The universal enveloping algebra of free Lie algebra is the tensor algebra on the Free Abelian Group?

Question

Let $A$ be a set, and let $M_A$ be the free abelian group generated by $A$. Let $L(A)$ be the free Lie algebra generated by $A$. I am reading On Injective Homomorphisms For Pure Braid Groups, And Associated Lie Algebras By F. R. Cohen And Stratos Prassidis (arXiv:math/0404278v1) and here is an extract:

Proof. Let $M_A$ be free abelian group generated by $A$ with $a \in A$, and $A$ of cardinality at least $2$. The universal enveloping algebra of $L[A]$ is the tensor algebra on $M_A$, $T[M_A]$, and the standard Lie algebra homomorphism: $$j : L[A] \to T[M_A]$$ is injective ([1], and [9], p. 168).

I tried to look for the references but 1 was not English and it seems to me that 9 was not relevant. Why is the universal enveloping algebra of $L(A)$ the tensor algebra on $M_A$? And why is $j$ one-to-one?

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For reference:

• F. R. Cohen, S. Prassidis. On injective homomorphisms for pure braid groups, and associated Lie algebras. J. Algebra 298 (2006), no. 2, 363–370.
• [1] N. Bourbaki, Groupes et algèbres de Lie, Hermann, Paris, 1972 (chapitres 2–3).
• [9] N. Jacobson, Lie Algebras, Interscience Tracts in Pure Appl. Math., vol. 10, Wiley–Interscience, New York, 1962.

Answer 1

This is because of universal properties: for every associative algebra $X$, there is a sequence of isomorphisms, natural in $X$: $$\begin{align} \hom_{\mathsf{Alg}}(\mathcal U(L[A]), X) & \cong \hom_{\mathsf{Lie}}(L[A], X) \\ & \cong \hom_{\mathsf{Set}}(A, X) \\ & \cong \hom_{\mathsf{Ab}}(M_A, X) \\ & \cong \hom_{\mathsf{Alg}}(T[M_A], X) \end{align}$$ (where I didn't write the forgetful functors on $X$). Thus because of the Yoneda lemma, $$\mathcal U(L[A]) \cong T[M_A].$$

Injectivity of $j$ follows from the Poincaré–Birkhoff–Witt theorem (this is written explicitly in the published version of the article), which says that over a field and for any Lie algebra $\mathfrak{g}$, the canonical map $\mathfrak{g} \to \mathcal{U}(\mathfrak{g})$ is injective. Apply this to $L[A]$ and use the isomorphism.

Answer 2

This is a purely categorical fact:

1) The free Lie algebra functor $\text{Set}\to\text{Lie}_{\mathbb k}$ is left adjoint to the forgetful functor $\text{Lie}_{\mathbb k}\to\text{Set}$.

2) The universal enveloping algebra functor $\text{Lie}_{\mathbb k}\to\text{Alg}_{\mathbb k}$ is left adjoint to the forgetful functor $\text{Alg}_{\mathbb k}\to\text{Lie}_{\mathbb k}$.

Hence, the composition of both is left adjoint to the forgetful functor $\text{Alg}_{\mathbb k}\to\text{Set}$.

However, you can also factor this forgetful functor as the composition of $\text{Alg}_{\mathbb k}\to\text{Ab}\to\text{Set}$, and so a left adjoint is also given by the composition of the free abelian group functor $\text{Set}\to\text{Ab}$ and the ${\mathbb k}$-tensor algebra functor $\text{Ab}\to\text{Alg}_{\mathbb k}$.

By uniqueness of adjoints, your statement follows. The uniqueness of adjoints is a consequence of the Yoneda lemma, and writing out the details you will indeed end up with Najib's solution, so the arguments are essentially the same.