[proofplan] The prescribed formula is forced by the desired images of the two summands, so the only possible map is $\Phi(h,k)=i(h)+j(k)$. The unique decomposition hypothesis makes this map a vector-space isomorphism. It remains to compare the semidirect-product bracket with the bracket in $\mathfrak a$; the homomorphism properties of $i$ and $j$ handle the pure terms, and the compatibility relation handles the mixed terms. [/proofplan] [step:Construct the prescribed linear map and prove it is a vector-space isomorphism] Define the $\mathbb{F}$-[linear map](/page/Linear%20Map) \begin{align*} \Phi:\mathfrak h \rtimes_\rho \mathfrak k &\to \mathfrak a \\ (h,k) &\mapsto i(h)+j(k). \end{align*} The map is linear because $i$ and $j$ are Lie algebra homomorphisms, hence $\mathbb{F}$-linear maps. We prove that $\Phi$ is bijective. Let $x \in \mathfrak a$. By hypothesis, there exist $h \in \mathfrak h$ and $k \in \mathfrak k$ such that $x=i(h)+j(k)=\Phi(h,k)$, so $\Phi$ is surjective. If $\Phi(h,k)=0$, then \begin{align*} i(h)+j(k)=i(0_{\mathfrak h})+j(0_{\mathfrak k}). \end{align*} The uniqueness of the decomposition in $\mathfrak a$ gives $h=0_{\mathfrak h}$ and $k=0_{\mathfrak k}$. Thus $\ker \Phi=\{(0_{\mathfrak h},0_{\mathfrak k})\}$, so $\Phi$ is injective. Therefore $\Phi$ is a vector-space isomorphism. [/step] [step:Compare the bracket in $\mathfrak a$ with the semidirect-product bracket] Let $h_1,h_2 \in \mathfrak h$ and $k_1,k_2 \in \mathfrak k$. Since the bracket in $\mathfrak a$ is bilinear, \begin{align*} [\Phi(h_1,k_1),\Phi(h_2,k_2)]_{\mathfrak a} &=[i(h_1)+j(k_1),i(h_2)+j(k_2)]_{\mathfrak a} \\ &=[i(h_1),i(h_2)]_{\mathfrak a} +[i(h_1),j(k_2)]_{\mathfrak a} \\ &\quad +[j(k_1),i(h_2)]_{\mathfrak a} +[j(k_1),j(k_2)]_{\mathfrak a}. \end{align*} Because $i$ and $j$ are Lie algebra homomorphisms, \begin{align*} [i(h_1),i(h_2)]_{\mathfrak a} &=i([h_1,h_2]_{\mathfrak h}), \\ [j(k_1),j(k_2)]_{\mathfrak a} &=j([k_1,k_2]_{\mathfrak k}). \end{align*} The mixed compatibility hypothesis gives \begin{align*} [j(k_1),i(h_2)]_{\mathfrak a}=i(\rho(k_1)(h_2)). \end{align*} Using skew-symmetry of the bracket in $\mathfrak a$ and then the same compatibility hypothesis, \begin{align*} [i(h_1),j(k_2)]_{\mathfrak a} = -[j(k_2),i(h_1)]_{\mathfrak a} = -i(\rho(k_2)(h_1)). \end{align*} Substituting these four identities into the bilinear expansion gives \begin{align*} [\Phi(h_1,k_1),\Phi(h_2,k_2)]_{\mathfrak a} &= i([h_1,h_2]_{\mathfrak h}) -i(\rho(k_2)(h_1)) +i(\rho(k_1)(h_2)) \\ &\quad +j([k_1,k_2]_{\mathfrak k}) \\ &= i([h_1,h_2]_{\mathfrak h}+\rho(k_1)(h_2)-\rho(k_2)(h_1)) +j([k_1,k_2]_{\mathfrak k}) \\ &= \Phi\bigl([h_1,h_2]_{\mathfrak h}+\rho(k_1)(h_2)-\rho(k_2)(h_1),[k_1,k_2]_{\mathfrak k}\bigr) \\ &= \Phi([(h_1,k_1),(h_2,k_2)]_{\mathfrak h\rtimes_\rho\mathfrak k}). \end{align*} Thus $\Phi$ preserves brackets, so $\Phi$ is a Lie algebra homomorphism. Since $\Phi$ is also bijective, it is a Lie algebra isomorphism. [guided] We must check bracket preservation on arbitrary elements of the semidirect product. Take $h_1,h_2 \in \mathfrak h$ and $k_1,k_2 \in \mathfrak k$. The map $\Phi$ sends these elements to \begin{align*} \Phi(h_1,k_1)=i(h_1)+j(k_1), \qquad \Phi(h_2,k_2)=i(h_2)+j(k_2). \end{align*} Expanding the bracket in $\mathfrak a$ by bilinearity gives \begin{align*} [\Phi(h_1,k_1),\Phi(h_2,k_2)]_{\mathfrak a} &=[i(h_1)+j(k_1),i(h_2)+j(k_2)]_{\mathfrak a} \\ &=[i(h_1),i(h_2)]_{\mathfrak a} +[i(h_1),j(k_2)]_{\mathfrak a} \\ &\quad +[j(k_1),i(h_2)]_{\mathfrak a} +[j(k_1),j(k_2)]_{\mathfrak a}. \end{align*} The pure $\mathfrak h$ and pure $\mathfrak k$ terms are controlled by the fact that $i$ and $j$ are Lie algebra homomorphisms: \begin{align*} [i(h_1),i(h_2)]_{\mathfrak a} &=i([h_1,h_2]_{\mathfrak h}), \\ [j(k_1),j(k_2)]_{\mathfrak a} &=j([k_1,k_2]_{\mathfrak k}). \end{align*} The mixed terms are exactly where the action $\rho$ enters. The hypothesis says that for every $h \in \mathfrak h$ and $k \in \mathfrak k$, \begin{align*} [j(k),i(h)]_{\mathfrak a}=i(\rho(k)(h)). \end{align*} Applying this with $(k,h)=(k_1,h_2)$ gives \begin{align*} [j(k_1),i(h_2)]_{\mathfrak a}=i(\rho(k_1)(h_2)). \end{align*} For the other mixed term, the order is reversed, so we first use skew-symmetry of the Lie bracket in $\mathfrak a$ and then apply the same compatibility relation: \begin{align*} [i(h_1),j(k_2)]_{\mathfrak a} = -[j(k_2),i(h_1)]_{\mathfrak a} = -i(\rho(k_2)(h_1)). \end{align*} Putting these four pieces back into the expansion, we obtain \begin{align*} [\Phi(h_1,k_1),\Phi(h_2,k_2)]_{\mathfrak a} &= i([h_1,h_2]_{\mathfrak h}) -i(\rho(k_2)(h_1)) +i(\rho(k_1)(h_2)) \\ &\quad +j([k_1,k_2]_{\mathfrak k}) \\ &= i([h_1,h_2]_{\mathfrak h}+\rho(k_1)(h_2)-\rho(k_2)(h_1)) +j([k_1,k_2]_{\mathfrak k}). \end{align*} By the definition of $\Phi$, this last expression is \begin{align*} \Phi\bigl([h_1,h_2]_{\mathfrak h}+\rho(k_1)(h_2)-\rho(k_2)(h_1),[k_1,k_2]_{\mathfrak k}\bigr). \end{align*} By the definition of the bracket on $\mathfrak h \rtimes_\rho \mathfrak k$, the ordered pair inside $\Phi$ is precisely \begin{align*} [(h_1,k_1),(h_2,k_2)]_{\mathfrak h\rtimes_\rho\mathfrak k}. \end{align*} Therefore \begin{align*} [\Phi(h_1,k_1),\Phi(h_2,k_2)]_{\mathfrak a} = \Phi([(h_1,k_1),(h_2,k_2)]_{\mathfrak h\rtimes_\rho\mathfrak k}). \end{align*} Since this holds for arbitrary $(h_1,k_1)$ and $(h_2,k_2)$, $\Phi$ preserves the Lie bracket. Together with the vector-space bijectivity proved above, this makes $\Phi$ a Lie algebra isomorphism. [/guided] [/step] [step:Prove uniqueness from the prescribed values on all pairs] Let \begin{align*} \Psi:\mathfrak h \rtimes_\rho \mathfrak k \to \mathfrak a \end{align*} be a Lie algebra homomorphism satisfying \begin{align*} \Psi(h,k)=i(h)+j(k) \end{align*} for all $h \in \mathfrak h$ and $k \in \mathfrak k$. Then for every $(h,k) \in \mathfrak h \rtimes_\rho \mathfrak k$, \begin{align*} \Psi(h,k)=i(h)+j(k)=\Phi(h,k). \end{align*} Hence $\Psi=\Phi$. Therefore the Lie algebra isomorphism with the stated property is unique. [/step]