Can the obvious "product" of complete atomistic Boolean algebras be realized as a categorial product? Let $X$ and $Y$ denote sets, and $\eta_X,\eta_Y : X,Y \rightarrow X+Y$ denote the natural injections to the disjoint union. Then intuitively, the "product" of the Boolean algebras $2^X$ and $2^Y$ should be the Boolean algebra $2^{X + Y}$ with projection maps $\eta_X^*, \eta_Y^* : 2^{X + Y} \rightarrow 2^X,2^Y$.
Its clear that projection maps preserve the operations of complete Boolean algebra. Can this construction be viewed as a categorial product in some appropriate (concrete) category? I've had a hard time finding information about products of complete Boolean algebras in general (and complete lattices / semilattices, too), both with and without the stipulation "atomistic."
Thank you.
 A: The product Boolean algebra could more easily be described (I believe, someone please correct me if I'm mistaken) as $2^X\times 2^Y$ with the operations being:


*

*$\langle x,y\rangle\vee\langle z,w\rangle =\langle x\cup z,y\cup w\rangle$

*$\langle x, y\rangle' = \langle X\setminus x,Y\setminus y\rangle$

*$1\in 2^X\times 2^Y = \langle X,Y\rangle$

*etc. 
The obvious projections are also Boolean homomorphisms. This is the ordinary categorical product in $\mathbf{Pos}$, as well as (I believe) in the category of Boolean algebras. You can translate this back into your construction above with  pairs being subsets of $X+Y$ and the projections being given by the inverse image on the injections.
Edit: It's a good idea to show this is a product. Obviously any Boolean homomorphism into the above alleged project will yield two homomorphisms (one to $2^X$, one to $2^Y$) when composed with the obvious projections. So it suffices to show the converse: that any pair of Boolean homomorphisms from some BA $Z$ produces a unique morphism to the alleged product.
The correct mappings for the $0$ nd $1$ of morphisms from $Z$ are obvious, so we check that it does the right things for the Boolean operations of meet, join, and complement. I will show just two of these as above. 
For $f:Z\to 2^X$ and $g:Z\to 2^Y$, and elements $z,w\in Z$, we want $h: Z\to 2^X\times 2^Y$ with $h(z)\vee h(w)$ being carried to $f(z)\cup f(w)$ and $g(z)\cup g(w)$ by the respective projections. It's clear that $h: z\mapsto\langle f(z),g(z)\rangle$ does this. We also want $h(z)'$ to be carried by the projections to $X\setminus f(z)$ and $Y\setminus g(z)$; it's easy to see that the action above gives us this, too. Uniqueness is straightforward.
