What are the simplicies?

Question

Let $X$ be a bisimplicial set: that means that $X$ is a simplicial object in simplicial sets. Let $\Delta^n$ be the collection of simplicial sets given by $$ \Delta^n(m):=\Delta(m,n), $$ so we could say $\Delta^n:=\Delta(\underline{\phantom m}, n)$. How would I simply describe $$ |X|=\int^{n\in\Delta}\Delta^n\otimes X_n $$ as a simplicial set? In particular, what is the set of simplices $|X|_n$ for each $n$?

Answer 1

I'm assuming that in $\Delta^n\otimes X_n$ the tensor product is the ordinary product, as it is on nLab.

This functor is a colimit, and as a colimit, I think you'll struggle to say anything interesting about its simplices except by just taking the construction of the colimit as a coequalizer, and bashing it out.

So let's do that. Taking the definition gives you that $i$-simplices of $|X|$ are the quotient of the set of all pairs $(a :i\to n, b_n)$ of $i$-simplices of $\Delta^n\times X_n$ by the smallest equivalence relation generated by $(a,f^*b_m)\sim (f\circ a,b_m)$ where $b_m\in X_m(i)$ and $f:n\to m$ is a morphism in $\Delta$.

Note then that we always have $(a,b_n) \sim (\mathrm{id}_i,a^*b_n)$. Thus the $i$ simplices will be $X_i(i)$ for all $i$. I assume therefore that $|X|$ should be the composite of $X$ with the diagonal functor $D:\Delta^{\text{op}}\to \Delta^{\text{op}}\times \Delta^{\text{op}}$.

To prove this, we could do the following $$ \newcommand\of[1]{\left({#1}\right)} \newcommand\Set{\mathbf{Set}} \newcommand\sSet{\mathbf{sSet}} \begin{align} \sSet(|X|,S) &= \sSet\of{ \int^{n\in\Delta} \Delta^n\times X_n ,S } \\ &= \int_{n\in\Delta} \sSet(\Delta^n\times X_n,S) \\ &= \int_{n\in\Delta} \int_{m\in\Delta} \Set\of{ X_n(m)\times \Delta(m,n), S(m) } \\ &= \int_{n\in\Delta} \int_{m\in\Delta} \Set\of{ \Delta(m,n), \Set\of{ X_n(m), S(m) } } \\ &= \int_{n\in\Delta} \Set\of{ X_n(n), S(n) } \\ &= \sSet(X\circ D,S). \end{align} $$