Theorem. Let $A$ be a finite-dimensional $k$-algebra, and let $S$ be an $A$-module. The structural algebra homomorphism $A \to \operatorname{End}_k(S)$ is surjective if and only if $S$ is a simple $A$-module and $\operatorname{End}_A(S) \cong k$.
Proof. After replacing $A$ by $A/I$, where $I$ is the annihilator of $S$, we may assume that the structural map $A \to \operatorname{End}_k(S)$ is injective. Note that then $A$ is simple, since the annihilator of a simple module is a maximal ideal. If this map is also surjective, then $A$ is isomorphic to a matrix algebra over $k$, hence split, and $S$ is a simple $A$-module, hence satisfes $\operatorname{End}_k(S) \cong k$. Since we know already that $A$ is simple, the converse follows from 1.13.2 and comparing dimensions.
I'm having trouble understanding the proof. Is there anything off with the structure of the proof? How is $A$ simple in line 2 of the proof as $S$ is not a simple module yet.
Any help would be appreciated!
Theorem 1.13.2 A $k$-algebra $A$ is simple Artinian if and only if $A \cong M_n(D)$ for some positive integer $n$ and some division algebra $D$ over $k$. In that case, $A$ has, up to isomorphism, a unique simple module $S$, and moreover the following hold:
(i) We have an isomorphism of left $A$-modules $A \cong S_n$; in particular, $A$ is semisimple.
(ii) We have $D \cong \operatorname{End}_A(S)^{op}$.
(iii) We have $A \cong \operatorname{End}_{D^{op}} (S)$