Bourbaki

Theorem 1

Tag 0A2B, permanent. Cite it as [Bourbaki, Tag 0A2B].

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

Let E be a cocommutative bigebra. (a) The bigebra morphism fE:U(P(E))→Ef_E : U(P(E)) \to E is injective. (b) If there exists on E a filtration compatible with its bigebra structure (no. 3, Definition 2), the morphism fEf_E is an isomorphism. (In case (b), the bigebra E is therefore identified with the enveloping bigebra of the Lie algebra of its primitive elements.) Let cEc_E (resp. εE\varepsilon_E) be the coproduct (resp. counit) of E. We write g=P(E)g = P(E); let (ei)i∈I(e_i)_{i \in I} be a basis of the vector K-space g, where the indexing set I is totally ordered, and let (eα)α∈N(I)(e_\alpha)_{\alpha \in \mathbf{N}^{(I)}} be the basis introduced in the preceding no. We write Xα=fE(eα)X_\alpha = f_E(e_\alpha) for α∈N(I)\alpha \in \mathbf{N}^{(I)}. By (15) and (16), we have:

εE(X0)=1,εE(Xα)=0for ∣α∣≥1, \varepsilon_E(X_0) = 1, \quad \varepsilon_E(X_\alpha) = 0 \quad \text{for } |\alpha| \geq 1,
cE(Xα)=∑β+γ=αXβ⊗Xγfor α∈N(I), c_E(X_\alpha) = \sum_{\beta + \gamma = \alpha} X_\beta \otimes X_\gamma \quad \text{for } \alpha \in \mathbf{N}^{(I)},
since fEf_E is a cogebra morphism. We show that fEf_E is injective. This results from the following lemma:

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

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