Let $A$ be an algebra. We know that $Soc M =\oplus \_A Soc M\_{\alpha}$ and $Rad M =\oplus \_A Rad M\_{\alpha}$ if $M= \oplus \_A M\_{\alpha}$ as $A$-modules.


Now let $M,N$ be $A$-modules, $\Lambda:=End\_A(N)$, whether $Soc\_{\Lambda}(Hom\_A(N,M)) =Hom\_A(N,Soc\_A(M))$ and $Rad\_{\Lambda}(Hom\_A(N,M)) =Hom\_A(N, Rad\_A(M))$ hold(here $Soc\_A$ means the socle functor acts on $A$-modules, so does $Rad\_A$ )?



Whether hom functor and socle (radical) are commutative?
Take a symmetric Nakayama algebra with Kupisch series [3,3]. (see for example the book by assem simson skowronski in the chapter about nakayama algebras for the meaning) Then $Hom(e\_0 A, soc(e\_1A))=Hom(e\_0 A, S\_1)=0$, but $Hom(e\_0A , e\_1A)$ is nonzero and thus has nonzero socle. 


Next try: Let $A=K[x]/(x^2)$ with simple modules $S$ and $N=A \oplus S$ and $M=S$.
Then $Hom(N,rad(M))=0$. But $Hom(N,M)$ is a projective indecomposable $End(N)$-module which is not simple and thus has nonzero radical.


