Let $\frak{g}$ be a semisimple Lie algebra and $U(\frak{g})$ its universal enveloping algebra. We have a decomposition $\frak{g}=\frak{n}^-\oplus\frak{b}$ where $\frak{b}=\frak{h}\oplus\frak{n}^+$. By PBW we have an isomorphism of vector spaces $$U({\frak{g}})\simeq U({\frak{n}^-})\otimes_k U({\frak{b}}) $$
Now if we extend scalars on the right then we can regard $U({\frak{n}^-})\otimes_k U({\frak{b}})$ as a left $U(\frak{n}^-)$-module. Then the above isomorphism is actually an isomorphism of left $U(\frak{n}^-)$-modules right? Or am I missing something?
The inclusion map $ι$ from $^-$ to $$ induces a homomorphism of algebras $\operatorname{U}(ι)$ from $\operatorname{U}(^-)$ to $\operatorname{U}()$, which allows us to regard $\operatorname{U}(^-)$ as a subalgebra of $\operatorname{U}()$. We can similarly regard $\operatorname{U}()$ as a subalgebra of $\operatorname{U}()$.
The above isomorphism af vector spaces is then given by $$ Φ \colon \operatorname{U}(^-) ⊗_ \operatorname{U}() \to \operatorname{U}() \,, \quad x ⊗ y \mapsto x y \,. $$ The $\operatorname{U}(^-)$-module structure on the left-hand side of this isomorphism is given by $$ x' ⋅ (x ⊗ y) = (x' x) ⊗ y \,, $$ and the module structure on the right-hand side is given by the multiplication of $\operatorname{U}()$. Therefore, $$ Φ(x' ⋅ (x ⊗ y)) = Φ((x' x) ⊗ y) = x' x y = x' ⋅ xy = x' ⋅ Φ(x ⊗ y) \,. $$ This shows that $Φ$ is indeed a homomorphism – and thus isomorphism – of $\operatorname{U}(^-)$-modules. (It as also a homomorphism of right $\operatorname{U}()$-modules, and therefore an isomorphism of $\operatorname{U}(^-)$-$\operatorname{U}()$-bimodules.)