Proposition 1

Let $\sigma$ be an $\alpha$-mapping of $g$ into the associative algebra $L$ with unit element. There exists one and only one homomorphism $\tau$ of $U$ into $L$, mapping 1 to 1, such that $\sigma = \tau \circ \sigma_0$, where $\sigma_0$ denotes the canonical mapping of $g$ into $U$.

Let $\tau'$ be the unique homomorphism of $T$ into $L$ which extends $\sigma$ and maps 1 to 1. Then, for $x, y$ in $g$,

hence $\tau'$ is zero on $J$ and defines on passing to the quotient a homomorphism $\tau$ of $U$ into $L$, mapping 1 to 1, such that $\sigma = \tau \circ \sigma_0$. The uniqueness of $\tau$ is immediate since $\sigma_0(g)$ and 1 generate the algebra $U$.

Let $g'$ be another Lie algebra over $K$, $U'$ its enveloping algebra and $\sigma'_0$ the canonical mapping of $g'$ into $U'$. Let $\phi$ be a homomorphism of $g$ into $g'$. Then $\sigma'_0 \circ \phi$ is an $\alpha$-mapping of $g$ into $U'$; hence there exists one and only one homomorphism $\tilde{\phi}$ of $U$ into $U'$ mapping 1 to 1 and such that the diagram

is commutative. This homomorphism maps the elements of $U$ whose constant term is zero to elements of $U'$ whose constant term is zero. If $g''$ is another Lie algebra over $K$ and $\phi'$ is a homomorphism of $g'$ into $g''$, then $(\phi' \circ \phi)^{\sim} = \tilde{\phi}' \circ \tilde{\phi}$.