1
$\begingroup$

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
$\endgroup$
  • $\begingroup$ 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 $\mathbb{R}^2$. $\endgroup$ – Eric Towers Jan 22 at 4:35
  • $\begingroup$ @EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle. $\endgroup$ – Lord Shark the Unknown Jan 22 at 4:46
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Jan 22 at 4:52
  • $\begingroup$ @EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary. $\endgroup$ – Lord Shark the Unknown Jan 22 at 4:54
  • $\begingroup$ @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 $\partial M\rightarrow M$. $\endgroup$ – Hempelicious Jan 22 at 5:07
1
$\begingroup$

"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.

$\endgroup$
  • $\begingroup$ How do you know $r_1$ is continuous? $\endgroup$ – Hempelicious Jan 22 at 20:44
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Jan 23 at 7:18
  • $\begingroup$ 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? $\endgroup$ – Hempelicious Jan 23 at 20:26
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Jan 23 at 20:48
  • $\begingroup$ 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! $\endgroup$ – Hempelicious Jan 23 at 21:23
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Eric Towers Jan 24 at 3:08
  • $\begingroup$ @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. $\endgroup$ – Hempelicious Jan 24 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.