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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 '11 at 23:57
    
@Chris: This does not prove anything. Not every forgetful functor has to preserve products. –  Martin Brandenburg Jan 11 '11 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 '11 at 9:24
    
@Martin: But an equivalence preserves products. –  Andrej Bauer Jan 11 '11 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 '11 at 18:44

1 Answer 1

up vote 11 down vote accepted

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 '11 at 2:42
    
Indeed, a nice proof! –  Martin Brandenburg Jan 12 '11 at 7:57
    
Is it true that $\hom(*,X)\cong|X|$? Dually it would mean that $\hom(B,2)\cong$(dual of $B$); but to get the whole dual one needs all Boolean algebra homomorphisms whereas we only have complete homomorphisms in our category. Now a complete homomorphism $B\to2$ has both adjoints, i. e. the corresponding ultrafilter is principal. I believe this means that $\hom(*,X)$ only consists of isolated points of $X$. –  მამუკა ჯიბლაძე Mar 28 at 19:24
    
@მამუკაჯიბლაძე I think everything I said after the first sentence is correct, if we understand the category of compact Hausdorff extremally disconnected spaces to have as its morphisms all continuous maps. So I think this means Chris Heunen's description of the dual category (which I hadn't checked carefully myself, I admit) is incomplete, according to your argument: the morphisms aren't all continuous maps but something else. (Or -- what is less likely -- the OP meant complete Boolean algebras and all Boolean algebra maps between them.) It would help if I knew a source for Chris's comment. –  Todd Trimble Mar 28 at 22:12
    
Yes you are right, in principle it could be that the full subcategory of all Boolean algebras was meant. I've asked that, hopefully the OP will answer. Still it is also interesting to find out whether products exist wrt complete homomorphisms. –  მამუკა ჯიბლაძე Mar 29 at 3:19

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