Suppose $R$ is a ring and $P$ is a projective generator of right $R$-modules. If $M$ is a left $A$-module, show that the natural map $M \to \text{Hom}_R(P, P \otimes_R M)$ that sends $m \mapsto (p \mapsto p \otimes m)$ is an isomorphism.
Above is a question that arose while I was studying Morita theory. Morita theory states that $P \otimes -$ is an equivalence of $R$-modules and $\text{End}_R(P)$ modules.
Here is my attempt. Since $P$ is a projective generator, there is some $n \in \mathbb{N}$ such that the following sequence is exact:
$$0 \to \ker (f) \to P^n \to M \to 0.$$
Since the tensor product is right exact, we thus have the following exact sequence
$$P \otimes \ker (f) \to P^n \to P \otimes M \to 0$$
which, since $P$ is projective (i.e. the $\text{Hom}(P,-)$ is exact), gives us the exact sequence
$$\text{Hom}(P, P \otimes \ker (f)) \to \text{Hom}(P, P^n) \to \text{Hom}(P, P \otimes M) \to 0.$$
However, I am not sure where to go from here. I know that, per Morita theorem, $P \otimes_R -$ is an equivalence of categories between left $R$-modules and left $\text{End}(P)$-modules. Could you state how I can intuitively understand Morita theorem? Taking $P = R^n$, we get an equivalence of categories between $R$-modules and $M_n(R)$ modules - is there another way to see the equivalence in this special case, to help me better understand Morita theorem?
As this question was what I found while trying to prepare for an exam, there may be an error in the statement of the problem, and I'd appreciate you if you could catch a mistake if there is any. I also would thank you if you can suggest any other problems that are not too deep but would help me advance my upractice and understanding of the Morita theorem.