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!