Enveloping algebra of a direct sum

Define $\varphi :\mathfrak{g}\oplus \mathfrak{h}\to (U(\mathfrak{g})\otimes U(\mathfrak{h}){)}_{\mathrm{L}\mathrm{i}\mathrm{e}}$ by $\varphi (x,y)={\iota }_{\mathfrak{g}}(x)\otimes 1+1\otimes {\iota }_{\mathfrak{h}}(y)$. Same-summand commutators give the respective Lie brackets, while the two tensor factors commute, so [L1] makes $\varphi$ a Lie map. By [L2] it extends uniquely to an algebra map $\mathrm{\Phi }:U(\mathfrak{g}\oplus \mathfrak{h})\to U(\mathfrak{g})\otimes U(\mathfrak{h})$.

Let $\alpha :U(\mathfrak{g})\to U(\mathfrak{g}\oplus \mathfrak{h})$ and $\beta :U(\mathfrak{h})\to U(\mathfrak{g}\oplus \mathfrak{h})$ be induced by the summand inclusions. Their generator images commute by [L1] and the canonical enveloping relation, hence all of $\alpha (U(\mathfrak{g}))$ commutes with all of $\beta (U(\mathfrak{h}))$. Therefore $(a,b)\mapsto \alpha (a)\beta (b)$ is bilinear and [L3] gives a linear map $\mathrm{\Psi }:U(\mathfrak{g})\otimes U(\mathfrak{h})\to U(\mathfrak{g}\oplus \mathfrak{h})$.

Commutation of the two images gives $\mathrm{\Psi }((a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }}))=\alpha (a{a}^{\mathrm{\prime }})\beta (b{b}^{\mathrm{\prime }})=\alpha (a)\beta (b)\alpha (a^{\mathrm{\prime }})\beta (b^{\mathrm{\prime }})$, so $\mathrm{\Psi }$ is a unital algebra homomorphism.

The composite $\mathrm{\Psi }\mathrm{\Phi }$ fixes the canonical images of $(x,0)$ and $(0,y)$, hence is the identity on $U(\mathfrak{g}\oplus \mathfrak{h})$ by uniqueness in [L2]. The composites $\mathrm{\Phi }\alpha$ and $a\mapsto a\otimes 1$ agree on ${\iota }_{\mathfrak{g}}(\mathfrak{g})$, and similarly $\mathrm{\Phi }\beta (b)=1\otimes b$; thus $\mathrm{\Phi }\mathrm{\Psi }$ fixes every pure tensor $(a\otimes 1)(1\otimes b)=a\otimes b$, hence is the identity.

Therefore $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse unital algebra isomorphisms, including when either summand is zero.