Let $R$ be a ring. If $ 0 \rightarrow X \xrightarrow{f} Y \xrightarrow{g} Z \rightarrow 0 $ is a short exact sequence of (left) modules over $R$, then are $ A:= \{ F \in \mathrm{Hom}_{R}(Y,X) : F \circ f =0_X \} $ and $ B:= \{ G \in \mathrm{Hom}_{R}(Z,Y) : g \circ G =0_Z \} $ necessarily isomorphic as Abelian groups?
And in the case when $R$ is commutative and the aforementioned $A,B$ are isomorphic, are they also isomorphic as $R$-modules?
So far I understand that if $0 \to X \to Y \to Z \to 0$ is a sequence of finite-dimensional vector spaces over some field $R=F$, then all the above statements are true, e.g. by dimension-counting. However, I'm not sure about the general case. Would anyone have any suggestions or hints on how to think about this question?
Edit: sorry actually this should be true in general for a split exact sequence.
$\newcommand{\Hom}{\operatorname{Hom}}$By the homomorphism theorem for $R$-modules (i.e. the universal property for cokernels) we have a canonical bijection $$ \phi\colon \Hom_R(Z,X) \longrightarrow A = \{F\in \Hom_R(Y,X) \mid F\circ f = 0\}, \\h\longmapsto h\circ g $$ and by the universal property of the kernel we have a canonical bijection $$ \psi \colon \Hom_R(Z,X) \longrightarrow B = \{G\in \Hom_R(Z,Y) \mid g\circ G = 0\},\\ h\longmapsto f\circ h. $$ These are obviously $R$-linear isomorphisms. Hence, we obtain an $R$-linear isomorphism $$ \psi\circ \phi^{-1} \colon A\xrightarrow{\cong} B. $$