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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 $\mathbb{C}$?

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

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    $\begingroup$ That very well may be, but we sure have lottery sums! And everybody loves the lottery! $\endgroup$ – Asaf Karagila Oct 4 at 11:17
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Oct 4 at 11:19
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Oct 4 at 11:20
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Oct 4 at 11:22
  • $\begingroup$ Yes yes, I get that. I was just making an ethical remark regarding forcing... $\endgroup$ – Asaf Karagila Oct 4 at 11:23

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