Birkhoff-Witt theorem and left adjoint of the embedding functor

Question

The Birkhoff-Witt Theorem asserts that every Lie algebra $\mathfrak g$ over the field $K$ may be embedded in an associative $K$-algebra $U\mathfrak g$ in such way that the Lie bracket $[x,y]$ coincide with $xy-yx$ in $U\mathfrak g$, $x,y\in\mathfrak g$, and such that to every associative $K$-algebra $A$ and every $K$-linear map $f:\mathfrak g\to A$ with \begin{equation*} f[x,y]=f(x)f(y)-f(y)f(x),\qquad x,y\in\mathfrak g \end{equation*} there exists a unique $K$-algebra homomorphism $f^*:U\mathfrak g\to A$ extending $f$. I want to express this theorem in the language adjoint functors.

Let $\mathfrak L_K$ be the category of Lie algebras over $K$ and $\mathfrak A_K$ be category of associated $K$-algebras over $K$. Since every $K$-algebra is also a Lie algebra, we have a natural (full) embedding $E:\mathfrak A_K\to\mathfrak L_K$. I guess the Birkoff-Witt theorem is equivalenct to say that $E$ has a left adjugant $U$.

If $E$ has a left adjoint $U:\mathfrak L_K\to\mathfrak A_K$, denote the adjugant by $\eta$ and the unit by $\epsilon:1\to EU$. Then if $A$ is an associative $K$-algebra and $f:\mathfrak g\to A=EA$ is a homomorphism of Lie algebras over $K$, then we have $f^*:=\eta^{-1}f\in\mathfrak A_K(U\mathfrak g,A)$. By definition, $f^*\circ\epsilon_{\mathfrak g}=Ef^*\circ\epsilon_{\mathfrak g}=\eta(f^*)=f$. On the other hand, it's also clear that $f^*$ is the only morphism such that $Ef^*\circ\epsilon_{\mathfrak g}=f$.

However, I have got some trouble to prove that $\epsilon$ is injective (and thus $\mathfrak g\to U\mathfrak g$ is an embedding). I think it is just impossible to show that $\epsilon$ is injective only by category theory, say I may need something from Lie algebra. But I'm not familiar with Lie algebras.

Any advice will be appreciated!

Answer 1

There are several mistakes here.

1. This is not what the PBW theorem says. You can see a correct statement on Wikipedia. The statement that the canonical map $\mathfrak{g} \to U(\mathfrak{g})$ is an embedding is a corollary of PBW but the full theorem is stronger.

2. $\mathfrak{g} \mapsto U(\mathfrak{g})$ is the left adjoint of the forgetful functor from associative algebras to Lie algebras given by taking the commutator bracket. This follows more or less by definition and does not require the PBW theorem which is harder.

3. The forgetful functor is not full; for example it sends every commutative algebra to an abelian Lie algebra with trivial bracket, and there are many more Lie algebra morphisms between these than commutative algebra morphisms.

You're correct that it's not possible to show that $\mathfrak{g} \to U(\mathfrak{g})$ is always an embedding using pure category theory (for example it is false for Lie algebras over arbitrary commutative rings); this is a corollary of PBW.