This is a question in elementary geometric topology, of which I know little. It has to do with the result that geometric realizations of simplicial sets and geometric realizations of abstract simplicial complexes coincide up to homeomorphism, and with how many subdivisions are needed to effect the argument.

Here is a sketch that the realization of a simplicial set is the realization of a simplicial complex. There are functors

$$Face: [\Delta^{op}, Set] \to Pos$$

$$Nerve: Pos \to [\Delta^{op}, Set]$$

where $Face(X)$ is the poset whose elements are simplices of $X$, ordered by $x \leq y$ if $x$ is a nondegenerate face of $y$. The composite $Nerve \circ Face$ is the barycentric subdivision of $X$. This subdivision is a regular simplicial set, whose realization is a regular CW complex. Associated to a regular CW complex is a simplicial complex, also called its subdivision, whose vertices are open cells in the CW decomposition, and where simplices are finite chains $e_1 < \ldots < e_n$ where each $e_i$ is a proper face of $e_{i+1}$ ($e_i$ is contained in the closure of $e_{i+1}$). There is a theorem that the realization of this simplicial complex is homeomorphic to the space obtained by gluing together the regular CW complex.

This line of argument almost feels like overkill to me: we subdivide once, realize, and then subdivide again to get to the simplicial complex. I'd like to simplify this if possible. Define a functor

$$Flag: Pos \to SimpComplex$$

which takes a poset $P$ to the simplicial complex whose vertices are elements of $P$, and whose simplices are finite "flags" $x_1 < \ldots < x_n$. (In passing, I'll note that there is a fourth functor

$$U: SimpComplex \to Pos$$

which takes a simplicial complex to the poset of simplices ordered by inclusion; the composite $Flag \circ U$ is the barycentric subdivision functor on simplicial complexes.) My question is:

Let $X$ be any simplicial set. Is the realization of the simplicial complex $Flag \circ Face(X)$ homeomorphic to the realization of $X$?

I am not hugely confident that the answer is "yes", and even suspect there are standard counterexamples, but it seems to be true for simple examples.

**Edit:** I have a feeling that I botched the description of the subdivision of simplicial sets, but I'll let the experts correct me if needed.

oneof the faces (specifically, thelastone) of a non-degenerate simplex is allowed to be degenerate, and there should be no identifications among the remaining faces. Then I guess it's true that a subdivision is regular. $\endgroup$ – Charles Rezk Nov 21 '10 at 18:591more comment