Proposition 10

Tag 0A2A, permanent. Cite it as [Bourbaki, Tag 0A2A].

Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

The mapping $\lambda \mapsto f_{\lambda}$ is an isomorphism of the algebra $U'$ onto the algebra of formal power series $K[[X_i]]_{i \in I}$ .

Because $(e_{\alpha})$ is a basis of U, the mapping $\lambda \mapsto f_{\lambda}$ is K-linear and bijective. On the other hand, for $\lambda, \mu$ in $U'$ ,

\begin{align*} f_{\lambda \mu} &= \sum_{\alpha} \langle \lambda \mu, e_{\alpha} \rangle X^{\alpha} = \sum_{\alpha} \langle \lambda \otimes \mu, c(e_{\alpha}) \rangle X^{\alpha} \\ &= \sum_{\alpha} \langle \lambda \otimes \mu, \sum_{\beta + \gamma = \alpha} e_{\beta} \otimes e_{\gamma} \rangle X^{\alpha} \tag{by (16)} \\ &= \sum_{\beta, \gamma} \langle \lambda, e_{\beta} \rangle \langle \mu, e_{\gamma} \rangle X^{\beta + \gamma} = f_{\lambda} f_{\mu}, \end{align*}

which shows that $\lambda \mapsto f_{\lambda}$ is an algebra isomorphism and completes the proof.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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