Proposition 12
Tag 0A9O, permanent. Cite it as [Bourbaki, Tag 0A9O].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
Let G, H be two Lie groups and a morphism of G into H. Then is a bigebra morphism of into .
This follows from Proposition 6 and Differentiable and Analytic Manifolds, R, 13.5.1.
Let G be a Lie group. The restrictions of the convolution and the coproduct to U(G) define a bigebra structure on U(G). We have U(G)^{\vee} = U(G). If is a Lie group morphism, we denote by U() the mapping of U(G) into U(H); this is a bigebra morphism. If is another Lie group morphism, then U() = U() \circ U(). If is an immersion (resp. a submersion), U() is injective (resp. surjective) by Differentiable and Analytic Manifolds, R, 13.2.3. In particular, if H is a Lie subgroup of G, U(H) is identified with a subalgebra of U(G), the coproduct on U(H) being the restriction of the coproduct on U(G). If H is open in G, then U(H) = U(G). If G_1, G_2 are Lie groups, U(G_1 \times G_2) is identified with U(G_1) \times U(G_2). The primitive elements of U(G) are those of T_e(G) (Differentiable and Analytic Manifolds, R, 13.5.3).
Again let be a Lie group morphism. If gr U(G) is identified TS(T_e(G)) and gr U(H) with TS(T_e(H)), then gr U() is identified with TS(T_e()) (Differentiable and Analytic Manifolds, R, 13.3.5). We apply this to the isomorphism of G onto G^{\vee}; then T_e() = -1 and hence
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Nothing.