# Every connected, open surface with the infinitely generated fundamental group is the interior of some non-compact surface with boundary

This question is based on the discussion in the comment section of my previous post. I used the fact: "Every connected, open surface with the finitely generated fundamental group is the interior of some compact surface" to prove open annulus covers any orientable non-simply-connected surface without boundary. My current question is the following:

$$\textbf{Problem:}$$ Is this also true: Every connected, open surface with the infinitely generated fundamental group is the interior of some non-compact surface with boundary.

$$\textbf{Attempt:}$$ I hope this is not in general true. I am trying to use $$\text{Theorem 3}$$ of Ian Richard's classification theorem on open surfaces. It says:

Every non-compact surface without boundary can be obtained by first removing a closed totally disconnected subset $$X$$ of $$\Bbb S^2$$, then removing the interior of a finite or countable number of non-overlapping closed discs, and after that suitably identifying the boundaries of these discs in pairs(to produce either handle or cross cap).

Now, if $$X$$ has an isolated point $$p$$, then removing $$X$$ from $$\Bbb S^2$$ is the same as removing $$X\backslash\{p\}$$ from $$\Bbb D:=\{z\in \Bbb C:|z|<1\}$$ using one-point compactification. Now, every other operation in $$\text{Theorem 3}$$ are happening in a smaller open disc $$\{z\in \Bbb C:|z|<\varepsilon\}$$ for some $$0<\varepsilon<1$$. So, considering the boundary $$\{z\in \Bbb C:|z|=1\}$$ of $$\Bbb D$$ we are done in this special case.

Some examples of connected, open surface with the infinitely generated fundamental group are $$\Bbb R^2\backslash\text{cantor set}$$, Loch ness Monster surface, Jacob Ladder surface. Note that all infinite-type surface(with or without boundary) are homotopically equivalent to $$\displaystyle\bigvee_{n\in\Bbb N}\Bbb S^1$$.

$$\bullet$$ Is my reasoning in the $$\textbf{Attempt}$$ section correct?

$$\bullet$$ Is $$\Bbb S^2\backslash \text{cantor set}$$ an example of an open surface for which the statement of $$\textbf{Problem}$$ false?

$$\bullet$$ Vague Question: Can we make some decision on the $$\textbf{Problem}$$ using some kind of cohomology or other useful end-functors?

Thanks for your interest.

• How would such a thing look for the Jacob's Ladder surface? Jan 31, 2021 at 16:18
• Sorry, which thing I do not understand. Jan 31, 2021 at 16:20
• A manifold with boundary which has interior that surface. Jan 31, 2021 at 16:24
• @ConnorMalin: Remove open unit disk $D$ from $R^2\cup \{\infty\}$ and a sequence of disjoint disks $D_n$ of radii $1/n$ converging to a point $p$ on the boundary of $D$. Then attach annuli connecting $\partial D_n$ to $\partial D_{n+1}$ for each odd $n$. Lastly, remove $p$. Jan 31, 2021 at 16:34
• Jacob Ladder Let $\Bbb D_r:=\{z\in\Bbb C:|z|<r\}$ for $r>0$. Consider small closed discs inside $\Bbb D_2$ and converging to the boundary of $\{|z|=1\}$ and then identifying the boundaries of these discs in pairs to get handels. Jan 31, 2021 at 16:36

## 1 Answer

I assume that when you say "with boundary," you mean "with nonempty boundary." Indeed, every (connected) open surface is homeomorphic to the interior of a surface with nonempty boundary. Let me explain how to do this in the case of the complement to the standard Cantor set $$C$$ in $$S^2=R^2\cup \{\infty\}$$, $$C\subset [0,1]$$, $$C$$ contains $$1$$. Consider the open unit disk $$D\subset R^2$$ centered at $$(2,0)$$. Take the surface with nonempty boundary $$X\subset R^2\cup \{\infty\}$$ obtained by removing from $$S^2-C$$ the disk $$D$$. Then $$X\setminus \partial X$$ equals $$S^2-(C\cup cl(D))$$. I claim that the latter is homeomorphic to $$S^2-C$$. Indeed, consider the quotient space $$Z$$ of $$S^2$$ obtained by collapsing $$cl(D)$$ to a single point. It is easy to see that $$Z$$ homeomorphic to $$S^2$$. The restriction of the quotient map $$q: S^2\to Z$$ to $$C$$ is 1-1, hence, by compactness of $$C$$, a homeomorphism to its image. Hence, the pair $$(Z,q(C))$$ is homeomorphic to $$(S^2,C)$$. Claim follows.

The general case can be proven using the above argument in conjunction with the quoted sentence from Richards' paper. (There is some work to be done but you can probably do this yourself.)

In fact, one can prove more (using a refined version of the above argument):

Theorem. Let $$S$$ be an open connected surface with boundary. Then there exists a surface with boundary $$F$$ whose interior is homeomorphic to $$S$$ and such that $$\partial F$$ is dense in the end-compactification $$\bar{F}$$ of $$F$$ (i.e. the closure of $$\partial F$$ in $$\bar{F}$$ contains $$e(F)= \bar{F}\setminus F$$).

• So, $X=(\Bbb S^2-\text{standard cantor set})-\{z\in\Bbb C:|z-2|<1\}$? Jan 31, 2021 at 16:29
• @User873110: Of course. Jan 31, 2021 at 16:30