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.

## Statement

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.

## Proof

(Hungerford 210)

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:

Let .

By universal property of tensor product there exists a unique homomorphism such that , where .

Therefore, for any generator of , hence is the identity map.

Similarly

so is the identity so that is an isomorphism.