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.

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: 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 $(k1)$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 $(k1)$simplicial space $Y$. Realizing first in the first simplicial coordinate we are seeing the $(k1)$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 pointset 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. 


[This answer is mostly a long comment to Peter May's answer.] Edit: I have corrected some arrows which were pointing the wrong way.$\newcommand{\real}[1]{\lvert #1\rvert} \newcommand{\Map}{\operatorname{Map}} \newcommand{\Top}{\mathrm{Top}} \newcommand{\sSet}{\mathrm{sSet}} \newcommand{\diag}{\operatorname{diag}} \newcommand{\To}{\longrightarrow} \newcommand{\DeltaOp}{\Delta^{\mathrm{op}}}$ 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 multisimplicial space $Y$ is absolutely fulcral. I am essentially dealing with some technical details. Let $X$ be a space. Recall that the object $Y:(\DeltaOp)^{\times(k1)}\to\Top$ from Peter's answer is a multisimplicial 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 $(k1)$fold multisimplicial 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 $(\DeltaOp)^{\times(k1)}\to\sSet$ 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 $(\DeltaOp)^{\times(k1)}\to\sSet$ as $k$fold multisimplicial sets instead. As Peter observes, $SY$ is then just the $k$fold multisimplicial 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 wellknown 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. 

