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Let $X$ be a (potentially infinite) 2-dimensional simplicial complex. Then each link at a vertex $x\in X$ is a graph.

Question. If $X$ is simply connected and each link is 2-connected (in the sense of graph theory), is it true that $X$ can be written as the (not necessarily finite or disjoint) union of sub-complexes, each of which is homeomorphic to a sphere or Euclidean plane?

If it makes any difference we may assume that the complex is locally finite.

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  • $\begingroup$ How do you define a link in a CW complex which is not a simplicial (or polyhedral) complex? $\endgroup$
    – Moishe Kohan
    Commented Nov 26 at 13:41
  • $\begingroup$ @MoisheKohan Does it help if I say that the CW complex is regular? I guess two vertices in the link are connected by an edge if there is a 2-cell glued along the corresponding edges of $X$. If this helps I will edit the question. $\endgroup$
    – M. Winter
    Commented Nov 26 at 13:59
  • $\begingroup$ Yes, you should edit your question, but what you wrote in the comment is not quite enough: You should first define vertices in the link. $\endgroup$
    – Moishe Kohan
    Commented Nov 26 at 14:04
  • $\begingroup$ @MoisheKohan I modified the question to be about simplicial complexes only. I hope this clarifies most problems with links. $\endgroup$
    – M. Winter
    Commented Nov 26 at 14:10
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    $\begingroup$ Ok, did you check Bing's "house with two rooms"? This tends to be a counter-example to naive conjectures about 2-dimensional contractible complexes. $\endgroup$
    – Moishe Kohan
    Commented Nov 26 at 14:13

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I checked: Bing's "house with two rooms" (see for instance here for a nice picture) is an example: It is a contractible but not collapsible finite 2-dimensional complex. Since it is contractible, it contains no subspaces homeomorphic to surfaces (without boundary). No matter how you triangulate it, links of vertices are 2-connected: The are homeomorphic to the circle, the $\theta$-graph and square with a diagonal added.

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    $\begingroup$ Thank you! I would phrase it like this: since Bing's house is contractible it contains no spheres. Since it is compact it contains no planes (which have no boundary, but are not forbidden due to being contractible). $\endgroup$
    – M. Winter
    Commented Nov 27 at 11:18

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