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This question is related with this one. For simplicial complex (which we have to assume is ordered as explained in the answer of the linked question) we have a construction of geometric realization which is defined as all formal convex combinations of vertices. For such simplicial complex we have the construction of the associated simplicial set (see the linked question) and for such simplicial set there is also the construction of geometric realization. In the linked question it was asked whether the homology of simplicial complex is isomorphic with the homology of the associated simplicial set. One can show that both simplicial homologies are isomorphic with the singular homology of geometric realizations. One could hope that it is possible to prove that simplicial homologies are isomorphic by proving that their geometric realizations are homotopy equivalent. So this is my question:

Given an ordered simplicial complex $K$ consider the associated simplicial set $Ss(K)$. Consider geometric realizations: $|K|$ and $|Ss(K)|$. Is it true that $|K|$ and $|Ss(K)|$ are homotopy equivalent?

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Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then the the thick realization $||X||$ of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$, the promotion $Ss(K)$ of $K$ to a simplicial set has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So, $$ |K| = ||Ss(K)|| \simeq |Ss(K)|\, . $$.

Edit: the second paragraph isn't correct as stated. It isn't true that $||Ss(K)|| = |K|$ identically as Neil points out. I am going to keep this post here until I can find a correction.

Correction: If $X$ is a simplicial set, we can consider the poset $P_X$ of its non-degenerate simplices. The nerve of this poset is the barycentric subdivision of $X$. If $X = Ss(K)$, where $K$ is as above, then $P_X$ coincides with the poset of simplices of $K$ under inclusion.

Therefore, with $X= Ss(K)$, the nerve of the poset $P_X$ coincides with the barycentric subdivision of $K$, so by homotopy invariance of the subdivision, we see that $|X| \simeq |P_X| \simeq |K|$.

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    $\begingroup$ I don't think this is right. Take $K$ to be a single point. Then $Ss(K)$ has a single $k$-simplex for each $k$, and $||Ss(K)||=\coprod_k\Delta_k^o$ as sets, whereas $|Ss(K)|$ is a single point. I think that $|K|$ is always homeomorphic to $|Ss(K)|$ rather than $||Ss(K)||$. $\endgroup$
    – Neil Strickland
    Commented Dec 14, 2017 at 11:14
  • $\begingroup$ Neil, you are right. I thought $Ss(\ast)$ is given by the functor which sends $[0]$ to $\ast$ and $[n]$ to the emptyset for $n>0$. But that isn't what truebaran has in mind: $Ss(\ast)$ is given by $[n] \mapsto \ast$ for every $n$. $\endgroup$
    – John Klein
    Commented Dec 14, 2017 at 12:44
  • $\begingroup$ Neil, I realize now where I went wrong: if $K$ is an ordered simplicial complex, the version of $Ss(K)$ I was using was the semi-simplicial set whose $n$ simplices are given by the sequences of strict orderings $x_0 < x_1 < \cdots < x_n$ where $x_k$ is a vertex. This semi-simplicial set has the same realization as that of $K$. The MO user truebaran is using sequences $x_0 \le x_1 \le \cdots \le x_n$. $\endgroup$
    – John Klein
    Commented Dec 14, 2017 at 13:05
  • $\begingroup$ Thank you for your answer. Could you please explain in some more detail why we get that the nerve of $P_X$ is the barycentric subdivision of X (and how to understand this sentence: $X$ is an abstract simplicial set). As far as I understood, $P_X$ being a poset is a small category, therefore its nerve can be defined. The nerve is a simplicial set thus we can speak about its geometric realization... But I don't understand the part with barycentric subdivision. $\endgroup$
    – truebaran
    Commented Dec 17, 2017 at 0:07
  • $\begingroup$ @truebaran the explanation can be found here: ncatlab.org/nlab/show/subdivision $\endgroup$
    – John Klein
    Commented Dec 17, 2017 at 13:13

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