It is well-known that over a quasi-Frobenius ring $R$ any f.g. right module $M$ is reflexive, in the sense that whenever we take $M^*=Hom_R(M,R_R)$ as a left $R$-module, the modules $M$ and $M^{**}$ are isomorphic. Also, an $R$-module $M$ is f.g. if and only if $M^*$ is so. In the proofs of the above alleged statements it is seen that the dual operator $^*$ is a contravariant exact functor from the category of right $R$-modules $\mathcal M_R$ to that of left ones $_R\mathcal M$. From what we said it is inferred that the "double-dual" functors are naturally equivalent to the identity functors on $\mathcal M^{fg}_R$ and $_R^{fg}\mathcal M$, the subcategories of respective f.g. modules.
Now, for any submodule $A⊆M$ we set $A^{\perp}=\{f\in M^*:f(A)=0\}$ (a submodule of $M^*$), and for any submodule $I⊆M^*$ let $I^{\perp}=⋂_{i∈I}\ker(i)$ (a submodule of $M$).
My question is how could we deduce the equalities $A^{\perp \perp}=A$ and $I^{\perp \perp}=I$ from the above-cited natural equivalence to the identity functors?
Thanks for any cooperation.
The idea is as follows: There is a commutative diagram $$\begin{array}{c} A & \rightarrow & A^{**} \\ \downarrow && \downarrow \\ M & \rightarrow & M^{**}. \end{array}$$ The horizontal maps are isomorphisms. Now show that the image of $A \to A^{**} \to M^{**} \to M$ is precisely $A^{\perp\perp}$ (but on the other hand this is the image of $A \to M$, i.e. $A$).