be a local diffeomorphism, via the standard definition. (See Jack Lee's answer above). – Asaf Shachar Oct 25 '16 at 12:59