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Suppose we have a directed system of inclusions of compact surfaces with boundary $$S_1 \hookrightarrow S_2 \hookrightarrow S_3 \cdots $$ such that all of the surfaces $\{S_k\}$ have uniformly bounded topology. That is, there exists some constant $C > 0$ independent of $k$ such that the genus and number of boundary components of $S_k$ is bounded above by $C$.

Then the surfaces have some well-defined direct limit surface $S$. I was wondering how pathological the topology of $S$ can become?

As of yet, I have been unable to produce any example that does not yield a punctured compact surface with finitely many punctures.

Venturing towards a possible proof of this statement, the uniform bound on the topology should imply via a quick Morse-theoretic argument that the surfaces $\{S_k\}$ become diffeomorphic for large $k$ and the inclusions are deformation retracts.

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  • $\begingroup$ So long as the inclusion $f: S_k \hookrightarrow S_{k+1}$ has $f(\partial S_k) \cap \partial S_{k+1}$ a union of connected components of $\partial S_{k+1}$, all you do at each stage is glue on a cylinder, so you get exactly what you say. In general I imagine you can also cook up arbitrarily nasty punctures on the boundary; certainly the upper half plane is such a colimit, but I think so is the disc minus a Cantor set from the boundary. $\endgroup$
    – mme
    Commented Jan 3, 2020 at 0:44
  • $\begingroup$ How would you construct the disk minus a Cantor set? For me this looks like the infinite type surface that looks like an infinite binary tree, which seems to be impossible since we're never adding any boundary components for large $k$. $\endgroup$
    – Rohil Prasad
    Commented Jan 3, 2020 at 5:45
  • $\begingroup$ You have to be more careful about boundary components, or else what you say about deformation retracts is false. You can have an example where each odd surface $S_{2n+1}$ is a disc, each even surface $S_{2n}$ is an annulus, $S_{2n}$ is obtained from $S_{2n-1}$ by attaching a pair-of-pants (a 3-holed sphere) to the unique component of $\partial S_{2n-1}$, and $S_{2n+1}$ is obtained from $S_{2n}$ by attaching a disc to one component of $\partial S_{2n}$. $\endgroup$
    – Lee Mosher
    Commented Jan 3, 2020 at 5:57
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    $\begingroup$ @Rohil Minus a Cantor set on the boundary. I can try to draw a picture of the suggestion tomorrow. Lee has pointed out a subtlety I missed above. $\endgroup$
    – mme
    Commented Jan 3, 2020 at 6:54
  • $\begingroup$ Does this picture clarify the idea? I have drawn all of the discs $S_i$ as topological submanifolds of the unit disc, each nested in the next. You can see how the portion of the boundary not contained in $S_i$ undergoes the middle-thirds construction of the Cantor set. photos.app.goo.gl/WEmkhNCQgeepSpYX7 I would be surprised if the answer was much different if you demanded smooth embeddings, though it would make those corners a little more irritating. $\endgroup$
    – mme
    Commented Jan 3, 2020 at 22:14

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