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

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 $$\mathbb{C}$$?

• What does it mean "complete embedding" here? Are them embeddings preserving all meets and all joins? – Evgeny Kuznetsov Oct 3 at 18:32
• Yes that is correct. – Toby Meadows Oct 3 at 19:30

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.

• That very well may be, but we sure have lottery sums! And everybody loves the lottery! – Asaf Karagila Oct 4 at 11:17
• Yes, it seems to me that lottery sums are very close to coproducts, but not in this category. I guess if you are allowed to embed into a cone, then it might actually be a coproduct. – Joel David Hamkins Oct 4 at 11:19
• I mean, morally speaking, lottery sums are "kind of coproducts" for forcing notions. That's why dating a set theorist is like winning the lottery, or so I was told. – Asaf Karagila Oct 4 at 11:20
• Yes, I agree. But the natural map of $\mathbb{B}$ into the lottery sum $\mathbb{B}\oplus\mathbb{C}$ is not a complete embedding, but only goes into a cone. – Joel David Hamkins Oct 4 at 11:22
• Yes yes, I get that. I was just making an ethical remark regarding forcing... – Asaf Karagila Oct 4 at 11:23