[proofplan] We prove first that the kernel of a Lie algebra homomorphism is stable under bracketing with arbitrary elements of the domain, which is exactly the ideal condition. For the converse, the only possible bracket on the quotient is forced by the requirement that the quotient map be a homomorphism. We verify that this forced formula is well-defined on cosets, then check that bilinearity, alternation, and the Jacobi identity descend from $\mathfrak g$. [/proofplan]

[step:Show that the kernel is a Lie ideal] Let $\varphi: \mathfrak g \to \mathfrak h$ be a Lie algebra homomorphism. Since $\varphi$ is $k$-linear, its kernel \begin{align*} \ker \varphi := \{u \in \mathfrak g : \varphi(u)=0\} \end{align*} is a $k$-linear subspace of $\mathfrak g$. Let $u \in \ker \varphi$ and $x \in \mathfrak g$. Since $\varphi$ preserves Lie brackets, \begin{align*} \varphi([x,u]_{\mathfrak g}) = [\varphi(x),\varphi(u)]_{\mathfrak h} = [\varphi(x),0]_{\mathfrak h} = 0. \end{align*} Thus $[x,u]_{\mathfrak g} \in \ker \varphi$. Therefore $[\mathfrak g,\ker \varphi]_{\mathfrak g} \subset \ker \varphi$, so $\ker \varphi$ is an ideal of $\mathfrak g$. [/step]

[step:Define the only possible quotient bracket] Let $\mathfrak a \subset \mathfrak g$ be an ideal, and let $\pi: \mathfrak g \to \mathfrak g/\mathfrak a$ denote the quotient map \begin{align*} \pi(x)=x+\mathfrak a. \end{align*} If $\pi$ is to be a Lie algebra homomorphism, then for every $x,y \in \mathfrak g$ the bracket on $\mathfrak g/\mathfrak a$ must satisfy \begin{align*} [\pi(x),\pi(y)]_{\mathfrak g/\mathfrak a} = \pi([x,y]_{\mathfrak g}). \end{align*} Hence the only possible definition is \begin{align*} [x+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a} := [x,y]_{\mathfrak g}+\mathfrak a. \end{align*} It remains to prove that this formula is well-defined and gives a Lie algebra structure. [/step]

[step:Prove that the quotient bracket is well-defined]Suppose $x,x',y,y' \in \mathfrak g$ satisfy \begin{align*} x+\mathfrak a = x'+\mathfrak a, \qquad y+\mathfrak a = y'+\mathfrak a. \end{align*} Then there exist $a,b \in \mathfrak a$ such that $x'=x+a$ and $y'=y+b$. Using bilinearity of the bracket on $\mathfrak g$, \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, $[x,b]_{\mathfrak g} \in \mathfrak a$ and $[a,y]_{\mathfrak g} \in \mathfrak a$. Since $\mathfrak a$ is a linear subspace and $a,b \in \mathfrak a$, the ideal condition also gives $[a,b]_{\mathfrak g} \in \mathfrak a$. Therefore \begin{align*} [x',y']_{\mathfrak g}-[x,y]_{\mathfrak g} \in \mathfrak a, \end{align*} so \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 the chosen representatives.[/step]

[guided]The possible difficulty is that an element of $\mathfrak g/\mathfrak a$ is a coset, not a single vector. Therefore the formula \begin{align*} [x+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a} = [x,y]_{\mathfrak g}+\mathfrak a \end{align*} is valid only if changing representatives does not change the resulting coset. Assume \begin{align*} x+\mathfrak a = x'+\mathfrak a, \qquad y+\mathfrak a = y'+\mathfrak a. \end{align*} By the definition of equality of cosets in a quotient [vector space](/page/Vector%20Space), there are elements $a,b \in \mathfrak a$ such that \begin{align*} x'=x+a, \qquad y'=y+b. \end{align*} We compare the two possible bracket representatives. Bilinearity of the Lie bracket on $\mathfrak g$ gives \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*} Now each error term lies in $\mathfrak a$. Indeed, $b \in \mathfrak a$ and $\mathfrak a$ is an ideal, so $[x,b]_{\mathfrak g} \in \mathfrak a$. Also $a \in \mathfrak a$, so $[a,y]_{\mathfrak g} \in \mathfrak a$. Finally, because $a,b \in \mathfrak a \subset \mathfrak g$, the ideal condition gives $[a,b]_{\mathfrak g} \in \mathfrak a$. Since $\mathfrak a$ is a linear subspace, the sum of these three error terms is in $\mathfrak a$. Hence \begin{align*} [x',y']_{\mathfrak g}-[x,y]_{\mathfrak g} \in \mathfrak a. \end{align*} This is exactly the statement that the two brackets determine the same coset: \begin{align*} [x',y']_{\mathfrak g}+\mathfrak a = [x,y]_{\mathfrak g}+\mathfrak a. \end{align*} So the quotient bracket is well-defined.[/guided]

[step:Verify the Lie algebra axioms on the quotient] The bracket on $\mathfrak g/\mathfrak a$ is bilinear because the bracket on $\mathfrak g$ is bilinear and the quotient operations are defined cosetwise. Explicitly, for $\lambda \in k$ and $x_1,x_2,y \in \mathfrak g$, \begin{align*} [\lambda(x_1+\mathfrak a)+(x_2+\mathfrak a),y+\mathfrak a]_{\mathfrak g/\mathfrak a} &= [\lambda x_1+x_2,y]_{\mathfrak g}+\mathfrak a \\ &= \lambda[x_1,y]_{\mathfrak g}+[x_2,y]_{\mathfrak g}+\mathfrak a \\ &= \lambda[x_1+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a} + [x_2+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a}. \end{align*} Linearity in the second variable is identical, using bilinearity of $[\cdot,\cdot]_{\mathfrak g}$ in the second argument. For alternation, let $x \in \mathfrak g$. Since $[x,x]_{\mathfrak g}=0$, \begin{align*} [x+\mathfrak a,x+\mathfrak a]_{\mathfrak g/\mathfrak a} = [x,x]_{\mathfrak g}+\mathfrak a = 0+\mathfrak a. \end{align*} For the Jacobi identity, let $x,y,z \in \mathfrak g$. Then \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 a, \end{align*} where the last equality is the Jacobi identity in $\mathfrak g$. Therefore $\mathfrak g/\mathfrak a$ is a Lie algebra over $k$. [/step]

[step:Check that the quotient map is a homomorphism and prove uniqueness] For all $x,y \in \mathfrak g$, the definition of the quotient bracket gives \begin{align*} [\pi(x),\pi(y)]_{\mathfrak g/\mathfrak a} = [x+\mathfrak a,y+\mathfrak a]_{\mathfrak g/\mathfrak a} = [x,y]_{\mathfrak g}+\mathfrak a = \pi([x,y]_{\mathfrak g}). \end{align*} Since $\pi$ is already a $k$-[linear map](/page/Linear%20Map) of vector spaces, this proves that $\pi$ is a Lie algebra homomorphism. Now suppose $[\cdot,\cdot]'$ is any Lie bracket on $\mathfrak g/\mathfrak a$ for which $\pi$ is a Lie algebra homomorphism. For arbitrary cosets $x+\mathfrak a$ and $y+\mathfrak a$, \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*} Thus $[\cdot,\cdot]'$ agrees with the bracket constructed above on every pair of elements of $\mathfrak g/\mathfrak a$. Hence the Lie algebra structure on $\mathfrak g/\mathfrak a$ making $\pi$ a homomorphism is unique. [/step]