Barycentric subdivision and 1-coskeletalization Let

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*$sd : sSet \to sSet$ 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

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*$sd X$ 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 $cosk_1(sd(cosk_1(X)))$?
 A: 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 $\mathrm{cosk}_1(\mathrm{sd}(X))$.
