I am aware that this question has been asked in other ways such as here:
(1) Show that $R \otimes_R M\cong M$ is isomorphic to $M$, for every left $R$-module $M$, $R$ a ring
(2) Prove that $R \otimes_R M \cong M$
And a related (more general) question is here:
(3) Show that $Hom_R(R^n,M) \cong M^n$ for R-modules
However, I some of the subtleties still seem a little hand-wavy to me, and I would very much appreciate if anyone could shed some light on them. In particular, my approach to try proving this was very similar to what is briefly mentioned (but not actually given in detail) in link (3) above. I said let $f: Hom_{R-Mod}(R,M) \rightarrow M$ be defined by $f(\phi) = \phi(1)$, for $\phi \in Hom_{R-Mod}(R,M)$. I proved f was an injective R-module homomorphism, but proving surjectivity is where I am not quite sure I am satisfied. It is relatively easy to prove that f is surjective onto the group M, but why does this imply it is surjective onto the the R-module M?
I suppose I am a little unclear about just what the "R-module structure" actually IS. For instance, in Aluffi's book, this structure is given as a ring homomorphism $R \rightarrow End_{Ab}(M)$, but yet this doesn't seem to matter when dealing with the surjectivity question? In the event that the order of the group M is finite, say |M|=n, is the set underlying the module M also of the same cardinality n? Why don't we care about what the size of the set $End_{Ab}(M)$ is (since this is what is actually giving M its module structure)?
I'm sorry if this question seems trivial, but I just can't seem to grasp why exactly we can "gloss over" what seems to be the the essential difference between M as a group and as a module.
Thanks for any help!