[proofplan] We first verify that the two algebraic objects appearing in the statement are legitimate: the kernel is an ideal, so the quotient Lie algebra $\mathfrak g/\ker\varphi$ is defined, and the image is a Lie subalgebra of $\mathfrak h$. We then define the induced map on cosets by sending $x+\ker\varphi$ to $\varphi(x)$ and prove that this definition is independent of the representative. Finally, we check that the induced map is a bijective Lie algebra homomorphism and that its defining property forces uniqueness. [/proofplan] [step:Show that the kernel is an ideal of $\mathfrak g$] Let \begin{align*} \ker\varphi := \{x \in \mathfrak g : \varphi(x)=0_{\mathfrak h}\} \end{align*} denote the kernel of $\varphi$. Since $\varphi$ is linear, $\ker\varphi$ is a linear subspace of $\mathfrak g$. To prove that $\ker\varphi$ is an ideal, let $x \in \mathfrak g$ and $k_0 \in \ker\varphi$. Since $\varphi$ preserves Lie brackets, \begin{align*} \varphi([x,k_0]_{\mathfrak g}) = [\varphi(x),\varphi(k_0)]_{\mathfrak h} = [\varphi(x),0_{\mathfrak h}]_{\mathfrak h} = 0_{\mathfrak h}. \end{align*} Thus $[x,k_0]_{\mathfrak g} \in \ker\varphi$. Therefore $\ker\varphi$ is an ideal of $\mathfrak g$. [guided] We need the quotient $\mathfrak g/\ker\varphi$ to be a Lie algebra, and for that the kernel must be an ideal, not merely a linear subspace. Define \begin{align*} \ker\varphi := \{x \in \mathfrak g : \varphi(x)=0_{\mathfrak h}\}. \end{align*} Because $\varphi$ is a [linear map](/page/Linear%20Map) of $k$-vector spaces, its kernel is a linear subspace of $\mathfrak g$. Now we verify the ideal condition. Let $x \in \mathfrak g$ be arbitrary and let $k_0 \in \ker\varphi$. The defining property of $k_0$ is $\varphi(k_0)=0_{\mathfrak h}$. Since $\varphi$ is a Lie algebra homomorphism, it preserves brackets, so \begin{align*} \varphi([x,k_0]_{\mathfrak g}) = [\varphi(x),\varphi(k_0)]_{\mathfrak h}. \end{align*} Substituting $\varphi(k_0)=0_{\mathfrak h}$ and using bilinearity of the Lie bracket in $\mathfrak h$, we get \begin{align*} \varphi([x,k_0]_{\mathfrak g}) = [\varphi(x),0_{\mathfrak h}]_{\mathfrak h} = 0_{\mathfrak h}. \end{align*} Hence $[x,k_0]_{\mathfrak g} \in \ker\varphi$. Since this holds for every $x \in \mathfrak g$ and every $k_0 \in \ker\varphi$, the subspace $\ker\varphi$ is an ideal of $\mathfrak g$. [/guided] [/step] [step:Show that the image is a Lie subalgebra of $\mathfrak h$] Let \begin{align*} \operatorname{im}\varphi := \{\varphi(x) : x \in \mathfrak g\} \end{align*} denote the image of $\varphi$. Since $\varphi$ is linear, $\operatorname{im}\varphi$ is a linear subspace of $\mathfrak h$. Let $u,v \in \operatorname{im}\varphi$. Then there exist $x,y \in \mathfrak g$ such that $u=\varphi(x)$ and $v=\varphi(y)$. Since $\varphi$ preserves Lie brackets, \begin{align*} [u,v]_{\mathfrak h} = [\varphi(x),\varphi(y)]_{\mathfrak h} = \varphi([x,y]_{\mathfrak g}) \in \operatorname{im}\varphi. \end{align*} Thus $\operatorname{im}\varphi$ is closed under the Lie bracket of $\mathfrak h$, and therefore $\operatorname{im}\varphi$ is a Lie subalgebra of $\mathfrak h$. [/step] [step:Define the induced map on cosets and prove it is well defined] Since $\ker\varphi$ is an ideal of $\mathfrak g$, the quotient [vector space](/page/Vector%20Space) $\mathfrak g/\ker\varphi$ carries the quotient Lie bracket \begin{align*} [x+\ker\varphi, y+\ker\varphi]_{\mathfrak g/\ker\varphi} := [x,y]_{\mathfrak g}+\ker\varphi. \end{align*} Define a map \begin{align*} \overline{\varphi}: \mathfrak g/\ker\varphi &\to \operatorname{im}\varphi \\ x+\ker\varphi &\mapsto \varphi(x). \end{align*} We prove that this definition is independent of the representative. Suppose $x,y \in \mathfrak g$ satisfy \begin{align*} x+\ker\varphi = y+\ker\varphi. \end{align*} Then $x-y \in \ker\varphi$, so $\varphi(x-y)=0_{\mathfrak h}$. By linearity of $\varphi$, \begin{align*} \varphi(x)-\varphi(y) = \varphi(x-y) = 0_{\mathfrak h}. \end{align*} Hence $\varphi(x)=\varphi(y)$, and $\overline{\varphi}$ is well defined. [guided] The proposed formula is forced: if a coset is written as $x+\ker\varphi$, it should map to the value $\varphi(x)$. The only possible problem is that the same coset may have more than one representative, so we must prove that the value does not change when the representative changes. Because $\ker\varphi$ is an ideal of $\mathfrak g$, the quotient $\mathfrak g/\ker\varphi$ is a Lie algebra with bracket \begin{align*} [x+\ker\varphi, y+\ker\varphi]_{\mathfrak g/\ker\varphi} := [x,y]_{\mathfrak g}+\ker\varphi. \end{align*} Define \begin{align*} \overline{\varphi}: \mathfrak g/\ker\varphi &\to \operatorname{im}\varphi \\ x+\ker\varphi &\mapsto \varphi(x). \end{align*} The codomain is indeed $\operatorname{im}\varphi$, because $\varphi(x)$ belongs to $\operatorname{im}\varphi$ for every $x \in \mathfrak g$. Now suppose $x,y \in \mathfrak g$ represent the same coset: \begin{align*} x+\ker\varphi = y+\ker\varphi. \end{align*} By the definition of equality in a quotient vector space, this means \begin{align*} x-y \in \ker\varphi. \end{align*} The definition of the kernel gives $\varphi(x-y)=0_{\mathfrak h}$. Since $\varphi$ is linear, \begin{align*} \varphi(x)-\varphi(y) = \varphi(x-y) = 0_{\mathfrak h}. \end{align*} Therefore $\varphi(x)=\varphi(y)$. Thus the formula for $\overline{\varphi}$ gives the same value for every representative of a coset, so $\overline{\varphi}$ is well defined. [/guided] [/step] [step:Verify that the induced map is a Lie algebra homomorphism] Let $a,b \in k$ and let $x,y \in \mathfrak g$. Using the quotient vector space operations and the linearity of $\varphi$, \begin{align*} \overline{\varphi}\bigl(a(x+\ker\varphi)+b(y+\ker\varphi)\bigr) &= \overline{\varphi}\bigl((ax+by)+\ker\varphi\bigr) \\ &= \varphi(ax+by) \\ &= a\varphi(x)+b\varphi(y) \\ &= a\overline{\varphi}(x+\ker\varphi)+b\overline{\varphi}(y+\ker\varphi). \end{align*} Thus $\overline{\varphi}$ is linear. Using the quotient Lie bracket and the bracket-preserving property of $\varphi$, \begin{align*} \overline{\varphi}\bigl([x+\ker\varphi,y+\ker\varphi]_{\mathfrak g/\ker\varphi}\bigr) &= \overline{\varphi}([x,y]_{\mathfrak g}+\ker\varphi) \\ &= \varphi([x,y]_{\mathfrak g}) \\ &= [\varphi(x),\varphi(y)]_{\mathfrak h} \\ &= [\overline{\varphi}(x+\ker\varphi),\overline{\varphi}(y+\ker\varphi)]_{\mathfrak h}. \end{align*} Therefore $\overline{\varphi}$ is a Lie algebra homomorphism from $\mathfrak g/\ker\varphi$ to $\operatorname{im}\varphi$. [/step] [step:Prove that the induced homomorphism is bijective] First, $\overline{\varphi}$ is surjective onto $\operatorname{im}\varphi$. If $u \in \operatorname{im}\varphi$, then there exists $x \in \mathfrak g$ such that $u=\varphi(x)$, and hence \begin{align*} u = \varphi(x) = \overline{\varphi}(x+\ker\varphi). \end{align*} Next, $\overline{\varphi}$ is injective. Let $x+\ker\varphi \in \mathfrak g/\ker\varphi$ satisfy \begin{align*} \overline{\varphi}(x+\ker\varphi)=0_{\mathfrak h}. \end{align*} Then $\varphi(x)=0_{\mathfrak h}$, so $x \in \ker\varphi$. Therefore \begin{align*} x+\ker\varphi = \ker\varphi, \end{align*} which is the zero element of $\mathfrak g/\ker\varphi$. Thus the kernel of $\overline{\varphi}$ is zero, and $\overline{\varphi}$ is injective. Since $\overline{\varphi}$ is a bijective Lie algebra homomorphism, it is a Lie algebra isomorphism. [/step] [step:Establish uniqueness of the induced isomorphism] Let \begin{align*} \psi: \mathfrak g/\ker\varphi \to \operatorname{im}\varphi \end{align*} be a Lie algebra homomorphism satisfying \begin{align*} \psi(x+\ker\varphi)=\varphi(x) \end{align*} for every $x \in \mathfrak g$. Then for every coset $x+\ker\varphi \in \mathfrak g/\ker\varphi$, \begin{align*} \psi(x+\ker\varphi) = \varphi(x) = \overline{\varphi}(x+\ker\varphi). \end{align*} Thus $\psi$ and $\overline{\varphi}$ agree on every element of $\mathfrak g/\ker\varphi$, so $\psi=\overline{\varphi}$. Hence the Lie algebra isomorphism with the stated property is unique. [/step]