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Let $\cal C, \cal D$ be model categories. Hovey says in his monograph "Model Categories" that the homotopy category $\operatorname{Ho}(\cal C \times D)$ is isomorphic to $\operatorname{Ho}(\cal C) \times \operatorname{Ho} (\cal D)$, and that this is true for any (finite I assume) number of model categories.

Is this true in general? Let $\{ \mathcal C_i \}_{i \in I}$ be a family of categories, let $W_i \subseteq \operatorname{Mor} \cal C_i$ be families of morphisms and let $\mathcal C = \prod_{i \in I} \cal C_i$. Are the localisations $$\mathcal C\left[ \prod_{i \in I} W_i ^{-1}\right] \cong \prod_{i \in I} \mathcal C_i\left[W_i^{-1}\right]?$$

I can find a functor from the left to the right in general.

Is it at least true for infinite families of model categories?

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Continued at math.stackexchange.com/questions/464991 where I also gave a complete answer for the case that $I$ is finite. –  Martin Brandenburg Aug 15 '13 at 14:51
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up vote 12 down vote accepted

It is true for arbitrary products of model categories (or just cofibration categories) as proven in Theorem 7.1.1 of http://arxiv.org/abs/math/0610009v4.

It is also true for finite products of arbitrary relative categories as discussed in this answser: Localizing an arbitrary additive category. (For more details on this and related arguments see http://nforum.mathforge.org/discussion/4769/connected-components-preserve-finite-products/.)

Addendum: This doesn't hold for infinite products of arbitrary relative categories. Consider a sequence of relative categories $\mathcal{C}_0, \mathcal{C}_1, \mathcal{C}_2, \ldots$ such that $\mathcal{C}_i$ is saturated and has objects $X_i$ and $Y_i$ that are weakly equivalent, but the shortest zig-zag of weak equivalences witnessing this is at least $i$ arrows long. Then $(X_0, X_1, \ldots)$ and $(Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(\mathcal{C}_i)$ but not as objects of $\mathrm{Ho}(\prod_i \mathcal{C}_i)$ since this would imply that the length of the shortest zig-zag connecting $X_i$ to $Y_i$ is bounded.

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