I have following problem: let $R$, $A$ be rings, $M$ a R-Module. I have to prove that $\tau: \operatorname{Hom}_{Ab}(M,A) \to \operatorname{Hom}_{R-Mod}(M,\operatorname{Hom}_{Ab}(M,A)) $ is an bijection for $\tau$ which is defined as $\tau(f)(m) = (r \to f(rm))$ where $f \in \operatorname{Hom}_{Ab}(M,A), r \in R, m \in M$.
I tried this by finding an inverse $\beta$ and showing $\tau \beta = id, \beta \tau = id$, whereby I defined $\beta$ as follows:
Let $s \in Hom_{R-Mod}(M,\operatorname{Hom}_{Ab}(M,A))$, then $\beta(s) = (m \to s(m)(1))$.
While proving $\tau \beta = id$ there occurred a problem: By definitions of $\tau $ and $\beta $ we get $\tau \beta(s) =(m \to \tau (\beta(s))(m) = (r \to \beta(s)(rm)=s(rm)(1)))$. Therefore $\tau \beta(s)(m) =(r \to s(rm)(1))$.
If $\tau \beta = id$ would hold, then $(r \to s(m)(r)) = s(m) = id(s)(m) = \tau \beta(s)(m) =(r \to s(rm)(1))$. So for all $r \in R, m \in M$ there should be $s(m)(r) =s(rm)(1))$, but that's not clear to me. Since $s \in \operatorname{Hom}_{R-Mod}(M,\operatorname{Hom}_{Ab}(M,A))$ we have $ s(rm)(1)) = r* s(m)(1)$, but $s(m) \in \operatorname{Hom}_{Ab}(M,A) $ and not from $\operatorname{Hom}_{R-Mod}(M,A)$, so I i.g. don't see any possibility to "insert" r from (rm) to (1).
