# Left-Exact Functors

We saw (in theorem 1 here) that the localization functor is exact, which gave us a whole slew of nice properties, including preservation of submodules, quotient modules, finite intersection/sum, etc. However, exactness is often too much to ask for.

Throughout this article, *A* and *B* are fixed rings and is a covariant additive functor.

Definition.We say F is

left-exactif it takes a short exact sequence of A-modulesto an exact sequence of B-modules

Immediately we can weaken the condition.

Lemma 1.F is left-exact if and only if it takes an exact sequence of A-modules

to an exact sequence of B-modules

**Proof**

(⇐) is obvious. For (⇒) given an exact , we extend it to an exact . [Recall that is the cokernel of *g*.] We can split this exact sequence into short exact ones:

which gives an exact sequence as desired. ♦

**Note**

Left-exact functors are not as nice as exact ones but we still get some useful results out of them. For example, if is injective, so is so for a submodule , we can consider as a submodule of .

Also, for any , since is exact, applying *F* gives an exact and thus .

# Hom Functors Are Left-Exact

The main result we wish to show is

Proposition 1.For any A-module M, the functor is a left-exact functor.

**Proof**

Take the exact sequence . We need to show that

is exact. It is easy to show is injective (easy exercise). Next, we have

Conversely suppose satisfies . Then . Since *f* is injective it follows that for each , we have for a unique . This map is clearly *A*-linear so . ♦

Here is one application of this result.

Corollary 1.Let ; for each A-module, let . Then

**Proof**

Indeed the functor is naturally isomorphic to . Now apply the above. ♦

**Exercise A**

Find a surjective for which is not surjective. This shows that the functor is not exact in general.

# Converse Statement

Next we have the following converse.

Proposition 2.Suppose and are A-linear maps such that for any A-module M,

is exact. Then is exact.

**Proof**

First we show *f* is injective. Let so that is exact. By left-exactness of Hom we have an exact:

.

Since is injective, and in particular .

Next we show . Setting gives:

so This gives .

Finally we set . The following sequence:

is exact. The inclusion gives and so for some . Then . ♦

# Another Left-Exactness

Now suppose is a *contravariant* additive functor.

Definition.We say F is

left-exactif it takes a short exact sequence of A-modulesto an exact sequence of B-modules

Again we have:

Lemma 2.F is left-exact if and only if it takes an exact sequence of A-modules

to an exact sequence of B-modules

**Proof**

Exercise. ♦

Now we show, as before:

Proposition 3.For any A-module M, the functor is a left-exact functor.

**Proof**

Take the exact sequence . We need to show that

is exact for any *A*-module *M*. Injectivity of is obvious. Also so .

It remains to show . Pick such that so that . Hence *h* factors through . Since , we have for some . ♦

# Another Converse Statement

Finally, the reader should expect the following.

Proposition 4.Suppose and are A-linear maps such that for any A-module M,

is exact. Then is exact.

**Proof**

We leave it as an exercise to show: *g* is surjective, .

It remains to show ; let and take the exact sequence

If is the canonical map, so we have for some . Thus . ♦

As an application, let us prove the following.

Corollary 2.Let be an ideal. If is an exact sequence of A-modules, then we get an exact sequence of -modules:

**Proof**

Let . It suffices to show that for any *B*-module *Q*, the sequence

is exact. But since gives the *B*-module induced from *M*, the above sequence is naturally isomorphic to:

which we know is exact because the Hom functor is left-exact. ♦

This leads to the following definition.

Definition.Suppose is a covariant additive functor. We say F is

right-exactif it takes a short exact sequence of A-modulesto an exact sequence of B-modules

**Exercise B**

Prove that *F* is right-exact if and only if it takes an exact sequence of *A*-modules to an exact sequence of *B*-modules

Thus, we have shown that the functor

is right-exact. This will be generalized in future chapters.

**Exercise C**

- Prove that the functor is not exact in general.
- Use a direct proof to show is right-exact.

In summary, we have covered the following concepts: