Does $\pi_0$ preserve infinite products? The functor $\pi_0\colon \operatorname{sSets}\to \operatorname{Sets}$ has value on a simplicial set $X$ the coequalizer of the two boundary morphisms $d_0,d_1\colon X_1\to X_0$.
It is easy to see that $\pi_0$ preserves finite products.

Does $\pi_0$ preserve infinite products?

Please let me add a follow-up question.

Does $\pi_0$ preserve infinite products of Kan complexes? If this is not true, is there also a counterexample with simply connected spaces and countable products?


$\pi_0$ does not preserve infinite products. Let $X$ be the nerve of the following category:
$$\bullet \leftarrow \bullet \rightarrow \bullet \leftarrow \bullet \rightarrow \bullet \leftarrow \bullet \rightarrow \cdots$$
$X$ is contractible but not a Kan complex. On the other hand, $X^{\mathbb{N}}$ is not contractible: the problem is that there is no uniform upper bound to the lengths of paths between points in $X$. 
The problem goes away if you restrict to Kan complexes: indeed, if $K$ is a Kan complex, then the image of $K_1 \to K_0 \times K_0$ is an equivalence relation; in particular, two points in $K$ are in the same connected component if and only if there is a path of length 1 between them. Thus $\pi_0$ preserves all products of Kan complexes.
