Let $X$ be a topological space, covered by a collection of open sets $\{U_\alpha\}$ (or more generally a cover by subspaces whose interiors cover $X$). Consider the singular simplicial set $S(X)_\bullet$ and the simplicial subset $S^{\mathfrak{A}}(X)_\bullet$ whose $n$-simplices are those maps from the topological $n$-simplex into $X$ that factor through one of the $U_\alpha$. We know (by essentially the excision theorem) that they have the same homology. Is it true that the inclusion is, though, a homotopy equivalence of simplicial sets?
I suspect I've heard this before in a class, and it seems highly plausible, but my memory is not great, and in any case it would be nice to have a reference for this fact as a less "ad hoc" way of thinking about the construction. My guess is that the subdivision operator should give the homotopy as it gives a chain homotopy in the usual proof of excision, but I'm not very sure.