TITLE: Coproducts of complete Boolean algebras
QUESTION [12 upvotes]: 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.

REPLY [16 votes]: 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.