TITLE: Do infinite products commute with trivial cofibrations, for simplicial sets?
QUESTION [5 upvotes]: I'm reading Voevodsky and Morel's book '$\mathbb{A}^1$-homotopy theory of schemes'. In Remark 3.1.15, it says that for any simplicial fibrant sheaf $F$ and open sets $U\subseteq V$, $F(V)\to F(U)$ is a fibration.
Prove by definition. We have a bifunctor
$$\begin{array}{ccccc}sSet&\times&Shv(Sm/k)&\to&sShv(Sm/k)\\(S&,&F)&\mapsto&S\times F\end{array},$$
where $(S\times F)(X)_n=S_n\times F(X)$. Consider the coequalizer
$$\Lambda^n_k\times U\rightrightarrows\Lambda^n_k\times V\coprod\triangle^n\times U\to C.$$
Then there is a map $i:C\to \triangle^n\times V$ and the question is reduced to the RLP of $F$ w.r.t $i$. So I want to prove $i$ is a trivial cofibration.
It's obviously a cofibration but I'm stuck at proving it's a weak equivalence. It suffices to prove that the functor $-\times F:sSet\to sShv(Sm/k)$ is a left Quillen functor since we could then use the pushout diagram of $C$. So we are going to prove that trivial cofibrations commute with infinite products, by passing to stalks...
I think we have to prove that geometric realization functor commutes with infinite products, as least up to a weak equvalence. Is this true?
Thanks a lot!

REPLY [9 votes]: To answer the title question, here's an example which I think I learned from some standard reference but I forget which.

Let $N$ be the following graph, considered as a 1-skeletal simplicial set: the vertices are the natural numbers $\mathbb N$, and there is an edge from $n$ to $n+1$ for each $n \in \mathbb N$; there are no other nondegenerate cells.
Let $i: \Delta[0] \to N$ be the inclusion of your favorite point.

Then clearly $N$ is weakly contractible, so that $i$ is an anodyne extension (i.e. an acyclic cofibration in the usual Kan-Quillen model structure). Then consider

The power $N^\omega$, i.e. the product of $N$ with itself $\omega$ many times;
The inclusion $i^\omega: \Delta[0] = \Delta[0]^\omega \to N^\omega$.

The thing to notice is that $\pi_0(N^\omega)$ is nontrivial (I think it's the cardinality of the continuum): a point in $N^\omega$ is a sequence of natural numbers, and two sequences are connected by a path only if their sequence of differences is bounded!
The upshot is that $i^\omega$ is not a weak equivalence, even though $i$ is an anodyne extension. So anodyne extensions are not closed under products.
We can play a similar game by taking $j: N \to \bar N$ to be any fibrant replacement of $N$: again $j^\omega$ can't be a weak equivalence. This shows that even the class of anodyne extensions into Kan complexes is not stable under products.
Also, instead of $N$, we can use its subdivision $sd(N)$, which has a point for each nonnegative half-integer, and 1-cells from $n+\frac 1 2$ to $n$ and $n+1$ for each $n \in \mathbb N$. The same conclusion applies: your favorite anodyne extension $\Delta[0] \to sd(N)$ fails to be preserved by powers. This is illustrative because $sd(N)$ has some nice properties that $N$ doesn't have -- it's the nerve of a poset (in particular it's the nerve of a category (in particular it's a quasicategory)). Thus we see that the power of an anodyne extension may fail to be an anodyne extension, even when it's the nerve of a map of posets (in particular even when it's the nerve of a functor (in particular when it's a map of quasicategories)).