Theorem 1

Let $\alpha : L(X) \to A(X)$ be the unique Lie algebra homomorphism extending the canonical injection of $X$ into $A(X)$ (§ 2, no. 2, Proposition 1). Let $\sigma : L(X) \to U(L(X))$ be the canonical mapping of $L(X)$ into its enveloping algebra and let $\beta : U(L(X)) \to A(X)$ be the unique unital algebra homomorphism such that $\beta \circ \sigma = \alpha$ (Chapter I, § 2, no. 1, Proposition 1). Then: (a) $\alpha$ is injective and $\alpha(L(X))$ is a direct factor submodule of $A(X)$. (b) $\beta$ is bijective.

Let B be a unital K-algebra and $\phi$ a mapping of X into B; by Proposition 1 of § 2, no. 2, there exists a Lie algebra homomorphism $\psi : L(X) \to B$ such that $\psi | X = \phi$; by Proposition 1 of Chapter I, § 2, no. 1, there exists a unital algebra homomorphism $\theta : U(L(X)) \to B$ such that $\theta \circ \sigma = \psi$ and hence such that $(\theta \circ \sigma) | X = \phi$. As $\sigma(X)$ generates the unital algebra $U(L(X))$, the homomorphism $\theta$ is the unique unital algebra homomorphism satisfying the latter condition. This shows that the ordered pair $(U(L(X)), \sigma | X)$ is a solution of the same universal mapping problem as $A(X)$; taking $\phi$ to be the canonical injection of X into $A(X)$, we deduce that $\beta$ is an isomorphism, which proves (b).

Finally, as $L(X)$ is a free K-module (§ 2, no. 11, Corollary to Theorem 1), $\sigma$ is injective and $\sigma(L(X))$ is a direct factor submodule of $U(L(X))$ (Chapter I, § 2, no. 7, Corollary 3 to Theorem 1). By (b), this proves (a).