Why does $F:P\otimes−:Mod_R\to Mod_R$ define an equivalence of categories? Let $P$ be a finitely generated projective R-module of rank 1 which is not free.
Let $F:P\otimes_R−:Mod_R\to Mod_R$ be the functor.
We have that $R$ is free and $F(R)=P\otimes_R R\simeq P$ is not free.
If $F$ did define a equivalence of category, we would prove that "Free" is not a categorical property.
But why does $F$ define a equivalence of categories? (I suppose that the fact that $P$ is projective plays an important role).
Let $p$ be the generator of $P$.
The natural transformation $\alpha_M:M\to M\otimes_R P$ defined as $\alpha_M(m)= m\otimes p$ may not define a natural isomorphism: $\alpha_M$ may not be an ismorphism for every $M$.

Thanks to the comment by rschwieb, it is easy to see that 'free' is not a categorical property.
In relation to my specific question about the functor $F$, I have just found a theorem by Morita that explains the reason why $F$ defines a equivalence of categories. I have found it in Rotman's book "An introduction to Homological Algebra", page 269. All these ideas are developed in Morita Theory.
The general idea would be the following:
We need to define another functor $G = Hom_R(P,-)$. Since $F$ and $G$ are adjoint, they define two natural transformation $FG\to 1_R$ and $1_R\to GF$.
Since $P$ is a small projective generator of $Mod_R$, we have that these natural transformation are also isomorphism.
All the details can be found in the proof:
Theorem 5.55. Let $R$ be a ring and let $P$ be a small projective generator of $Mod_R$. If $S = End_R(P)$, then there is an isomorphism $F : Mod_S \to Mod_R$ given by $M \to M \otimes_S P$.
