I've been trying to understand the proof of the following statement:
An injective map of $\mathfrak{g}$-representations of a semisimple Lie algebra splits.
I'm supposed to show this considering the $\mathfrak{g}$-representations $S= \lbrace \phi \in \text{Hom}(V,U) : \phi \text{ is a scalar on } U \rbrace$ and $T=\lbrace \phi \in \text{Hom}(V,U) : \phi(U)=0 \rbrace$. Supposedly, we consider the quotient $T/S$, but this doesn't make any sense to me - surely $T \subset S$? Then, theoretically, $T/S$ is a trivial representation, and this is in the subspace of invariants $T/S^{\mathfrak{g}}$, and so as the map $T \rightarrow T/S$ is surjective and a map of $\mathfrak{g}$-representations, then there is (by properties of the Casimir operator) a $t \in T$ mapping to this. Then, does this mean $t \circ i$ then is scalar multiplication on $U$. I'm very confused.
Any help would be appreicated!
By Weyl's theorem, every finite-dimensional $\mathfrak{g}$-module of a semisimple Lie algebra $\mathfrak{g}$ is semisimple, i.e., every submodule of it has a complement. Hence all extensions of modules are trivial, i.e., $$ 0=ext(B,A)=H^1(\mathfrak{g},Hom(B,A)). $$ In other words, every short exact sequence of $\mathfrak{g}$-modules $0\rightarrow A\rightarrow C\rightarrow B\rightarrow 0$ splits (here the map $A\rightarrow C$ is injective).