# Retract of noncompact surface to its boundary?

Suppose $$M$$ is a connected, noncompact 2-manifold, and its boundary $$\partial M$$ is a circle. What's the simplest way to show there is a retraction $$r: M\rightarrow \partial M$$?

Here are some examples of such surfaces:

1. Probably the easiest example is the closed unit disk, minus the origin.
2. A more complicated example is the closed unit disk minus a Cantor set.
3. A totally different example is to join two infinite-genus tori, and attach a cylinder to that:
• What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $\mathbb{R}^2$. – Eric Towers Jan 22 at 4:35
• @EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle. – Lord Shark the Unknown Jan 22 at 4:46
• @LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating. – Eric Towers Jan 22 at 4:52
• @EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary. – Lord Shark the Unknown Jan 22 at 4:54
• @EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $\partial M\rightarrow M$. – Hempelicious Jan 22 at 5:07

## 2 Answers

"Simple" depends on what you've studied. Under one valuation ...

Let $$M$$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $$\partial M \cong S^1$$. Show there is a retraction $$r: M \rightarrow \partial M$$.

Let $$E$$ be the set of ideal points in the Freudenthal end point compactification of $$M$$ and let $$M''$$ be the Freudenthal compactification of $$M$$. Let $$e \in E$$ and $$M' = M'' \smallsetminus \{e\}$$, the Freudenthal end point compactification of $$M$$, ignoring the end corresponding to $$e$$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $$E$$ is not countable. $$M'$$ is a connected, noncompact, 2-manifold with circle boundary $$\partial M = \partial M'$$, and one end. We will abuse notation and call this end $$e$$.

Under a nested compact exhaustion of $$M'$$, there is an annular neighborhood of the end $$e$$. (It is an open disk neighborhood of $$e$$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48$${}^\circ$$2x for more.) Let $$A$$ be a core $$S^1$$ of this annulus and $$p_1$$ be a point of $$A$$. Let $$B = \partial M' \cong S^1$$ and $$p_2$$ be a point of $$B$$. There is a simple path, $$p$$, in $$M'$$ connecting $$p_1$$ and $$p_2$$ (which we would use to change the basepoint of the fundamental group from one to the other). $$B$$ has an open collar homotopic to an annulus closed on the $$B$$ boundary component and open on the other. $$p$$ has an open bicollar homeomorphic to $$p \times (0,1)$$. Let $$S$$ be the union of these three open collars. ($$S$$ is a regular neighborhood of a $$1$$-skeleton of the boundary, the end $$e$$, and a path between the two in $$M'$$.)

Observe that $$M' \smallsetminus S$$ is a connected, noncompact 2-manifold with circle boundary. We construct $$r$$ in two steps. Let $$r_1$$ be the continuous map holding $$A$$, $$B$$, and $$p$$ fixed and contracting $$M' \smallsetminus S$$ to a point. Let $$D$$ be the open unit disk and $$r_1(M')+D$$ be $$D$$ injectively identified to the boundary of $$r_1(M')$$. The nullhomotopic homotopy classes were parallel to $$A$$ and $$B$$, and were rendered nullhomotopic by transport across $$D$$, so $$r_1(M')+D$$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $$\mathbb{R}^2$$. Therefore, $$r_1(M') +D \cong \mathbb{R}^2$$. Constructing the retract $$r_2:\mathbb{R}^2 \smallsetminus\{(0,0)]\} \rightarrow B$$ is a standard exercise. Then $$\left. r_2 \right|_{r_1(M')}$$ is a domain restriction of a continuous function, so is continuous, and $$r = \left. r_2 \right|_{r_1(M)} \circ \left. r_1 \right|_{M}$$ is the desired map.

In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.

• How do you know $r_1$ is continuous? – Hempelicious Jan 22 at 20:44
• @Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' \smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' \smallsetminus S$, which is $M' \smallsetminus S$ union an open regular neighborhood of $\partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point. – Eric Towers Jan 23 at 7:18
• I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'\setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'\setminus S$. What does $r_1$ look like in this case? – Hempelicious Jan 23 at 20:26
• @Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk. – Eric Towers Jan 23 at 20:48
• Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me! – Hempelicious Jan 23 at 21:23

Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.

One thing I don't like about this proof is you either need some smooth structure on $$M$$, or you need to verify the "supposes" in the next paragraph.

Suppose we have an embedding $$\rho:\mathbb{H}\rightarrow M$$, where $$\mathbb{H}=[0,\infty)$$. Let $$x\in\partial M$$ and further suppose this embedding is such that $$\begin{equation*} \rho(\mathbb{H})\cap\partial M = \rho(0)=x \end{equation*}$$ Finally, suppose that there's a neighborhood of $$\rho(\mathbb{H})$$ that looks like $$\mathbb{H}\times (-1,1)$$.

Then we can cut $$M$$ along $$\rho(\mathbb{H})$$ to get a new manifold $$N$$, where $$\partial N\cong\mathbb{R}$$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $$\rho(\mathbb{H})$$, and middle is $$\partial M$$ cut at $$x$$. Note that middle is homemorphic to $$[0,1]$$.

We define a map $$f:\partial N\rightarrow [0,1]$$ on each piece: $$\begin{equation*} f(z) = \begin{cases} 0 & z\in\textit{left}\\ z & z\in\textit{middle}\\ 1 & z\in\textit{right}\\ \end{cases} \end{equation*}$$

By the Tietze extension theorem, this extends to a map $$F: N\rightarrow [0,1]$$. If we compose that with the quotient map $$\pi:[0,1]\rightarrow S^1$$, we get a map $$\pi F:N\rightarrow S^1$$. Because this map agrees on left and right, it descends to a map $$\begin{equation*} \widetilde{\pi F}:M\rightarrow S^1 \end{equation*}$$

which is the identity on $$\partial M$$. That means it is our desired retraction.

• This $\rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x \in \partial M$ exists. My argument replaces putting $\rho(\infty)$ on the end with putting $\rho(\text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $\rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $\rho$ has an open tubular neighborhood. – Eric Towers Jan 24 at 3:08
• @EricTowers: yes, $\rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful. – Hempelicious Jan 24 at 4:09