[proofplan]
First we prove that the kernel of a Lie algebra homomorphism is a linear subspace and is stable under bracketing with arbitrary elements of the domain. Then, for an ideal $\mathfrak a \subset \mathfrak g$, we define the bracket on cosets by taking representatives and prove that this definition is independent of the chosen representatives. Once well-definedness is established, bilinearity, alternation, and the Jacobi identity descend from $\mathfrak g$, and the quotient map is automatically a Lie algebra homomorphism. Finally, uniqueness follows because any Lie bracket making the quotient map a homomorphism must agree on all cosets with the displayed formula.
[/proofplan]
[step:Show that the kernel is a linear subspace stable under the Lie bracket]
Let $0_{\mathfrak h}$ denote the zero vector of $\mathfrak h$. Define
\begin{align*}
\ker \varphi := \{x \in \mathfrak g : \varphi(x) = 0_{\mathfrak h}\}.
\end{align*}
Since $\varphi: \mathfrak g \to \mathfrak h$ is $k$-linear, $\ker \varphi$ is a $k$-linear subspace of $\mathfrak g$: if $u,v \in \ker \varphi$ and $\lambda,\mu \in k$, then
\begin{align*}
\varphi(\lambda u + \mu v)
=
\lambda \varphi(u) + \mu \varphi(v)
=
\lambda 0_{\mathfrak h} + \mu 0_{\mathfrak h}
=
0_{\mathfrak h}.
\end{align*}
It remains to verify the ideal condition. Let $x \in \mathfrak g$ and let $y \in \ker \varphi$. Because $\varphi$ is a Lie algebra homomorphism, it preserves brackets:
\begin{align*}
\varphi([x,y]_{\mathfrak g})
=
[\varphi(x),\varphi(y)]_{\mathfrak h}.
\end{align*}
Since $y \in \ker \varphi$, we have $\varphi(y) = 0_{\mathfrak h}$. Bilinearity of the bracket on $\mathfrak h$ gives
\begin{align*}
[\varphi(x),\varphi(y)]_{\mathfrak h}
=
[\varphi(x),0_{\mathfrak h}]_{\mathfrak h}
=
0_{\mathfrak h}.
\end{align*}
Therefore $\varphi([x,y]_{\mathfrak g}) = 0_{\mathfrak h}$, so $[x,y]_{\mathfrak g} \in \ker \varphi$. Hence $\ker \varphi$ is an ideal of $\mathfrak g$.
[/step]
[step:Define the quotient bracket and prove it is independent of representatives]
Assume now that $\mathfrak a$ is an ideal of $\mathfrak g$. Let $\mathfrak g / \mathfrak a$ denote the quotient [vector space](/page/Vector%20Space), and let
\begin{align*}
\pi: \mathfrak g &\to \mathfrak g / \mathfrak a \\
x &\mapsto x + \mathfrak a
\end{align*}
be the quotient map. Define a candidate bracket
\begin{align*}
[\cdot,\cdot]_{\mathfrak g / \mathfrak a}: (\mathfrak g / \mathfrak a) \times (\mathfrak g / \mathfrak a) &\to \mathfrak g / \mathfrak a \\
(x+\mathfrak a, y+\mathfrak a) &\mapsto [x,y]_{\mathfrak g} + \mathfrak a.
\end{align*}
We prove that this rule is well-defined. Let $x,x',y,y' \in \mathfrak g$ satisfy
\begin{align*}
x+\mathfrak a = x' + \mathfrak a,
\qquad
y+\mathfrak a = y' + \mathfrak a.
\end{align*}
Then $x'-x \in \mathfrak a$ and $y'-y \in \mathfrak a$. Define $a := x'-x \in \mathfrak a$ and $b := y'-y \in \mathfrak a$, so $x' = x+a$ and $y' = y+b$. By bilinearity of the bracket in $\mathfrak g$,
\begin{align*}
[x',y']_{\mathfrak g}
&=
[x+a,y+b]_{\mathfrak g} \\
&=
[x,y]_{\mathfrak g}
+
[x,b]_{\mathfrak g}
+
[a,y]_{\mathfrak g}
+
[a,b]_{\mathfrak g}.
\end{align*}
Because $\mathfrak a$ is an ideal, $[x,b]_{\mathfrak g} \in \mathfrak a$, $[y,a]_{\mathfrak g} \in \mathfrak a$, and $[a,b]_{\mathfrak g} \in \mathfrak a$. Since Lie brackets are alternating, $[a,y]_{\mathfrak g} = -[y,a]_{\mathfrak g}$, hence $[a,y]_{\mathfrak g} \in \mathfrak a$. As $\mathfrak a$ is a linear subspace, the sum
\begin{align*}
[x,b]_{\mathfrak g}
+
[a,y]_{\mathfrak g}
+
[a,b]_{\mathfrak g}
\end{align*}
belongs to $\mathfrak a$. Therefore
\begin{align*}
[x',y']_{\mathfrak g} - [x,y]_{\mathfrak g} \in \mathfrak a,
\end{align*}
which means
\begin{align*}
[x',y']_{\mathfrak g}+\mathfrak a
=
[x,y]_{\mathfrak g}+\mathfrak a.
\end{align*}
Thus the bracket on $\mathfrak g/\mathfrak a$ is independent of representatives.
[/step]
[step:Verify that the quotient bracket satisfies the Lie algebra axioms]
We now prove that $\mathfrak g/\mathfrak a$ with the bracket just defined is a Lie algebra over $k$.
First, bilinearity follows from bilinearity of the bracket on $\mathfrak g$. For $x_1,x_2,y \in \mathfrak g$ and $\lambda,\mu \in k$,
\begin{align*}
[\lambda(x_1+\mathfrak a)+\mu(x_2+\mathfrak a), y+\mathfrak a]_{\mathfrak g/\mathfrak a}
&=
[(\lambda x_1+\mu x_2)+\mathfrak a, y+\mathfrak a]_{\mathfrak g/\mathfrak a} \\
&=
[\lambda x_1+\mu x_2,y]_{\mathfrak g}+\mathfrak a \\
&=
\lambda[x_1,y]_{\mathfrak g}+\mu[x_2,y]_{\mathfrak g}+\mathfrak a \\
&=
\lambda[x_1+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a}
+
\mu[x_2+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a}.
\end{align*}
Linearity in the second variable is proved by the same bilinearity computation in the second argument.
Second, alternation descends from $\mathfrak g$. For every $x \in \mathfrak g$,
\begin{align*}
[x+\mathfrak a,x+\mathfrak a]_{\mathfrak g/\mathfrak a}
=
[x,x]_{\mathfrak g}+\mathfrak a
=
0_{\mathfrak g}+\mathfrak a,
\end{align*}
which is the zero vector of $\mathfrak g/\mathfrak a$.
Third, the Jacobi identity descends from $\mathfrak g$. For $x,y,z \in \mathfrak g$,
\begin{align*}
&[x+\mathfrak a,[y+\mathfrak a,z+\mathfrak a]_{\mathfrak g/\mathfrak a}]_{\mathfrak g/\mathfrak a}
+
[y+\mathfrak a,[z+\mathfrak a,x+\mathfrak a]_{\mathfrak g/\mathfrak a}]_{\mathfrak g/\mathfrak a} \\
&\qquad
+
[z+\mathfrak a,[x+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a}]_{\mathfrak g/\mathfrak a} \\
&=
[x,[y,z]_{\mathfrak g}]_{\mathfrak g}+\mathfrak a
+
[y,[z,x]_{\mathfrak g}]_{\mathfrak g}+\mathfrak a
+
[z,[x,y]_{\mathfrak g}]_{\mathfrak g}+\mathfrak a \\
&=
\left(
[x,[y,z]_{\mathfrak g}]_{\mathfrak g}
+
[y,[z,x]_{\mathfrak g}]_{\mathfrak g}
+
[z,[x,y]_{\mathfrak g}]_{\mathfrak g}
\right)+\mathfrak a \\
&=
0_{\mathfrak g}+\mathfrak a,
\end{align*}
where the final equality is the Jacobi identity in $\mathfrak g$. Therefore $\mathfrak g/\mathfrak a$ is a Lie algebra over $k$.
[/step]
[step:Show that the quotient map is a Lie algebra homomorphism]
The map $\pi: \mathfrak g \to \mathfrak g/\mathfrak a$ is $k$-linear by the construction of the quotient vector space. For $x,y \in \mathfrak g$, the definition of the quotient bracket gives
\begin{align*}
\pi([x,y]_{\mathfrak g})
=
[x,y]_{\mathfrak g}+\mathfrak a
=
[x+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a}
=
[\pi(x),\pi(y)]_{\mathfrak g/\mathfrak a}.
\end{align*}
Thus $\pi$ preserves Lie brackets, so $\pi$ is a Lie algebra homomorphism.
[/step]
[step:Prove uniqueness of the quotient Lie algebra structure]
Let $[\cdot,\cdot]'$ be any Lie bracket on the quotient vector space $\mathfrak g/\mathfrak a$ for which $\pi: \mathfrak g \to \mathfrak g/\mathfrak a$ is a Lie algebra homomorphism. Every element of $\mathfrak g/\mathfrak a$ has the form $\pi(x)=x+\mathfrak a$ for some $x \in \mathfrak g$. Therefore, for all $x,y \in \mathfrak g$,
\begin{align*}
[x+\mathfrak a,y+\mathfrak a]'
=
[\pi(x),\pi(y)]'
=
\pi([x,y]_{\mathfrak g})
=
[x,y]_{\mathfrak g}+\mathfrak a.
\end{align*}
Hence $[\cdot,\cdot]'$ agrees with the bracket constructed above on every ordered pair of elements of $\mathfrak g/\mathfrak a$. The quotient Lie algebra structure is therefore unique.
[/step]
