Suppose $I$ and $J$ are ideals of a Lie Algebra L. I know that we have the fact that:
$\frac{I+J}{J} \cong \frac{I}{I\cap J}$
Prove that the ideals of $\frac{L}{I}$ - the quotient algebra of L defined by $x + I$ $x \in L$ are of the form $\frac{K}{I}$ where K is an ideal of L containing I.
If $f\colon L\to L'$ is a surjective homomorphism of Lie algebras, then, for any ideal $I$ of $L$,
$$ f^\to(K)=\{f(x):x\in K\} $$
is an ideal of $L'$.
For any ideal $K'$ of $L'$, $$ f^\gets(K')=\{x\in L:f(x)\in K'\} $$ is an ideal of $L$ (here surjectivity is not needed), containing $\ker f$.
If an ideal $K$ of $L$ contains $\ker f$, then $f^\gets(f^\to(K))=K$. Moreover, for any ideal $K'$ of $L'$, $f^\to(f^\gets(K'))=K'$. The proofs of these two statements are easy.
Therefore there is a bijection, given by $f^\to$ and $f^\gets$ between the ideals of $L$ containing $\ker f$ and the ideals of $L'$.
Apply this to the canonical homomorphism $L\to L/I$, when $I$ is an ideal of $L$.