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 $\mathbb{C}$?



Is the category of atomless Boolean algebras with complete embeddings closed under coproducts?
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](https://en.wikipedia.org/wiki/Coproduct).


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. 


