Corollary 2

Let g, $g'$ be two Lie algebras, r and $r'$ their radicals and f a homomorphism of g onto $g'$. Then $r' = f(r)$.

As $f(r)$ is solvable, $f(r) \subset r'$. On the other hand, $g/r$ is semi-simple (§ 5, no. 2, Proposition 3), hence $g'/f(r)$, which is isomorphic to a quotient of $g/r$, is semi-simple (Lemma 1) and hence $f(r) \supset r'$ (§ 5, no. 2, Proposition 3).