Homotopy equivalence of geometric realizations 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? 

 A: 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 $|X| \simeq |P_X| \simeq |K|$.
