There are two version of the singular simplicial space of a topological space $X$, one discrete and one internal. At least if X is nice, both of them have homotopy equivalent geometric realizations (and are both equivalent to X itself). I want to know why?


Let $\Delta$ be a skeleton of the category of finite non-empty ordered sets. The objects of $\Delta$ are the ordered sets [n]. A simplicial object of a category C is a functor $X: \Delta^{op} \to C$. The category $\Delta$ can be realized as a sub-category of topological spaces (the category of n-simplices, $\Delta^n$) and (via left Kan extension) this gives rise to a geometric realization functor from simplicial sets to topological spaces. It lands in the nice category of CW-complexes, and is denoted $|X|$.

The same geometric realization formula works for simplicial spaces and defines a functor from simplicial spaces to topological spaces. It doesn't always land in CW-complexes.

By general non-sense there is a right adjoint to the realization which is the singular functor. It associates to a topological space the simplicial set given by $Sing(X):[n] \mapsto map(\Delta^n, X)$. The realization from simplicial spaces to topological space also has an adjoint which is given by the simplicial space $\underline{Sing}(X): [n] \mapsto \underline{map}(\Delta^n, X)$, where $\underline{map}$ denotes the mapping space with the compactly generated compact open topology (I'm assuming all spaces are compactly generated).

A simplcial set can be viewed as a discrete simplicial space and so we have two different singular functors and a natural map of simplicial spaces between them:

$Sing(X) \to \underline{Sing}(X)$

This gives, on geometric realization a map of spaces: $|Sing(X)| \to |\underline{Sing}(X)|$. When X is sufficiently nice this map is known to be a homotopy equivalence, and both spaces are homotopy equivalent to X.

Why is this the map $|Sing(X)| \to |\underline{Sing}(X)|$ a homotopy equivalence? Can this be deduced from some sort of connectivity estimate between the constituent spaces of these simplicial spaces?

I know there is an indirect way to prove this equivalence by comparing both spaces to X, but I'm interested in generalizing this result to some related constructions and so I would like a direct argument which uses the map between them. I'm willing to replaces "simplicial space" by bisimplicial set or to assume that the simplicial space is sufficiently "good". I'm hoping for something related to this question.

  • 2
    $\begingroup$ I think that if you want to think about bisimplicial sets, the codiagonal is the droid that you are looking for. $\endgroup$ – Lunasaurus Rex Nov 11 '11 at 2:58

There are maps $|Sing(X)| \to |\underline{Sing}(X)| \to X$ which realize to weak homotopy equivalences. The inclusion of the n-skeleton $|Sing(X)|^{(n)}| \to |Sing(X)|$ is n-connected, because this is always true for CW-complexes, and so the map $|Sing(X)|^{(n)} \to X$ is n-connected. You can't really do any better than this estimate because the n-skeleton has zero homology groups in degrees above n.

The simplicial space $\underline{Sing}(X)$ contains the sub-simplicial space of constant simplices $\Delta^n \to X$. This is homeomorphic to $X$ itself and so, if we write $cX$ for the constant simplicial space with value $X$, we get a map $cX \to \underline{Sing}(X)$. This inclusion $X \to \underline{map}(\Delta^n,X)$ is a homotopy equivalence because the simplex is contractible, so this map of simplicial spaces is levelwise a weak equivalence. The geometric realization of $cX$ is $X$ itself, and so is its n-skeleton for all n. An excision argument (which takes some work) will show that the same is true for the simplicial space (at least under good conditions), and so each of the skeleta of $|\underline{Sing}(X)|$ is homotopy equivalent to $X$.

Therefore the map on n-skeleta is n-connected. I realize that this is the "compare with $X$" game that you mentioned, but my point is that because the simplicial space $\underline{Sing}(X)$ is homotopically constant the comparison $|Sing(X)| \to |\underline{Sing}(X)|$ really is comparing with $X$. From the point of view of homotopy theory it's not arising from good levelwise structure of the map at all, but comes from the simplicial assemblage.

I don't know whether this helps your generalization.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.