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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: infinte genus tori with boundary
Hempelicious
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  • 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 '19 at 04:35
  • @EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle. – Angina Seng Jan 22 '19 at 04:46
  • @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 '19 at 04:52
  • @EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary. – Angina Seng Jan 22 '19 at 04:54
  • @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 '19 at 05:07
  • Do you believe the following to be simple: The simply connected noncompact 2-manifold without boundary is $\mathbb{R}^2$; and the Freudenthal end point compactification of a connected noncompact 2-manifold (with or without boundary) exists. – Eric Towers Jan 22 '19 at 07:09
  • @EricTowers: I think "simple" is subjective, and I'm not familiar with the approach you outline, so of course I'd like to see it! I am concerned that the result you mention about simply-connected surfaces is in fact difficult... – Hempelicious Jan 22 '19 at 17:43

2 Answers2

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"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.

Eric Towers
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  • How do you know $r_1$ is continuous? – Hempelicious Jan 22 '19 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 '19 at 07:18
  • 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 '19 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 '19 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 '19 at 21:23
  • @Hempelicious : Yes. Equivalently, I'm taking all the (potential) dragons outside of $S$ and mapping them to a point, inducing a map on the fundamental group that quotients out everything except the classes containing $pAp^{-1}$ and $B$. – Eric Towers Jan 24 '19 at 02:21
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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.

Hempelicious
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  • 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 '19 at 03:08
  • @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 '19 at 04:09