# Barycentric subdivision and 1-coskeletalization

Let

• $$sd : sSet \to sSet$$ denote barycentric subdivsion;

• $$cosk_1 : sSet \to sSet$$ denote 1-coskeletalization.

Question: Let $$X$$ be a graph or simplicial set. If the homotopy type of $$cosk_1(X)$$ is known, then what can be said about the homotopy type of $$cosk_1(sd(cosk_1(X)))$$?

I'm interested in this question primarily in the case where my simplicial sets are in fact abstract simplicial complexes. This special case of the question can be read as follows. Let

• $$sd X$$ denote the barycentric subdivision of an abstract simplicial complex $$X$$;

• $$cosk_1(X)$$, for an abstract simplicial complex $$X$$, denote the largest abstract simplicial complex with the same vertices and edges as $$X$$.

Question (bis): Let $$X$$ be a graph or abstract simplicial complex. If the homotopy type of $$cosk_1(X)$$ is known, then what can be said about the homotopy type of $$cosk_1(sd(cosk_1(X)))$$?

I don't know anything about simplicial sets, but I think the question for simplicial complexes is easy.

A complex is said to be "flag" (also known as a clique complex) if every subset $$S$$ for which every pair $$u, v\in S$$ form an edge is in fact a face. Evidently a flag complex is determined by a graph. So when you are asking about "the largest abstract simplicial complex with the same vertices and edges as $$X$$," you really just mean the flag complex determined by the graph (i.e., 1-skeleton) of $$X$$.

To answer your question, we may as well replace $$X$$ by the flag complex determined by its graph. Then you are asking about how $$X$$ compares to the flag complex determined by the graph of $$\mathrm{sd}(X)$$, where $$\mathrm{sd}(X)$$ is the barycentric subdivision of $$X$$. But it is well known that a barycentric subdivision is already flag (indeed, I think this is where the terminology "flag" comes from: the faces of the barycentric subdivision are precisely flags of faces of the original complex). So, in your language, we have $$\mathrm{sd}(X) = \mathrm{cosk}_1(\mathrm{sd}(X))$$ as simplicial complexes, and because $$\mathrm{sd}(X)$$ is homotopy equivalent (in fact, homeomorphic) to $$X$$, certainly $$X$$ has the same homotopy type as $$\mathrm{cosk}_1(\mathrm{sd}(X))$$.