Why are boundary points preserved by smooth maps?

Question

I'm reading some lecture notes on representation of surfaces groups.

Author defines a surface with boundary to be:

A surface with boundary is a metrisable space S together with a smooth atlas of charts with boundary.

To define the boundary of the "surface with boundary" object, we use the auxiliary lemma:

Let S be a surface with boundary and let $(U_1, \phi_1)$ $(U_2, \phi_2)$ be charts with boundary in an atlas of S.

Suppose $x \in \partial U_1 \cap U_2$ . in other words x is a boundary point of $U_1$ . Then $x \in \partial U_2$ .

The proof for this lemma is omitted, and I'm having trouble coming up with it myself.

Any help will be appreciated.

Thanks!

Answer 1

Let $\mathbb{H}^n=\{(x^1,\cdots,x^n)\in\mathbb{R}^n:x^1\ge 0\}$ denote the upper half space and $\partial\mathbb{H}^n=\{(0,x^2,\cdots,x^n)\in\mathbb{R}^n\}$ denote its boundary. Note that open subsets of $\mathbb{H}^n$ will not necessarily be open in $\mathbb{R}^n$.

The crux of the argument can be stated this way:

Theorem: Given any diffeomorphism $\varphi:U\to V$ where $U,V\subseteq\mathbb{H}^n$ are open subsets, $x\in U$ is a boundary point iff $\varphi(x)\in V$ is a boundary point, i.e. $\varphi(\partial\mathbb{H}^n\cap U)=\partial\mathbb{H}^n\cap V$.

There are a number of ways of proving this; here's one: we can say that a point $x\in\mathbb{H}^n$ has an inextendible curve if there is a smooth curve $\gamma:[0,a)\to\mathbb{H}^n$ such that $\gamma(0)=x$ and the domain of $\gamma$ cannot be extended to an open interval. Inextendible curves have a few important properties:

• Inextendible curves are a local property, in that the condition would be equivalent if we choose any open subset $U\subset\mathbb{H}^n$ containing $x$ and require the curve and its extension to map into $U$.
• $x\in\mathbb{H}^n$ has an inextendible curve iff $x\in\partial\mathbb{H}^n$.
• If $\varphi:U\to V$ is a diffeomorphism with $U,V\subseteq\mathbb{H}^n$ open subsets, then $x$ has an inextendible curve iff $\varphi(x)$ does.

These can be proven using the definitions of continuity and smoothness on subsets of $\mathbb{R}^n$.

In fact, boundary points are a topological invariant, in that we can modify the theorem to only require that $\varphi$ be a homeomorphism. This case can be proven using relative homology, or fundamental groups in the $n=2$ case.