I was struggling with this problem regarding the socle and the top of a finitely generated right-module $M$ over a finite dimensional $K$- algebra $A$. If you define the duality functor $DM = Hom_{k}(M,K)$, show that $Soc (DM )= DTop(M)$. My line of thinking was:
$DTop(M) \subseteq Soc (DM ) $: Consider the short exact sequence \begin{align}0 \rightarrow Rad(M) \rightarrow M \rightarrow Top(M) \rightarrow 0\end{align} Applying the duality gives \begin{align}0 \rightarrow DTop(M) \rightarrow DM \rightarrow DRad(M) \rightarrow 0\end{align} Since $DTop(M)$ is semisimple (because $Top(M)$ is), $DTop(M)\subseteq Soc(DM)$ by definition of $Soc(DM)$ as the sum of all simples.
$Soc (DM ) = DTop(M)$: since $M$ has is a finite dimensional vector space over $K$, it suffices to show $dim_K(Soc(DM)) \leq dim_K(DTopM)$. By duality, $Soc(DM) = DL$ with $L$ a semisimple right module. Then $DL \leq DM$ and applying duality on both sides gives that there is a surjective morphism $\xi: M \twoheadrightarrow L$. So that means $M _{/Ker \xi} \simeq L$ is semisimple. Since $RadM$ is the smallest submodule $L$ such that $M/L$ is semisimple, that means $RadM \subseteq Ker \xi$ and thus there is a surjective morphism $TopM \twoheadrightarrow L$ so there is an injective morphism $DL=Soc(DM) \hookrightarrow DTopM$