## Coproducts of complete Boolean algebras

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Does the category of complete Boolean algebras have binary coproducts?

Note that this category does not have countable coproducts. Indeed, the coproduct of countably many copies of the four element complete Boolean algebra would be the free complete Boolean algebra on countably many generators, and such an object does not exist.

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By Stone duality, the category of complete Boolean algebras is dually equivalent to the category of so-called Stonean spaces, i.e. compact, Hausdorff, extremally disconnected, topological spaces. The question then becomes whether the latter category has binary products. Products of compact spaces are compact, and products of Hausdorff spaces are Hausdorff. But binary products of extremally disconnected spaces need not be extremally disconnected. – Chris Heunen Jan 10 2011 at 23:57
@Chris: This does not prove anything. Not every forgetful functor has to preserve products. – Martin Brandenburg Jan 11 2011 at 0:05
Martin, I know, that's why I only added it as a comment. I just thought that it might lead to a counterexample. – Chris Heunen Jan 11 2011 at 9:24
@Martin: But an equivalence preserves products. – Andrej Bauer Jan 11 2011 at 17:59

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... and such an object does not exist assuming AC (dx.doi.org/10.1007/BF02757883) – Adam Jan 11 2011 at 18:44
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## 1 Answer

Chris Heunen's comment under the OP can be turned into a proof. Suppose the category of compact Hausdorff extremally disconnected spaces has binary products. Let $X \times Y$ denote the product in that category. If $|X|$ denotes the underlying set, then of course the canonical map

$$|X \times Y| \to |X| \times |Y|$$

is an isomorphism, because $|X| \cong \hom(\ast, X)$ where $\ast$ is the one-point space, i.e., the underlying set functor is representable and representables preserve products.

Chris observes that the ordinary product space $X \times_{Top} Y$ of two compact Hausdorff extremally disconnected spaces need not be extremally disconnected. However, under our supposition we would have a continuous comparison map

$$X \times Y \to X \times_{Top} Y$$

in $Top$ which is a bijection at the level of the underlying sets. Being a continuous bijection between compact Hausdorff spaces, it is a homeomorphism, and this contradicts Chris's observation.

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Nice! (and some more to fill up characters) – David Roberts Jan 12 2011 at 2:42
Indeed, a nice proof! – Martin Brandenburg Jan 12 2011 at 7:57
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