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Let $\textbf{C}$ be a category with products and co-products.

If $\textbf{C}$ is $\textbf{Set}$ or $\textbf{Top}$ I'm pretty sure the following "distributive law" holds: for all $X,Y,Z \in \operatorname{ob}(\textbf{C})$, $$(X \sqcup Y) \times Z \cong (X \times Z) \sqcup (Y \times Z).$$

Does this (or a similar law) hold for any such category $\textbf{C}$ or not?

Larry B.
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Paul Orland
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    One common class where this is true is the case of Cartesian closed categories. Since $-\times Z$ is a left adjoint in such categories, it preserves colimits and thus coproducts which is exactly distributivity. This includes Set but not Top. – Derek Elkins left SE Dec 29 '17 at 19:01
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    To be completely clear, Cartesian closure is not at all a necessary condition. – Derek Elkins left SE Dec 29 '17 at 19:14
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    Correct me if I'm wrong, but I think an easy way to see that this isn't true for all categories is that if it did, then the dual statement $(X\times Y)\sqcup Z\cong(X\sqcup Z)\times(Y\sqcup Z)$ would also hold. – Carmeister Dec 29 '17 at 21:18
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    @Carmeister: yes, that's right; this is implicit in my answer. – Qiaochu Yuan Dec 29 '17 at 21:39
  • A quite general family of categories being distributive are extensive categories, check them out (there's a paper by Carboni-Lack-Walters where various properties of extensivity and distributivity are studied) – TheMadcapLaughs Dec 31 '17 at 06:09

3 Answers3

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You always have a canonical arrow $(X\times Z)\sqcup (Y\times Z) \to (X\sqcup Y)\times Z$, obtained from the universal properties of $\sqcup$ and $\times$ (there is a canonical arrow $X\times Z\to X\to X\sqcup Y$ and a canonical arrow $X\times Z\to Z$, so there is a canonical arrow $X\times Z\to (X\sqcup Y)\times Z$, and similarly for $Y\times Z$). But this arrow is not an isomorphism in general.

For a very simple example, consider the lattice M3, considered as a poset (a category with at most one arrow between any two objects). Here is a picture:

M3

In a lattice, $\times$ is $\min$ and $\sqcup$ is $\max$, so we have $(x\sqcup y)\times z = 1\times z = z$, but $(x\times z)\sqcup (y\times z) = 0\sqcup 0 = 0$.

Actually, this is (almost) the universal obstruction: it happens that a lattice is distributive if and only if it contains no sublattice isomorphic to M3 or N5. N5 is another lattice that looks like this:

N5

Here's another example, this time in a concrete category: distributivity fails in the category of abelian groups. In that category, $\times = \sqcup = \oplus$, so we have $(X\sqcup Y)\times Z \cong X\oplus Y \oplus Z$, but $(X\times Z)\sqcup (Y\times Z) \cong X\oplus Y\oplus Z^2$.

Alex Kruckman
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A simple counterexample is $\text{Set}^{op}$. Asking for $\text{Set}^{op}$ to be distributive means asking for coproducts to distribute over products in $\text{Set}$, which is easy to rule out for finite sets using cardinality. Another simple counterexample is $\text{Vect}$, where products and coproducts agree, and where it's again easy to rule out for finite-dimensional vector spaces using dimension. (However, it is true that tensor products distribute over direct sums; $\text{Vect}$ is not cartesian closed but it is the next best thing, closed monoidal.)

An interesting example is $\text{CRing}^{op}$, the category of affine schemes. It's a nice exercise to show that this category is distributive, despite neither being cartesian closed nor (as far as I know) admitting a conservative functor to $\text{Set}$ preserving products and coproducts (taking the Zariski spectrum is not conservative and does not preserve products).

Qiaochu Yuan
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Any category $\mathcal{K}$ having a functor $F : \mathcal{K} \to \text{Set}$ which is conservative, preserves finite products and finite coproducts has the required property.

This property is quite common in fact and quite ofter the functor is the forgetful.

Ivan Di Liberti
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  • Thanks! Is this also a necessary condition? And do you have a proof or reference? – Paul Orland Dec 29 '17 at 18:29
  • No I am not claiming any necessity. For a proof, give a try on your own. In general the functor $F$ is precisely the forgetful one. – Ivan Di Liberti Dec 29 '17 at 18:31
  • I just looked at this definition of conservative https://ncatlab.org/nlab/show/conservative+functor and am not convinced the forgetful functor $\mathcal{F}:\textbf{Top} \rightarrow \textbf{Set}$ is conservative. An isomorphism of underlying sets needn't be an isomorphism of topological spaces right? – Paul Orland Dec 29 '17 at 18:34
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    You are right. Top is not a corollary. Absolutely right. – Ivan Di Liberti Dec 29 '17 at 18:35
  • The question of whether the existence of some kind of functor to Set is necessary and sufficient for distributivity strikes me as very similar to the Stone representation theorem for distributive lattices: A lattice is distributive if and only if it is isomorphic (as a lattice) to some lattice of subsets of a set (where meet is intersection and join is union). [Of course this theorem has a topological component involving coherent spaces, but let's ignore that.] I wonder whether there is a categorified version of this theorem? – Alex Kruckman Dec 29 '17 at 18:48
  • The existence of conservative functors to set was solved by Freyd in a lost paper but it is a very hard problem. Any locally small category has a conservative functor to Set. In general these kinds of problems are easier to handle when the category is wellpowered, because subobjects are a good candidate for such a functor. – Ivan Di Liberti Dec 29 '17 at 18:50
  • @IvanDiLiberti That's interesting - but what about the question of functors to Set that preserve products and coproducts? – Alex Kruckman Dec 29 '17 at 18:54
  • I do not think there is a clear result on the subject, I am just claiming it to be very hard. – Ivan Di Liberti Dec 29 '17 at 18:56