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On Page 13 of Bourbaki's Chapter I on Lie algebras, it is proven that if $\mathfrak g_1, \mathfrak g_2$ are Lie algebras and $\mathrm U$ denotes the functor from the category of Lie algebras over a field $F$ to the category of associative unital $F$-algebras taking the universal envelope algebra, then $$ \mathrm U(\mathfrak g_1 \oplus \mathfrak g_2) \simeq \mathrm U(\mathfrak g_1) \otimes_F \mathrm U(\mathfrak g_2). $$ Given an associative $F$-algebra $A$, denote by $[A]$ the corresponding Lie algebra over $F$ defined by the classical Lie bracket, namely $[x,y] = xy-yx$. This is functorial and $(\mathrm U,[-])$ form an adjoint pair. So I thought that the above was a consequence of the fact that $\mathrm U$ was left-adjoint and commuted with small colimits. However, neither of the above are coproducts, namely $\mathfrak g_1 \oplus \mathfrak g_2$ is not the coproduct in the category of Lie algebras and $\mathrm U(\mathfrak g_1) \otimes_F \mathrm U(\mathfrak g_2)$ is not the coproduct in the category of associative $F$-algebras.

So can anyone figure out categorically why this isomorphism is to be expected, or is it just a coincidence? I usually don't like coincidences. The most confusing part is that this isomorphism is constructed "naturally" in some way, namely we take the $F$-linear maps $\mathfrak g_i \to \mathrm U(\mathfrak g_i) \to \mathrm U(\mathfrak g_1) \otimes_F \mathrm U(\mathfrak g_2)$, use the fact that their images commute (which is precisely why $\mathfrak g_1 \oplus \mathfrak g_2$ is not the coproduct in the category of Lie algebras, i.e. the images do not always commute in general) and construct the map of Lie algebras $\mathfrak g_1 \oplus \mathfrak g_2 \to [\mathrm U(\mathfrak g_1) \otimes_F \mathrm U(\mathfrak g_2)]$ which then corresponds to a map $\mathrm U(\mathfrak g_1 \oplus \mathfrak g_2) \to \mathrm U(\mathfrak g_1) \otimes_F \mathrm U(\mathfrak g_2)$ by the $(\mathrm U,[-])$ adjunction.

The inverse map is also constructed naturally, namely by applying the functor $\mathrm U$ on the inclusions $\mathfrak g_i \to \mathfrak g_1 \oplus \mathfrak g_2$ and tensoring the resulting maps over $F$. It turns out that the two are inverse to each other, but I can't naturally figure out why, I have to dive into formulas with elements to see it.

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It is not a coincidence. The tensor product of algebras and the direct sum of Lie algebras are both "commutative coproducts": they are the universal objects admitting a map from two objects whose images commute (this condition makes sense for both algebras and Lie algebras, compatibly with the forgetful functor from algebras to Lie algebras).

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  • $\begingroup$ I understand this informally, but how does one remove the " " in "commutative coproducts"? Can I see them as some kind of colimits to use the fact that left adjoints commute with colimits? I can't think of a diagram which means "the images commute". $\endgroup$ – Patrick Da Silva Dec 7 '16 at 23:50
  • $\begingroup$ @Patrick: I don't know if commutative coproducts can literally be seen as colimits, but you don't need this in order to see the isomorphism in a natural way without working with elements. The point is that both the LHS and the RHS are both universal with respect to admitting a map from $\mathfrak{g}_1$ and $\mathfrak{g}_2$ whose images commute; that is, commutative coproducts are a convenient way of seeing why the LHS and the RHS represent the same functor. $\endgroup$ – Qiaochu Yuan Dec 8 '16 at 1:11
  • $\begingroup$ Ah ok, so instead of using adjoints I'd have to go with the good old Yoneda lemma. Thanks! $\endgroup$ – Patrick Da Silva Dec 8 '16 at 1:14

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