# Multisimplicial geometric realization

Does anyone know a reference or proof for the following? Let $k\geq 1$ and let $X$ be a space. There is a $k$-fold multisimplicial set whose simplices in degree $n_1,\ldots,n_k$ are the maps $\Delta^{n_1}\times\cdots\times\Delta^{n_k}\to X$. There is a map from the geometric realization of this multisimplicial set to X. I'd like to know that this map is a weak equivalence, or at least a homology isomorphism.

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Welcome to MO, Jim! –  Ryan Budney Mar 16 '13 at 23:27
Observations: $\Delta^{op}$ is a (homotopy) sifted category, so the diagonal map $\Delta^{op} \rightarrow \Delta^{op} \times \cdots \times \Delta^{op}$ is (homotopy) cofinal. It follows that computing homotopy colimits over the big guy is the same as computing homotopy colimits of the diagonal. So it's enough to show that the geometric realization of the diagonal is weakly equivalent to the space itself. But this simplicial set looks, at level $n$, like maps $\Delta^n \times \cdots \times \Delta^n \rightarrow X$. It's not obvious to me what to do from here... is the theorem even true? –  Dylan Wilson Mar 17 '13 at 0:06
A $k$-fold multisimplicial set determines a simplicial set, its "diagonal", whose $n$-simplices are the $(n_1,\dots ,n_k)$-multisimplices. The realization of the latter is homeomorphic to that of the former. –  Tom Goodwillie Mar 17 '13 at 3:06
I'm pretty sure Tom meant "the $(n,\ldots,n)$-multisimplices" instead of "the $(n_1,\ldots,n_k)$-multisimplices" in his comment. –  Ricardo Andrade Mar 17 '13 at 9:20
Yes, Ricardo, that's what I meant to say. Dylan, the realization of a multisimplicial set is homeomorphic (not just homotopy equivalent) to the realization of its diagonal. –  Tom Goodwillie Mar 18 '13 at 3:16

Hi Jim, maybe I can sketch an answer, assuming that I'm not doing something stupid. The question is related to another one on this toy, namely:

Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?

By adjunction, your multisimplicial set has as its set of $(n_1,\cdots,n_k)$-simplices the set of continuous maps from $\Delta^{n_{1}}$ to the mapping space $Map(\Delta^{n_{2}} \times \cdots \times \Delta^{n_{k}}, X)$. As $(n_2,\cdots,n_k)$ vary, these spaces form a $(k-1)$-simplicial space, which I'll denote by $Y$. All of the terms of $Y$ are homotopy equivalent to $X$ since the simplices are contractible, and it follows that the geometric realization of $Y$ is homotopy equivalent to $X$. (See the question cited above). Your original $k$-simplicial set is obtained by applying the total singular complex functor $S$ to each space of the $(k-1)$-simplicial space $Y$. Realizing first in the first simplicial coordinate we are seeing the $(k-1)$-simplicial space $|SY|$, and its realization is homeomorphic to the realization you first asked about. Clearly $|SY|$ is equivalent to $Y$, hence its realization is equivalent to $X$.

Edit: Following up with Ricardo, the Reedy condition on Y should hold modulo the sort of elementary point-set topology that I won't look at in public (never was any good at it). Let's look at a degeneracy operator $s_i\colon Map(\Delta^n,X)\to Map(\Delta^{n+1},X)$ for simplicity. The multisimplicial case is no different. It is induced by a collapse map $\sigma_i\colon \Delta^{n+1} \to \Delta^n$ which is right inverse to $\delta_i$. It follows that the image of $s_i$ is a deformation retract of $Map(\Delta^{n+1},X)$. If there is a continuous map $u\colon Map(\Delta^{n+1},X)\to I$ such that $u^{-1}(0) = Im (s_i)$, then the inclusion of $Im(s_i)$ is a cofibration by the standard NDR pair criterion. Etc. The point is that Reedy cofibrancy is no big deal in the present context.

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I absolutely apologize for prolonging this discussion into point-set topology. However, it seems to me that one requires non-trivial conditions on the space $X$ for the NDR condition that Peter states to hold ($X$ metrizable would certainly suffice). Perhaps that is what Peter means, but I acknowledge I am a bit dense. Let us consider the simplest case: the single degeneracy map $\newcommand{\Map}{\operatorname{Map}} s:X=\Map(\Delta^0,X)\to\Map(\Delta^1,X)=\Map(I,X)$. Assume there exists a map $f:\Map(I,X)\to I$ such that $f^{-1}(0)=s(X)$ is the image of the degeneracy. (to be continued) –  Ricardo Andrade Mar 19 '13 at 0:10
(continuation) Further, let $X=[0,1]^\delta$ be the unit interval (or any other set with at least two points, really) with the indiscrete topology: it has only two open subsets, so it is certainly not Hausdorff. Then $X$ deformation retracts to $\{0\}$, and the homotopy underlying one such deformation retraction gives a map $H:X\to\operatorname{Map}(I,X)$ such that $H(x)\in s(X)$ iff $x=0$. In conclusion, the composite $f\circ H: X\to I$ is a non-constant continuous map. This is not possible because $X$ is indiscrete and has more than two points. Please let me know if I made a mistake. –  Ricardo Andrade Mar 19 '13 at 0:13
I know the above example is not Hausdorff, so certainly not compactly generated Hausdorff, but it is the best I could come up with. However, it does not seem to me that assuming that X is compactly generated and Hausdorff would be sufficient to establish the NDR property either. –  Ricardo Andrade Mar 19 '13 at 0:19
I agree completely: I am lazy about point-set topology, and of course some (mild?) condition on a pair $(X,A)$, A closed, is needed to ensure the existence of a map u from X to I with A = u^{0}. It would be nice if some expert on point-set topology weighed in, but that is not the kind of thing I'm willing to spend time on for recreation. –  Peter May Mar 19 '13 at 1:56
@Peter: Fair enough. Thank you. –  Ricardo Andrade Mar 19 '13 at 2:02


I was unable to prove the map $X\to\real{Y}$ that Peter May uses in his answer is always a weak equivalence. Unfortunately, the answer Peter links to gives no details. The result would hold if $Y$ were Reedy cofibrant, but it does not seem to be.

So here is an alternative method which proves directly that the map $\real{SY}\to X$ is a weak equivalence, thus answering the question. It avoids any use of realizations of (multi) simplicial spaces, which are problematic by not always being homotopy invariant. Perhaps there is an easier method. In any case, I want to make clear that Peter's idea to use the multi-simplicial space $Y$ is absolutely fulcral. I am essentially dealing with some technical details.

Let $X$ be a space. Recall that the object $Y:(\DeltaOp)^{\times(k-1)}\to\Top$ from Peter's answer is a multi-simplicial space whose space of $(n_2,\cdots,n_k)$-simplices is the space of maps $\Map(\Delta^{n_2}\times\ldots\times\Delta^{n_k},X)$. Let $cX$ be the constant $(k-1)$-fold multi-simplicial space with value $X$. Then we have an obvious map $cX\to Y$ which is objectwise a homotopy equivalence — as Peter remarks — due to the contractibility of products of simplices.

Let $S:\Top\to\sSet$ be the singular complex functor taking a space $Z$ to the simplicial set whose set of $n$-simplices is the set of maps $\Top(\Delta^n,Z)$. Applying $S$ objectwise to $cX$ and $Y$, we get a map $S(cX)\to SY$ (where $SY$ actually stands for $S\circ Y$, and similarly for $cX$). Since $S$ preserves weak equivalences, then $S(cX)\to SY$ is objectwise a weak equivalence. Now we conclude the map on realizations $\real{S(cX)}\to\real{SY}$ is an equivalence. This follows from the fact that any functor $(\DeltaOp)^{\times(k-1)}\to\sSet$ is Reedy cofibrant: this is a straightforward generalization of proposition 15.8.7 of Hirschhorn; alternatively, it is a particular case of the main theorem (proposition 3.15) of the article "Reedy categories and the $\Theta$-construction" by Charles Rezk and Julia Bergner.

Let us now view functors $(\DeltaOp)^{\times(k-1)}\to\sSet$ as $k$-fold multi-simplicial sets instead. As Peter observes, $SY$ is then just the $k$-fold multi-simplicial set described in the question. So it suffices to prove that the natural map $\real{SY}\to X$ is a weak equivalence. But we already saw above that $\real{S(cX)}\to\real{SY}$ is a weak equivalence. Moreover, $\real{S(cX)}=\real{\diag(S(cX))}=\real{SX}$ as Tom Goodwillie observes in a comment above (together with the simple fact that $\diag(S(cX))=SX$). But $\real{SX}$ is well-known to be weakly equivalent to $X$ (the case $k=1$ of the question). Finally, this weak equivalence $\real{SX}\to X$ is the composite of the maps $$\real{SX}=\real{S(cX)}\To\real{SY}\To X$$ and the first map is a weak equivalence. In conclusion, the second map $\real{SY}\To X$ is also a weak equivalence, as desired.

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Ricardo, yes I was being lazy about Reedy cofibrant, but I can prove that I was aware of it. See my answer (and comments) to the question mathoverflow.net/questions/124687/… which was asked just two days ago. That question addresses the missing details in my answer to this one. (Minor: I think you reversed the directions of arrows between your third and fourth paragraphs.) –  Peter May Mar 18 '13 at 13:55
@Peter: Thank you. I actually saw that answer two days ago. I certainly did not mean to imply you were unaware of the technicalities. My comment/answer above was mostly for my own benefit and, to a small extent, perhaps that of some future reader. Also, thank you very much for pointing out the error in my answer. I will correct it by reversing the arrows when I get the chance. –  Ricardo Andrade Mar 18 '13 at 18:26