What is the natural morphism between an $R$ module and its bidual?

Question

Let $A$ be an $R$-module, then we can define the dual as $$A^*=Hom(A,R)$$ similarly $$A^{**}=Hom(Hom(A,R),R)$$ so as I understood it, an element $f\in A^{**}$ is a function $$f:Hom(A,R)\rightarrow R,~~~(g:A\rightarrow R)\mapsto f\circ g:A\rightarrow R$$but now our prof told us that there is a natural homomorphism from $A\rightarrow A^{**}$ but I somehow don‘t see which one. Could maybe someone show me this homomorphism and explain it to me?

Thanks for your help.

You can check this post: Motivation to understand double dual space. Though it is talking about vector spaces, the idea here does not rely on the algebraic properties of the underlying ring/field. — Bernard Pan, Jun 02 '22 at 20:05
@BernardPan sorry but I still do not see what the map is that we search or maybe I havent found it — user123234, Jun 02 '22 at 20:10
To every $x \in A$ associate $\mathrm{ev}_x: f \longmapsto f(x)$. — Aphelli, Jun 02 '22 at 20:21

score 3 · Accepted Answer · answered Jun 02 '22 at 20:22

$\DeclareMathOperator{\Hom}{Hom}$

An element of $A^{\ast\ast}$ is a function $\Phi: \Hom(A,R) \rightarrow R$. Hence to have a map $A \rightarrow A^{\ast\ast}$ you have to specify for every element $a\in A$ such a function $\Phi_a$. This itself has to be a map which turns homomorphisms $\phi:A\rightarrow R$ into elements of $R$.

So what is the most naive thing you can do with the datum of a function $\phi: A\rightarrow R$ and an element $a\in A$ (+ knowing that you have to turn the function $\phi$ into a value of $R$)?

Putting all of this together the canonical map is the one given by the assignment $$\begin{array}{rcl} A & \longrightarrow & \Hom(\Hom(A,R),R)\\ a & \mapsto & \left( \begin{array}{rcl} \Phi_a:\Hom(A,R) & \rightarrow & R\\ \phi & \mapsto & \phi(a) \end{array}\right) \end{array}$$

What is the natural morphism between an $R$ module and its bidual?

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