This is a relatively elementary proof (compared to others out there) of the fact that tensor is a right exact functor. Proof is taken from Hungerford, and reworded slightly. The key prerequisites needed are the universal property of quotient and of tensor product.
If is an exact sequence of left modules over a ring and is a right -module, then is an exact sequence of abelian groups. An analogous statement holds for an exact sequence in the first variable.
We split our proof into 3 parts: (i) ; (ii) ; and (iii) .
(i) Since is an epimorphism by hypothesis every generator of is of the form for some . Thus contains all generators of , hence .
(ii) Since we have and hence .
(iii) Let be the canonical epimorphism. From (ii), so (by universal property of quotient Theorem 1.7) there is a homomorphism such that We shall show that is an isomorphism. Then .
We show first that the map given by , where , is independent of the choice of . Note that there is at least one such since is an epimorphism. If , then and , hence for some . Since and , we have
Therefore is well-defined.
Verify that is middle linear:
By universal property of tensor product there exists a unique homomorphism such that , where .
Therefore, for any generator of , hence is the identity map.
so is the identity so that is an isomorphism.