[proofplan]
We prove the correspondence by checking both directions directly through the quotient map. First, an ideal $J \trianglelefteq \mathfrak g$ containing $I$ descends to an ideal $J/I$ of the quotient because the quotient bracket is computed by representatives. Conversely, an ideal $K$ of $\mathfrak g/I$ pulls back under the quotient homomorphism to an ideal $q^{-1}(K)$ of $\mathfrak g$ containing $I$. The two constructions are inverse to each other, and inclusion preservation follows from the definition of quotient subspaces.
[/proofplan]
[step:Show that an ideal containing $I$ descends to an ideal of $\mathfrak g/I$]
Let $J \trianglelefteq \mathfrak g$ be an ideal such that $I \subset J$. Define the quotient subspace
\begin{align*}
J/I := \{x+I \in \mathfrak g/I : x \in J\}.
\end{align*}
This is a $k$-linear subspace of $\mathfrak g/I$ because $J$ is a $k$-linear subspace of $\mathfrak g$ and $I \subset J$.
We verify the ideal property. Let $x+I \in \mathfrak g/I$ and let $y+I \in J/I$, with $x \in \mathfrak g$ and $y \in J$. The Lie bracket in the quotient Lie algebra is defined by
\begin{align*}
[x+I,y+I]_{\mathfrak g/I} := [x,y]_{\mathfrak g}+I.
\end{align*}
Since $J \trianglelefteq \mathfrak g$ and $y \in J$, we have $[x,y]_{\mathfrak g} \in J$. Hence
\begin{align*}
[x+I,y+I]_{\mathfrak g/I} = [x,y]_{\mathfrak g}+I \in J/I.
\end{align*}
Therefore $J/I \trianglelefteq \mathfrak g/I$.
[guided]
Let $J \trianglelefteq \mathfrak g$ be an ideal satisfying $I \subset J$. The quotient object attached to $J$ is the subspace
\begin{align*}
J/I := \{x+I \in \mathfrak g/I : x \in J\}.
\end{align*}
Because $J$ is a $k$-linear subspace of $\mathfrak g$ and $I \subset J$, this is a well-defined $k$-linear subspace of the quotient [vector space](/page/Vector%20Space) $\mathfrak g/I$.
We now check that $J/I$ is closed under bracketing with arbitrary elements of $\mathfrak g/I$. Take an arbitrary coset $x+I \in \mathfrak g/I$, where $x \in \mathfrak g$, and take an arbitrary coset $y+I \in J/I$, where $y \in J$. The quotient Lie bracket is computed by choosing representatives:
\begin{align*}
[x+I,y+I]_{\mathfrak g/I} := [x,y]_{\mathfrak g}+I.
\end{align*}
The point of assuming that $J$ is an ideal of $\mathfrak g$ is exactly that bracketing any element of $\mathfrak g$ with any element of $J$ lands back in $J$. Thus $[x,y]_{\mathfrak g} \in J$, and consequently
\begin{align*}
[x+I,y+I]_{\mathfrak g/I} = [x,y]_{\mathfrak g}+I \in J/I.
\end{align*}
Since the arbitrary element of $\mathfrak g/I$ and the arbitrary element of $J/I$ bracket back into $J/I$, we have proved $J/I \trianglelefteq \mathfrak g/I$.
[/guided]
[/step]
[step:Pull back an ideal of the quotient to an ideal containing $I$]
Let $K \trianglelefteq \mathfrak g/I$. Define the quotient map
\begin{align*}
q:\mathfrak g &\to \mathfrak g/I \\
x &\mapsto x+I.
\end{align*}
Define the inverse image subspace
\begin{align*}
J := q^{-1}(K)=\{x \in \mathfrak g : q(x) \in K\}.
\end{align*}
Since $q$ is $k$-linear and $K$ is a $k$-linear subspace of $\mathfrak g/I$, the set $J$ is a $k$-linear subspace of $\mathfrak g$. Since $q(x)=0+I \in K$ for every $x \in I$, we have $I \subset J$.
We verify that $J$ is an ideal of $\mathfrak g$. Let $a \in \mathfrak g$ and $b \in J$. Then $q(b) \in K$. Since $q$ is a Lie algebra homomorphism,
\begin{align*}
q([a,b]_{\mathfrak g}) = [q(a),q(b)]_{\mathfrak g/I}.
\end{align*}
Because $K \trianglelefteq \mathfrak g/I$ and $q(b) \in K$, the right-hand side belongs to $K$. Therefore $q([a,b]_{\mathfrak g}) \in K$, so $[a,b]_{\mathfrak g} \in q^{-1}(K)=J$. Hence $J \trianglelefteq \mathfrak g$.
[guided]
Now start with an ideal $K \trianglelefteq \mathfrak g/I$ of the quotient. We pull it back to $\mathfrak g$ using the quotient map
\begin{align*}
q:\mathfrak g &\to \mathfrak g/I \\
x &\mapsto x+I.
\end{align*}
Define
\begin{align*}
J := q^{-1}(K)=\{x \in \mathfrak g : q(x) \in K\}.
\end{align*}
Because $q$ is $k$-linear and $K$ is a $k$-linear subspace, the inverse image $J$ is a $k$-linear subspace of $\mathfrak g$: if $x_1,x_2 \in J$ and $\lambda_1,\lambda_2 \in k$, then
\begin{align*}
q(\lambda_1 x_1+\lambda_2 x_2)
= \lambda_1 q(x_1)+\lambda_2 q(x_2) \in K,
\end{align*}
so $\lambda_1 x_1+\lambda_2 x_2 \in J$. Also $I \subset J$, because every $x \in I$ satisfies $q(x)=0+I$, and $0+I \in K$ since $K$ is a subspace.
It remains to prove that $J$ is an ideal. Take $a \in \mathfrak g$ and $b \in J$. Since $b \in J$, the coset $q(b)$ lies in $K$. The quotient map $q$ is a Lie algebra homomorphism, so it preserves brackets:
\begin{align*}
q([a,b]_{\mathfrak g}) = [q(a),q(b)]_{\mathfrak g/I}.
\end{align*}
Because $K$ is an ideal of $\mathfrak g/I$, bracketing the arbitrary element $q(a) \in \mathfrak g/I$ with the element $q(b) \in K$ gives
\begin{align*}
[q(a),q(b)]_{\mathfrak g/I} \in K.
\end{align*}
Thus $q([a,b]_{\mathfrak g}) \in K$, which is exactly the condition that $[a,b]_{\mathfrak g} \in q^{-1}(K)=J$. Since this holds for every $a \in \mathfrak g$ and every $b \in J$, we conclude that $J \trianglelefteq \mathfrak g$.
[/guided]
[/step]
[step:Verify that the two constructions are inverse]
Let $J \trianglelefteq \mathfrak g$ satisfy $I \subset J$. Then
\begin{align*}
q^{-1}(J/I)
&= \{x \in \mathfrak g : q(x) \in J/I\} \\
&= \{x \in \mathfrak g : x+I \in J/I\}.
\end{align*}
If $x \in J$, then $x+I \in J/I$, so $x \in q^{-1}(J/I)$. Conversely, if $x+I \in J/I$, then there exists $y \in J$ such that $x+I=y+I$. Hence $x-y \in I \subset J$, and since $y \in J$ and $J$ is a subspace, $x=(x-y)+y \in J$. Therefore
\begin{align*}
q^{-1}(J/I)=J.
\end{align*}
Now let $K \trianglelefteq \mathfrak g/I$. We prove that
\begin{align*}
q^{-1}(K)/I=K.
\end{align*}
If $x+I \in q^{-1}(K)/I$, then $x \in q^{-1}(K)$, so $x+I=q(x) \in K$. Hence $q^{-1}(K)/I \subset K$. Conversely, if $z \in K$, then $z$ is a coset in $\mathfrak g/I$, so there exists $x \in \mathfrak g$ such that $z=x+I=q(x)$. Since $z \in K$, we have $x \in q^{-1}(K)$, and therefore $z=x+I \in q^{-1}(K)/I$. Thus $K \subset q^{-1}(K)/I$.
Therefore the assignments $J \mapsto J/I$ and $K \mapsto q^{-1}(K)$ are inverse bijections.
[/step]
[step:Check that inclusion is preserved]
Let $J_1,J_2 \trianglelefteq \mathfrak g$ satisfy $I \subset J_1$ and $I \subset J_2$. If $J_1 \subset J_2$, then every coset $x+I \in J_1/I$ has $x \in J_1 \subset J_2$, so $x+I \in J_2/I$. Hence
\begin{align*}
J_1/I \subset J_2/I.
\end{align*}
Conversely, assume $J_1/I \subset J_2/I$. Let $x \in J_1$. Then $x+I \in J_1/I$, so $x+I \in J_2/I$. Therefore there exists $y \in J_2$ such that $x+I=y+I$. Hence $x-y \in I \subset J_2$, and since $J_2$ is a subspace and $y \in J_2$, we get $x=(x-y)+y \in J_2$. Thus $J_1 \subset J_2$.
Combining both implications gives
\begin{align*}
J_1 \subset J_2 \iff J_1/I \subset J_2/I.
\end{align*}
This completes the proof of the correspondence and its inclusion-preserving property.
[/step]
