Is the category of atomless Boolean algebras with complete embeddings closed under coproducts?

Question

Consider the category whose objects are atomless Boolean algebras (not necessarily complete) and whose arrows are complete embeddings.

Does a coproduct exist in this category for any two atomless Boolean algebras $\mathbb{B}$ and $\mathbb{C}$?

Answer 1

This category does not have co-products. To see this, let $\newcommand\B{\mathbb{B}}\B$ be any atomless complete Boolean algebra with a nontrivial automorphism $\pi:\B\to\B$. For example, the forcing to add a Cohen real.

I claim that $\B$ has no co-product with itself in your category. Suppose toward contradiction that $\B\sqcup\B$ is the co-product, with complete embeddings $i,j:\B\to\B\sqcup\B$ realizing the co-product universal property.

Let $f_1:\B\to\B$ and $f_2:\B\to\B$ both be the identity embedding. By the universal property, there is $f:\B\sqcup\B\to\B$ making a commutative diagram. It follows that $i(b)$ and $j(b)$ are both carried by $f$ to $b$. Since $f$ is an embedding, this means in particular that $i(b)=j(b)$. Now replace $f_2$ with the automorphism $\pi$. By the universal property, there is again a complete embedding $f:\B\sqcup\B\to\B$ making the diagram commute. But now $f$ must take $i(b)$ both to $b$ and to $\pi(b)$, which is impossible if $\pi$ moves $b$.

So we don't have co-products.