A finite-dimensional algebra separates its split simple modules

Every submodule $N$ of a finite sum $W$ of simple modules has a complement that is a sum of simple modules of the listed types. Here is the induction proving this assertion. For $W=0$, take complement $0$. Write $W=S\oplus T$, with $S$ simple, and suppose the assertion proved for $T$. If $S\subseteq N$, then $N=S\oplus (N\cap T)$; a complement of $N\cap T$ in $T$ complements $N$. Otherwise $N\cap S=0$, and projection to $T$ identifies $N$ with a submodule $N^{\mathrm{\prime }}$ of $T$. Write $T=N^{\mathrm{\prime }}\oplus C$ by the induction hypothesis. There is an $A$-linear map $f:N^{\mathrm{\prime }}\to S$ with $N=\{(f(t),t):t\in N^{\mathrm{\prime }}\}$. Every $(s,t+c)$ equals $(f(t),t)+(s-f(t),c)$, uniquely, so $W=N\oplus (S\oplus C)$. These cases exhaust the possibilities since $N\cap S$ is a submodule of $S$.

For each member of a finite selection of the $S_i$, choose $0\mathrm{\ne }{s}_i\in S_i$. Simplicity gives $A{s}_i=S_i$. Images of a finite basis of $A$ span $S_i$. Scanning that list and retaining a vector precisely when it is outside the previous span gives an independent spanning list $e_{i1},\dots ,e_{i{d}_i}$, with $1\le d_i\le {\dim }_{k}A$. Only finitely many selections have been made.

For this finite selection put $W={⨁}_i{S}_i^{d_i}$ and $v=(e_{ij})$. If $Av\mathrm{\ne }W$, step 1.1 supplies a nonzero simple summand of a complement, isomorphic to some $S_l$, and projection onto it gives a nonzero $A$-linear $F:W\to S_l$ with $F(Av)=0$. Its restrictions to copies of $S_i$ vanish for $i\mathrm{\ne }l$ by Schur, and on the copies of $S_l$ are scalars $c_j$. Since $1v=v$, we have $0=F(v)={\sum }_j{c}_j{e}_{lj}$. Independence gives $c_j=0$ for all $j$, so $F=0$, a contradiction. Therefore $Av=W$.

Given any tuple of $k$-endomorphisms $u_i$, step 2.1 supplies $a\in A$ with $av=(u_i(e_{ij}))$. Then $a{e}_{ij}=u_i(e_{ij})$ for every basis vector, so $a$ induces $u_i$ on all of $S_i$. Thus the action map for every finite selection is surjective. Lifting a vector-space basis of the target gives an independent list in $A$ (apply the map to any relation). Consequently ${\sum }_i{d}_i^2\le {\dim }_{k}A$, in particular the size of the selection is at most ${\dim }_{k}A$.

If $I$ had more than ${\dim }_{k}A$ members, finite induction would select ${\dim }_{k}A+1$ distinct indices, contradicting step 3.1. Thus $I$ is finite and that step proves the asserted surjectivity. If $I$ is empty, the unique map onto zero is surjective; if $A=0$, no nonzero unital simple module exists and this is the only case.