[proofplan] We first prove directly that the derived subspace $[\mathfrak g,\mathfrak g]$ is an ideal, using the Jacobi identity to show that bracketing a generator $[x,y]_{\mathfrak g}$ with an arbitrary element remains in the span of brackets. This permits the [quotient Lie algebra](/theorems/8146) structure, and the quotient is abelian because every bracket $[x,y]_{\mathfrak g}$ becomes zero modulo $[\mathfrak g,\mathfrak g]$. Finally, if $\phi:\mathfrak g\to\mathfrak a$ maps into an abelian [Lie algebra](/page/Lie%20Algebra), then $\phi$ kills every bracket and hence kills $[\mathfrak g,\mathfrak g]$; this gives a well-defined induced map on the quotient, whose uniqueness and Lie algebra homomorphism property are checked from representatives. [/proofplan]

[step:Show that the derived subspace is an ideal] Let $D:=[\mathfrak g,\mathfrak g]$, so $D$ is the $F$-linear span of all elements $[x,y]_{\mathfrak g}$ with $x,y\in\mathfrak g$. We prove that $D$ is an ideal of $\mathfrak g$. It is an $F$-linear subspace by definition. Let $z\in\mathfrak g$ and let $[x,y]_{\mathfrak g}$ be a spanning generator of $D$. The Jacobi identity in $\mathfrak g$ gives \begin{align*} [z,[x,y]_{\mathfrak g}]_{\mathfrak g}+[x,[y,z]_{\mathfrak g}]_{\mathfrak g}+[y,[z,x]_{\mathfrak g}]_{\mathfrak g}=0. \end{align*} Using alternation, equivalently $[u,v]_{\mathfrak g}=-[v,u]_{\mathfrak g}$ for all $u,v\in\mathfrak g$, this becomes \begin{align*} [z,[x,y]_{\mathfrak g}]_{\mathfrak g}=[[z,x]_{\mathfrak g},y]_{\mathfrak g}+[x,[z,y]_{\mathfrak g}]_{\mathfrak g}. \end{align*} Both terms on the right are brackets of elements of $\mathfrak g$, so they belong to $D$. Hence $[z,[x,y]_{\mathfrak g}]_{\mathfrak g}\in D$. Now let $d\in D$. By definition of linear span, there exist an integer $m\geq 1$, scalars $c_1,\dots,c_m\in F$, and elements $x_1,\dots,x_m,y_1,\dots,y_m\in\mathfrak g$ such that \begin{align*} d=\sum_{i=1}^{m}c_i[x_i,y_i]_{\mathfrak g}. \end{align*} By bilinearity of the Lie bracket, \begin{align*} [z,d]_{\mathfrak g}=\sum_{i=1}^{m}c_i[z,[x_i,y_i]_{\mathfrak g}]_{\mathfrak g}. \end{align*} Each summand belongs to $D$ by the generator computation above, and $D$ is a linear subspace, so $[z,d]_{\mathfrak g}\in D$. Thus $D$ is an ideal of $\mathfrak g$. [/step]

[step:Compute the quotient bracket and prove that it is zero]Since $D=[\mathfrak g,\mathfrak g]$ is an ideal, the quotient Lie algebra $\mathfrak g/D$ is defined by [citetheorem:8146]. Let \begin{align*} q:\mathfrak g\to \mathfrak g/D \end{align*} be the quotient map, so $q(x)=x+D$ for $x\in\mathfrak g$. Take arbitrary cosets $x+D,y+D\in\mathfrak g/D$. By the definition of the quotient bracket, \begin{align*} [x+D,y+D]_{\mathfrak g/D}=[x,y]_{\mathfrak g}+D. \end{align*} But $[x,y]_{\mathfrak g}\in D$ by the definition of $D$, so \begin{align*} [x,y]_{\mathfrak g}+D=D. \end{align*} The coset $D$ is the zero element of the [vector space](/page/Vector%20Space) $\mathfrak g/D$. Hence \begin{align*} [x+D,y+D]_{\mathfrak g/D}=0_{\mathfrak g/D}. \end{align*} Since the two cosets were arbitrary, $\mathfrak g/D$ is abelian.[/step]

[guided]The quotient $\mathfrak g/D$ is a Lie algebra only after we know that $D$ is an ideal. That was the purpose of the previous step: it verifies the hypothesis needed to use the quotient Lie algebra construction [citetheorem:8146]. The quotient map is the [linear map](/page/Linear%20Map) \begin{align*} q:\mathfrak g\to \mathfrak g/D,\quad x\mapsto x+D. \end{align*} Now choose arbitrary elements of the quotient. Every element of $\mathfrak g/D$ has the form $x+D$ for some $x\in\mathfrak g$, so let $x+D$ and $y+D$ be two such cosets. The bracket in the quotient is defined by taking brackets of representatives: \begin{align*} [x+D,y+D]_{\mathfrak g/D}=[x,y]_{\mathfrak g}+D. \end{align*} The key point is that $D$ was chosen precisely to contain every bracket in $\mathfrak g$. Since $D$ is the linear span of all brackets, the particular bracket $[x,y]_{\mathfrak g}$ belongs to $D$. Therefore its coset modulo $D$ is the zero coset: \begin{align*} [x,y]_{\mathfrak g}+D=D=0_{\mathfrak g/D}. \end{align*} Thus every bracket in $\mathfrak g/D$ is zero. This is exactly the definition of an abelian Lie algebra, so $\mathfrak g/D$ is abelian.[/guided]

[step:Show that every map to an abelian Lie algebra kills the derived subspace] Let $\mathfrak a$ be an abelian Lie algebra over $F$, and let \begin{align*} \phi:\mathfrak g\to\mathfrak a \end{align*} be a Lie algebra homomorphism. For arbitrary $x,y\in\mathfrak g$, the homomorphism property gives \begin{align*} \phi([x,y]_{\mathfrak g})=[\phi(x),\phi(y)]_{\mathfrak a}. \end{align*} Since $\mathfrak a$ is abelian, $[\phi(x),\phi(y)]_{\mathfrak a}=0_{\mathfrak a}$. Hence \begin{align*} \phi([x,y]_{\mathfrak g})=0_{\mathfrak a} \end{align*} for all $x,y\in\mathfrak g$. Let $d\in D$. Choose $m\geq 1$, scalars $c_1,\dots,c_m\in F$, and elements $x_1,\dots,x_m,y_1,\dots,y_m\in\mathfrak g$ such that \begin{align*} d=\sum_{i=1}^{m}c_i[x_i,y_i]_{\mathfrak g}. \end{align*} Using linearity of $\phi$, \begin{align*} \phi(d)=\sum_{i=1}^{m}c_i\phi([x_i,y_i]_{\mathfrak g})=0_{\mathfrak a}. \end{align*} Thus $D\subset \ker\phi$. [/step]

[step:Construct the induced map on the quotient] Define a map \begin{align*} \bar{\phi}:\mathfrak g/D\to\mathfrak a,\quad x+D\mapsto \phi(x). \end{align*} We first prove that this definition is independent of the representative. Suppose $x+D=y+D$ in $\mathfrak g/D$. Then $x-y\in D$. Since $D\subset\ker\phi$, \begin{align*} \phi(x)-\phi(y)=\phi(x-y)=0_{\mathfrak a}. \end{align*} Therefore $\phi(x)=\phi(y)$, so $\bar{\phi}$ is well-defined. For $x,y\in\mathfrak g$ and $c\in F$, linearity follows from the vector space operations in the quotient and the linearity of $\phi$: \begin{align*} \bar{\phi}((x+D)+(y+D))=\bar{\phi}((x+y)+D)=\phi(x+y)=\phi(x)+\phi(y). \end{align*} Also, \begin{align*} \bar{\phi}(c(x+D))=\bar{\phi}(cx+D)=\phi(cx)=c\phi(x). \end{align*} Hence $\bar{\phi}$ is an $F$-linear map. Finally, for every $x\in\mathfrak g$, \begin{align*} (\bar{\phi}\circ q)(x)=\bar{\phi}(x+D)=\phi(x). \end{align*} Thus $\phi=\bar{\phi}\circ q$. [/step]

[step:Verify that the induced map preserves Lie brackets] Let $x+D,y+D\in\mathfrak g/D$. By the definition of the quotient bracket and by the definition of $\bar{\phi}$, \begin{align*} \bar{\phi}([x+D,y+D]_{\mathfrak g/D})=\bar{\phi}([x,y]_{\mathfrak g}+D)=\phi([x,y]_{\mathfrak g}). \end{align*} Since $\phi$ is a Lie algebra homomorphism, \begin{align*} \phi([x,y]_{\mathfrak g})=[\phi(x),\phi(y)]_{\mathfrak a}. \end{align*} Using the definition of $\bar{\phi}$ again, \begin{align*} [\phi(x),\phi(y)]_{\mathfrak a}=[\bar{\phi}(x+D),\bar{\phi}(y+D)]_{\mathfrak a}. \end{align*} Therefore \begin{align*} \bar{\phi}([x+D,y+D]_{\mathfrak g/D})=[\bar{\phi}(x+D),\bar{\phi}(y+D)]_{\mathfrak a}. \end{align*} Since the cosets $x+D$ and $y+D$ were arbitrary, $\bar{\phi}$ preserves Lie brackets. Together with linearity, this proves that $\bar{\phi}$ is a Lie algebra homomorphism. [/step]

[step:Prove uniqueness of the induced homomorphism] Let \begin{align*} \psi:\mathfrak g/D\to\mathfrak a \end{align*} be a Lie algebra homomorphism satisfying $\phi=\psi\circ q$. For any coset $x+D\in\mathfrak g/D$, \begin{align*} \psi(x+D)=\psi(q(x))=(\psi\circ q)(x)=\phi(x)=\bar{\phi}(x+D). \end{align*} Thus $\psi$ and $\bar{\phi}$ agree on every element of $\mathfrak g/D$, so $\psi=\bar{\phi}$. Therefore the induced Lie algebra homomorphism is unique, completing the proof of the universal property. [/step]