Let $R$ be a commutative ring and $A$ and $B$ two $R$-module. Suppose that $A$ is free of rank $n$ with basis $a\_1,\dots,a\_n$. Then there is an isomorphism $\Phi: Hom\_R(A,B) \to Hom\_R(A,R)\otimes\_R B$ defined by $\Phi(\sigma)=\sum\_{i=1}^n \psi\_i\otimes \sigma\_i$, where $\sigma\_i=\sigma(a\_i)$ and $\psi\_i$ is the map such that $\psi\_i(a\_j)=\delta\_{ij}$. Is there another way to describe $\Phi$?



$\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$
Such an isomorphism is worthless if it is written down with a choice of a basis, because then naturality is unclear (which is, of course, very important if you need this isomorphism not just as an isolated relation). There is a homomorphism of $R$-modules


$\alpha : A^\* \otimes B \to \hom(A,B)$


defined by $\phi \otimes b \mapsto \phi(-)b$, which is natural in $A$ and in $B$, which are arbitrary $R$-modules. It is a natural question when this is an isomorphism for all $B$, when we fix $A$. Note that the inverse will be, restricted to these $A,B$, also natural due to general reasons, although perhaps we need to make choices to write down the inverse (without making reference to $\alpha$)! Also remember the slogan "You can work locally, if you are given something globally".


Now both $(-)^\* \otimes B$ and $\hom(-,B)$ are functors which transform finite direct sums into finite products, and transform *split* cokernels into *split* kernels. In particular, the set of $A$s for $\alpha$ is an isomorphism is closed under these operations and since $R$ is an example, we see that every finitely generatd projective $R$-module is an example. Now, the converse is also true: If $A^\* \otimes - \cong \hom(A,-)$, then the right hand side is preserving *all* colimits (since this is true for the left hand side). Restricting to coequalizers shows that $A$ is projective, and restricting to filtered colimits shows that $A$ is finitely presented (in the categorical sense, thus also in the algebraic sense).


You can view this also as a special case of the Theorem of Eilenberg-Watts: If $A$ is finitely generated projective, then $\hom(A,-) : \text{Mod}(R) \to \text{Mod}(R)$ is a cocontinuous functor, thus is given by tensoring with a $R$-module, namely $\hom(A,R) = A^\*$.


