Proposition 5

The mapping $\phi$ of $T$ onto $G$ is an algebra homomorphism mapping 1 to 1 and is zero on the two-sided ideal generated by the tensors $x \otimes y - y \otimes x$ ($x \in g, y \in g$).

If $t \in T^n$ and $t' \in T^p$, then $\phi(t)\phi(t') = \phi(tt')$ by definition of the multiplication on $G$. Hence $\phi$ is an algebra homomorphism and clearly $\phi(1) = 1$. If $x, y$ are in $g$, then $x \otimes y - y \otimes x \in T^2$ and the canonical image of this element in $U_2$ is equal to that of $[x, y]$ and therefore belongs to $U_1$. Hence $\phi(x \otimes y - y \otimes x) = 0$, which proves the proposition.

Let $S$ be the symmetric algebra of the $K$-module $g$ and $\tau$ the canonical homomorphism of $T$ onto $S$. Proposition 5 proves that there exists a unique homomorphism $\omega$, called canonical, of the algebra $S$ onto the algebra $G$, mapping 1 to 1, such that $\phi = \omega \circ \tau$. We have $\omega(S^n) = \phi(T^n) = G^n$. Let $\tau_n$ be the restriction of $\tau$ to $T^n$, $\omega_n$ the restriction of $\omega$ to $S^n$, $\psi_n$ the canonical mapping of $T^n$ into $U_n$ and $\theta_n$ the canonical mapping of $U_n$ onto $G^n$. The definition of $\omega_n$ proves that the following diagram is commutative: