Let $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ be Lie algebras and
$$ \varphi : \mathfrak{g}_{1} \rightarrow \mathfrak{g}_{2} $$
be a homomorphism of Lie algebras. Somebody could tell me (proof) a necessary and sufficient condition on $\varphi$ to ensure that $\mbox{Im}\varphi$ be an ideal of $\mathfrak{g}_{2}$?
I would be happy with any suggestion. Thank you!
It is easy to give a sufficient condition, mamely that $\phi$ is surjective. In this case, $\phi(I)$ is an ideal in $\mathfrak{g}_2$ for any ideal $I$ in $\mathfrak{g}_1$, because $$ [\mathfrak{g}_2,\phi(I)]=[\phi(\mathfrak{g}_1),\phi(I)]=\phi([\mathfrak{g}_1,I])\subseteq \phi(I). $$
However, this is not a necessary condition. For, say, $\phi=0$ and $\mathfrak{g}_2\neq 0$, we have that $\phi(I)$ is an ideal, although $\phi$ is not surjective.