This is in context of a larger problem of showing that the dual of an invertible sheaf is invertible on a scheme.
I want to show that given a free R-module A of rank 1, the standard evaluation map given by
$A \otimes Hom_R(A,R) \rightarrow R \\ a \otimes f \mapsto f(a)$
is an isomorphism. Surjectivity is trivial but I'm having some trouble with injectivity.
I tried a proof by contradiction by assuming a non zero r and non zero f such that $f(r) = 0$ and show that since there would exist $q \in A$ s.t $f(q) = b$ for nonzero $b$. but after a lot of pushing symbols around, I didn't get anywhere.
Since A is of rank 1, I know it is sufficient to show that the same evaluation map from $R \otimes Hom_R(R,R) \rightarrow R$ is an isomorphism. But I ran into the same issue. Since the tensor product is the fiber product in the case of R-modules, I tried picking a clever R-module S and morphism from S to $R \otimes Hom_R(R,R)$ to see the injectivity of the above map. The obvious choices are clearly $R$ and $Hom_R(R,R)$ but I didn't see why they worked (If they do).
Edit: Any ring mentioned is a commutative ring with identity.
I'm still very inexperienced with this so any help would be appreciated!
Let's show that the evaluation map $R \otimes \operatorname{Hom}(R, R)$ is injective by producing an inverse explicitly.
The inverse will simply send $r \rightarrow r \otimes \text{id}_R$.
The composition of the maps then sends $$r \otimes f \rightarrow f(r) \rightarrow f(r) \otimes \text{id}_R$$
Note that the $f(r)$ can pass from left to right and then the $r$ can pass back to the left, i.e.
$$1 \otimes \text{id}_R f(r) = 1 \otimes f(r) = r \otimes f$$