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A simplicial complex is a PL structure for, and thus also homeomorphic to, a manifold if the link of every vertex is a simplicial sphere, for which there exists a definition. (I know that for high dimension, the definition might be untractable to check.) My question now is the same for simplicial sets.

Take for example the simplicial set with one vertex and one nondegenerate edge starting and ending at the vertex. Obviously, the geometrical realisation is $S^1$, but the link of the vertex is empty. In this case, the definition for a simplicial manifold fails.

I would be especially interested in a categorical definition, that is, one in terms of the simplicial set as a functor without much reference to its components.

Edit: I've clarified the question after a few years.

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    $\begingroup$ Your first sentence is false - there are simplicial complexes that are homeomorphic to manifolds but which contain vertices whose links are not spheres. Google the "double suspension theorem" for examples. $\endgroup$ Jan 27, 2014 at 19:14
  • $\begingroup$ I see, I ran into an intricacy here which I'm actually not interested in. Is it true that every simplicial complex that comes from a PL-manifold has a sphere as a link? Anyways, I've changed it to an "if" as opposed to an "iff", since the actual question is actually about simplicial sets. $\endgroup$ Jan 27, 2014 at 19:22
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    $\begingroup$ Yes, PL triangulations of manifolds always have spheres as links. $\endgroup$ Jan 27, 2014 at 19:50
  • $\begingroup$ The fact that in your $S^1$ example the link of the vertex is empty suggests that the "problem" is that you are using the "wrong" definition of link. $\endgroup$ Feb 6, 2014 at 22:56
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    $\begingroup$ Applying the barycentric subdivision functor twice produces a simplicial set that comes from a simplicial complex. One can then apply the usual criterion with links of vertices. $\endgroup$ Feb 23, 2014 at 20:18

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