[proofplan] The exact sequence identifies $i(U)$ as a $\mathfrak g$-submodule of $V$. By [Weyl's Complete Reducibility Theorem](/theorems/3755), this submodule has a $\mathfrak g$-[stable complement](/theorems/2279) $U_0$ in $V$. The quotient map $p$ restricts to a bijective $\mathfrak g$-module homomorphism from $U_0$ onto $W$, and the inverse of this restricted map gives the required splitting. [/proofplan] [step:Apply complete reducibility to split off the submodule $i(U)$] Since $i:U\to V$ is a $\mathfrak g$-module homomorphism, its image $i(U)\subset V$ is a $\mathfrak g$-submodule: for every $x\in\mathfrak g$ and $u\in U$, \begin{align*} x\cdot i(u)=i(x\cdot u)\in i(U). \end{align*} Because $V$ is finite-dimensional and $\mathfrak g$ is semisimple over an algebraically closed field of characteristic $0$, Weyl's Complete Reducibility Theorem applies to the finite-dimensional $\mathfrak g$-module $V$ and its submodule $i(U)$. Thus there exists a $\mathfrak g$-submodule $U_0\subset V$ such that \begin{align*} V=i(U)\oplus U_0. \end{align*} Here we are citing a result not yet in the wiki: Weyl's Complete Reducibility Theorem. [guided] The first point is to check that $i(U)$ is a legitimate input for complete reducibility. Since $i:U\to V$ is a $\mathfrak g$-module homomorphism, it intertwines the $\mathfrak g$-actions. Hence, for every $x\in\mathfrak g$ and every $u\in U$, \begin{align*} x\cdot i(u)=i(x\cdot u). \end{align*} The vector $x\cdot u$ lies in $U$, so $i(x\cdot u)$ lies in $i(U)$. Therefore $i(U)$ is stable under the $\mathfrak g$-action and is a $\mathfrak g$-submodule of $V$. Now we use the structural input. Weyl's Complete Reducibility Theorem says that every finite-dimensional module over a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $0$ is completely reducible. The hypotheses match our situation: $\mathfrak g$ is finite-dimensional and semisimple, $F$ is algebraically closed of characteristic $0$, and $V$ is finite-dimensional. Applying the theorem to the submodule $i(U)\subset V$, we obtain a $\mathfrak g$-submodule $U_0\subset V$ such that \begin{align*} V=i(U)\oplus U_0. \end{align*} This direct-sum decomposition is the algebraic replacement for choosing a complement to the kernel of $p$. Here we are citing a result not yet in the wiki: Weyl's Complete Reducibility Theorem. [/guided] [/step] [step:Show that $p$ restricts to an isomorphism from the complement onto $W$] Let \begin{align*} q:U_0&\to W\\ v&\mapsto p(v) \end{align*} be the restriction of $p$ to $U_0$. Since $p$ is a $\mathfrak g$-module homomorphism and $U_0$ is a $\mathfrak g$-submodule of $V$, the map $q$ is a $\mathfrak g$-module homomorphism. We first prove that $q$ is injective. If $v\in\ker q$, then $v\in U_0$ and $p(v)=0$, so $v\in U_0\cap\ker p$. Exactness gives $\ker p=\operatorname{im} i=i(U)$, hence \begin{align*} v\in U_0\cap i(U). \end{align*} Since $V=i(U)\oplus U_0$, the intersection $U_0\cap i(U)$ is $\{0\}$. Therefore $v=0$, so $\ker q=\{0\}$. We next prove that $q$ is surjective. Let $w\in W$. Since $p:V\to W$ is surjective, there exists $v\in V$ such that $p(v)=w$. Using the direct-sum decomposition $V=i(U)\oplus U_0$, choose the unique pair $(u,u_0)\in U\times U_0$ such that \begin{align*} v=i(u)+u_0. \end{align*} Because exactness gives $p\circ i=0$, we obtain \begin{align*} w=p(v)=p(i(u)+u_0)=p(i(u))+p(u_0)=0+q(u_0)=q(u_0). \end{align*} Thus $q$ is surjective. Hence $q:U_0\to W$ is an isomorphism of $\mathfrak g$-modules. [guided] The restriction of $p$ to the complementary summand is the map \begin{align*} q:U_0&\to W\\ v&\mapsto p(v). \end{align*} This is a $\mathfrak g$-module homomorphism because $p$ is one: for every $x\in\mathfrak g$ and $v\in U_0$, \begin{align*} q(x\cdot v)=p(x\cdot v)=x\cdot p(v)=x\cdot q(v). \end{align*} The condition $x\cdot v\in U_0$ is valid because $U_0$ is a $\mathfrak g$-submodule. To prove injectivity, take $v\in\ker q$. Then $v\in U_0$ and $p(v)=0$, so $v\in U_0\cap\ker p$. Exactness of the sequence says precisely that \begin{align*} \ker p=\operatorname{im} i=i(U). \end{align*} Therefore \begin{align*} v\in U_0\cap i(U). \end{align*} But the decomposition $V=i(U)\oplus U_0$ is a direct sum, so $U_0\cap i(U)=\{0\}$. Hence $v=0$, and $q$ is injective. To prove surjectivity, take an arbitrary $w\in W$. Since $p$ is surjective, there exists $v\in V$ with $p(v)=w$. The direct-sum decomposition gives a unique expression \begin{align*} v=i(u)+u_0 \end{align*} with $u\in U$ and $u_0\in U_0$. Applying $p$ and using linearity gives \begin{align*} w=p(v)=p(i(u)+u_0)=p(i(u))+p(u_0). \end{align*} Exactness gives $\operatorname{im} i\subseteq\ker p$, so $p(i(u))=0$. Therefore \begin{align*} w=q(u_0). \end{align*} Since $w\in W$ was arbitrary, $q$ is surjective. Thus $q:U_0\to W$ is a bijective $\mathfrak g$-module homomorphism, hence an isomorphism of $\mathfrak g$-modules. [/guided] [/step] [step:Invert the restricted map to construct the splitting] Let $j:U_0\to V$ denote the inclusion map. Since $U_0$ is a $\mathfrak g$-submodule of $V$, the map $j$ is a $\mathfrak g$-module homomorphism. Define \begin{align*} s:W&\to V\\ w&\mapsto j(q^{-1}(w)). \end{align*} Because $q:U_0\to W$ is an isomorphism of $\mathfrak g$-modules, its inverse $q^{-1}:W\to U_0$ is a $\mathfrak g$-module homomorphism, and therefore $s=j\circ q^{-1}$ is a $\mathfrak g$-module homomorphism. Finally, for every $w\in W$, \begin{align*} (p\circ s)(w)=p(j(q^{-1}(w)))=q(q^{-1}(w))=w. \end{align*} Thus $p\circ s=\operatorname{id}_W$, so the short exact sequence splits. [/step]