# Do infinite products commute with trivial cofibrations, for simplicial sets?

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!

This fact admits a much easier proof. To show that for any simplicial fibrant sheaf F and open sets U⊆V the map F(V)→F(U) is a fibration it suffices to show that F(V)→F(U) has a right lifting property with respect to horn inclusions.

Since F(V)→F(U) can be rewritten as Map(U→V,F), we can move U→V using the two-variable adjunction between ⨯ and Map, which boils down the problem to showing that F→1 has a right lifting property with respect to the pushout product of U→V and a horn inclusion.

Since F→1 is a fibration of simplicial sheaves and the pushout product axiom is satisfied for the categorical product, it suffices to observe that U→V is a cofibration (i.e., a monomorphism) of simplicial sheaves and a horn inclusion is an acyclic cofibration of simplicial sets.

(By the way, the geometric realization functor does not commute with infinite products, nor is this true if we derive both of them.)

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)).