Proposition 5

The isomorphism $i_{G,e} : \mathrm{gr}\ U(G) \to \mathrm{TS}(L)$ is an algebra isomorphism.

Let $p$ be the mapping $(t, t') \mapsto t \otimes t'$ of $U(G) \times U(G)$ into $U(G \times G)$. Let c be the mapping $(t, t') \mapsto t * t'$ of $U(G) \times U(G)$ into $U(G)$. Let m be the mapping $(g, g') \mapsto gg'$ of $G \times G$ into G. Then by Definition 1

Consider the diagram

where $q$ is the mapping derived from the canonical isomorphism of $\mathrm{TS}(L) \times \mathrm{TS}(L)$ onto $\mathrm{TS}(L \times L)$. By Differentiable and Analytic Manifolds, R, 13.4.6 and 13.3.5, the two squares of the diagram are commutative. Hence by (1) the diagram is commutative. Now $T(m): L \times L \to L$ maps $(x, y)$ to $x + y$ (§ 2, no. 1, Proposition 2 (ii)). By Algebra, Chapter IV, § 5, no. 6, Proposition 7, $\mathrm{TS}(T(m)) \circ q$ is therefore the multiplication of the algebra $\mathrm{TS}(L)$.