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This is a follow-up question to this one. I asked about the proof of $U(\mathfrak{g}\oplus \mathfrak{h})\cong U(\mathfrak{g})\otimes U(\mathfrak{h})$ using universal properties, where $U(\mathfrak{g})$ is the universal enveloping algebra of a Lie algebra $\mathfrak{g}$. The answer I was given contained the notion of "commutative coproduct": namely, it stated that tensor product is the commutative coproduct in the category of associative unital algebras, and $\oplus$ is the commutative coproduct in the category of Lie algebras. Roughly speaking, both $\oplus$ and $\otimes$ can be described as "coproduct, but only for pairs of morphisms from comultipliers to a third object, the actions of which "commute" on the third object."

While I checked all the implicit assumptions in the answer, and the definitions given in it appear to be sound and define a useful operation, I haven't been able to find any information on "commutative coproducts" whatsoever.

Beside that, the person answering my question had to use commutative coproduct, because the proof they gave originally assumed $\otimes$ is the regular coproduct in the category of associative unital $R$-algebras, while it is actually the coproduct in the category of commutative associative unital $R$-algebras. The interesting thing is that the person was able to modify their original answer into a correct one by using commutative coproduct instead of the regular one. So there appears to be some kind of adjunction between commutative coproducts in non-abelian categories and regular coproducts in abelian categories. Is it really the case, and where can I learn more about commutative coproducts?

Edit: I've cross-posted this on MathOverflow.

  • Hmm, all coproducts commute in the sense of there being a canonical symmetry morphism $X\cup Y\cong Y\cup X$ – FShrike Dec 26 '23 at 00:43
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    Please, read the question more carefully. Commutative coproduct is not a coproduct that is commutative, it is a coproduct for a subset of morphisms from its cofactors to a third object that "commute" in a certain sense (since these morphisms have different domains, you can't even outright say they commute) – Daigaku no Baku Dec 26 '23 at 02:39

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