This is a reformulation of a Mathematics Stack Exchange question that got no answer there, and, at the same time, a follow-up to another question on MSE.

The original problem was to prove $$U(\mathfrak{g}\oplus \mathfrak{h})\cong U(\mathfrak{g})\otimes U(\mathfrak{h})$$, where $$U(\mathfrak{g})$$ is the universal enveloping algebra of a Lie algebra $$\mathfrak{g}$$ using universal properties. To do that, the person answering the question made the following two statements:

1. $$\otimes$$ is the commutative coproduct in the category of (associative, unital) $$\mathbb{K}$$-algebras, where $$\mathbb{K}$$ is a field.
2. $$\oplus$$ is the commutative coproduct in the category of Lie algebras.

Both objects were defined as such that satisfy the diagram for coproduct only for a specific subset of morphisms from cofactors to the third object, namely:

1. $$R \otimes S$$ satisfies the coproduct diagram for every $$A$$ and every pair $$\sigma : R \to A, \tau : S \to A$$ such that $$\forall x \in R, y \in S: \sigma(x)\tau(y)=\tau(y)\sigma(x)$$ in $$A$$.
2. $$\mathfrak{g} \oplus \mathfrak{h}$$ satisfies the coproduct diagram for every $$\mathfrak{a}$$ and pair $$\tau: \mathfrak{g} \to \mathfrak{a}, \sigma : \mathfrak{h} \to \mathfrak{a}$$ such that $$\forall x \in \mathfrak{g}, y \in \mathfrak{h}: [\sigma(x), \tau(y)]=0$$ in $$\mathfrak{a}$$.

Using these statements, the person was able to show that $$U(\mathfrak{g}\oplus \mathfrak{h})\cong U(\mathfrak{g})\otimes U(\mathfrak{h})$$. What got my attention is that their first attempt at the proof was based on the assumption that $$\otimes$$ is just the coproduct in the category of associative unital algebras, but after an objection from my side that $$\otimes$$ is only a coproduct in the category of commutative associative unital rings, the person modified their original proof by introducing these new objects.

Naturally, I am curious if this is a coincidence, that an incorrect proof assuming an object is a coproduct for a non-abelian category, while in reality it is only a coproduct for the commutative part of the category, could be turned into a correct proof that uses commutative coproduct in non-abelian category. Is here some adjunction at play?

But what intrigues me the most is that I haven't been able to find a single mention of the commutative coproduct on the Web. So, what is this thing, generally, and where can I learn more about it?

• The standard tensor product of En algebras would be another example of a commutative coproduct. Maybe Lurie’s Higher Algebra has some discussion for more general operads? Dec 29, 2023 at 15:49

• The characterization of the commutating tensor product of associative algebras can be generalized to monoid objects in any braided monoidal category. Just take your definition in terms of elements and write it as a commutative diagram. You need a braiding to do this because the $$x$$ and $$y$$ switch order in the commutativity equation.
• Another kind of commuting tensor product is the Boardman-Vogt tensor product of operads, or the analogous tensor product of algebraic theories. If $$P$$ and $$Q$$ are operads or theories, then $$P\otimes Q$$ is another such with the property that $$(P\otimes Q)$$-algebras are objects with both a $$P$$-algebra structure and a $$Q$$-algebra structure, such that all the $$P$$-operations commute with all the $$Q$$-operations. Equivalently, this is an internal $$P$$-algebra in the category of $$Q$$-algebras, or an internal $$Q$$-algebra in the category of $$P$$-algebras.