Proposition 3

Let $h$ be an ideal of $g$, $p$ the canonical homomorphism of $g$ onto $g/h$ and $W$ the enveloping algebra of $g/h$. The homomorphism:

defined canonically by $p$ is surjective and its kernel is the ideal $R$ of $U$ generated by $\sigma(h)$.

Let $\sigma''$ be the canonical mapping of $g/\mathfrak{h}$ into W. The commutative diagram:

proves that $\tilde{p}$ is zero on $\sigma(\mathfrak{h})$ and hence on R. Let $\psi$ be the canonical homomorphism of U onto U/R. There exists a homomorphism $\phi$ of U/R

into W such that $\tilde{p} = \phi \circ \psi$. The mapping $\psi \circ \sigma$ of g into U/R is an $\alpha$-mapping and is zero on $\mathfrak{h}$ and hence defines an $\alpha$-mapping $\theta$ of $g/\mathfrak{h}$ into U/R such that $\theta \circ p = \psi \circ \sigma$. Then $\phi \circ \theta \circ p = \phi \circ \psi \circ \sigma = \sigma'' \circ p$. Hence $\phi \circ \theta = \sigma''$. There exists (no. 1, Proposition 1) one and only one homomorphism $\phi'$ of W into U/R mapping 1 to 1 and such that $\theta = \phi' \circ \sigma''$. Then $\phi' \circ \phi \circ \theta = \phi' \circ \sigma'' = \theta$ and $\phi \circ \phi' \circ \sigma'' = \phi \circ \theta = \sigma''$ and hence $\phi' \circ \phi$ and $\phi \circ \phi'$ are the identity mappings of U/R and W respectively. This completes the proof.

U/R is identified with W under the isomorphism $\phi$. Then the canonical mapping $\sigma''$ of $g/\mathfrak{h}$ into W is identified with $\theta$, that is with the mapping of $g/\mathfrak{h}$ into U/R derived from $\sigma$ by taking quotients.