# 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

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.
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