The goal of this post is to summarize some more-or-less standard facts about projective objects. A subtlety that arises here is that in abelian categories there are several conditions equivalent to being projective (we’ll list seven of them below) which are not equivalent in general. We’ll pay more attention than might be usual to this issue.

In particular, several times below we’ll give a list of conditions and a hypothesis under which they’ll be equivalent, and these conditions won’t all be equivalent in general. In these lists we’ll adopt the following convention: whenever we give a list of conditions and prove implications between them, the list will be organized so that proofs downward are easier and require fewer hypotheses, while proofs upward are harder and require more hypotheses. We’ll also prove more implications than we strictly need in order to see this more explicitly.

**Exact functors**

Here is some slightly nonstandard terminology: we’ll call an -enriched category a **linear category** for better agreement with the terminology that a -enriched category ( a commutative ring) is a -linear category, and similarly we’ll call an -enriched functor between such categories a **linear functor**. Recall that for a functor between additive categories this condition is equivalent to preserving finite biproducts and is often called being an additive functor.

For us, a functor is **right exact** if it preserves finite colimits. Dually, it is **left exact** if it preserves finite limits, and **exact** if it preserves both finite colimits and finite limits. This definition may not look familiar, but it’s equivalent to more familiar definitions in linear categories; on the other hand, unlike the more familiar definition in terms of short exact sequences, it continues to make sense in “nonlinear” categories.

**Proposition:** Let be a linear functor between linear categories with finite colimits. The following conditions are equivalent:

- is right exact.
- preserves coequalizers.
- preserves pushouts.
- preserves cokernels.

Dually, is left exact iff it preserves equalizers iff it preserves pullbacks iff it preserves kernels.

*Proof.* : by definition. (This makes no use of linearity except that for we need that have zero morphisms and that preserves them.)

: recall that in a category with finite colimits, all finite colimits can be built out of finite coproducts and coequalizers, and hence a functor between two such categories is right exact iff it preserves finite coproducts and coequalizers. (This again makes no use of linearity.)

Since a linear functor automatically preserves finite coproducts, is right exact iff it preserves coequalizers.

: we already know that finite colimits can be built out of finite coproducts and coequalizers, but it’s worth being explicit for pushouts. The pushout of two maps is equivalent to the coequalizer of

where the two arrows are the composites and . Hence if preserves coequalizers and finite coproducts, which as usual is automatic when is linear, then preserves pushouts.

: a cokernel is just a coequalizer where one of the maps is zero, so for a functor which preserves coequalizers to preserve cokernels it suffices that preserve zero morphisms. (This makes no use of linearity except for having and preserving zero morphisms.)

: a cokernel is also just a pushout where one of the objects is zero, so for a functor which preserves pushouts to preserve cokernels it suffices that preserves zero objects. (This makes no use of linearity except for having and preserving zero objects.)

: in a linear category, the coequalizer of two maps is the cokernel of the map , and hence a linear functor between linear categories which preserves cokernels also preserves coequalizers. (This is the first and only place where we’ve needed to subtract morphisms.)

These implications are not reversible in general. For example, the underlying set functor preserves coequalizers but not coproducts. It does not preserve coproducts since, for example, the natural map is not an isomorphism. But it does preserve coequalizers: if is a parallel pair of morphisms between two abelian groups, then their coequalizer is the quotient of by the relation in both sets and abelian groups.

(Note that restricting attention to abelian groups is crucial here: for groups, the coequalizer of a pair of maps is the quotient of by the normal closure of the subgroup generated by , and so the above argument fails for two reasons: first, the elements don’t form a subgroup in general, and second, they don’t form a normal subgroup in general.)

For functors between abelian categories there are some other equivalent conditions which are perhaps even more familiar, and which are the origin of the “right” part of “right exact.”

**Proposition:** Let be a linear functor between abelian categories. The following conditions are equivalent:

- is right exact.
- If is an exact sequence, then so is .

(**Edit, 2/18/16:** Previously this proposition claimed but did not prove that is right exact iff it preserves epimorphisms; in fact this is false, as Arpon pointed out in the comments.)

A subtlety in what follows is that there are various equivalent conditions for a sequence to be exact at an object which are equivalent in abelian categories but not in general. We won’t spend time on this, though; exact sequences are really only reasonable to talk about in abelian categories anyway.

*Proof.* : exactness at means that the cokernel of is trivial, and this follows since the cokernel of is trivial by exactness at and is right exact. Exactness at is equivalent to the condition that the cokernel of is , and this again follows since the cokernel of is by exactness at and is right exact.

: If is a morphism, then taking its cokernel gives an exact sequence

and by hypothesis, applying to this exact sequence gives an exact sequence

.

By exactness at , , and hence preserves cokernels.

**Projective objects**

There are several equivalent ways to define projective objects in an abelian category; six of them are provided by the above, while a seventh is new.

**Definition-Theorem:** An object in an abelian category is a **projective object** if it satisfies any of the following equivalent conditions:

- is right exact (hence exact).
- preserves coequalizers.
- preserves pushouts.
- preserves cokernels.
- If is a short exact sequence, then so is .
- preserves epimorphisms; explicitly, if is a morphism and is an epimorphism, then factors through .
- Every short exact sequence splits.

Dually, an object in an abelian category is an **injective object** if it satisfies any of the above equivalent conditions in the opposite category.

(It’s crucial for this theorem that is thought of as taking values in abelian groups rather than sets.)

*Proof.* is a linear functor between linear categories, so we know that the first four conditions are equivalent by the above. It is even a linear functor between abelian categories, so we know that the first five conditions are equivalent by the above. It remains to show that the sixth and seventh conditions are equivalent to the others. There are a few ways to do this.

: a morphism is an epimorphism iff its cokernel is trivial, which is preserved by hypothesis. (This argument shows more generally, with “cokernel” replaced by “cokernel pair,” that any functor that preserves pushouts preserves epimorphisms.)

: if is exact, then in particular is an epimorphism, hence so is the induced map . Taking a preimage of shows that there is a section to the epimorphism , which splits the short exact sequence.

: let be a map and be an epimorphism. Then factors through iff the projection map

from the pullback of and to has a section (by the universal property). In an abelian category, epimorphisms are stable under pullback, so is also an epimorphism, and hence it fits into a short exact sequence

which, by hypothesis, splits. Hence preserves epimorphisms.

: since is left exact, if is exact, then is exact in the middle. It is exact on the right iff the map remains an epimorphism, but this is true by hypothesis.

*Example.* In the abelian category of modules over a ring , every free module is a projective module. It’s easy to verify, for example, that every short exact sequence ending in a free module splits since one can choose a splitting on each generator using freeness (although for arbitrary free modules we need the axiom of choice to do this).

There’s a sense in which the above argument mostly does not take place in , but in . To formalize this, we’ll repackage this argument as follows:

- First, find a definition of projective object in an arbitrary category, not necessarily abelian, with the property that every set is a projective object (assuming the axiom of choice).
- Second, show that the free module functor sends projective objects to projective objects.

We have seven equivalent definitions to choose from. Of them, only definitions 1, 2, 3, and 5 make sense in an arbitrary category. The first three can be ruled out because they don’t have the correct behavior in : for example, the only set such that preserves even finite coproducts (let alone pushouts or finite colimits) is the one-element set. The definition we’ll use is Definition 5.

**Definition:** An object in a category is a **projective object** if preserves epimorphisms.

(Here it’s not crucial, in an abelian category, that is thought of as taking values in abelian groups; we can think of it as taking values in sets for the purposes of this definition. In general these two conditions may behave differently in an enriched category depending on whether one takes the ordinary hom or the enriched hom in the first condition, but we won’t deal with any such cases.)

It now follows, assuming the axiom of choice, that every set is projective. In fact this statement can be shown to be equivalent to the axiom of choice!

The statement that free modules are projective now follows from the following observation.

**Proposition:** Let be a functor with a right adjoint . If preserves epimorphisms, then preserves projective objects.

*Proof.* Let be a projective object. Then, by hypothesis,

.

The RHS is a composite of two functors that preserve epimorphisms, so it also preserves epimorphisms; hence the LHS also preserves epimorphisms.

**Corollary:** The free -module functor preserves projective objects; hence, assuming the axiom of choice, every free -module is projective.

*Proof.* has right adjoint the forgetful functor , so it suffices to show that the forgetful functor preserves epimorphisms. But the epimorphisms of -modules are precisely the morphisms with trivial cokernel, which in turn are precisely the morphisms which are surjective on underlying sets. One way to see this is to show, as we did above for abelian groups, that the forgetful functor preserves coequalizers.

**Corollary:** Let be a commutative ring and let be projective modules over . Then the tensor product is also projective.

*Proof.* The tensor-hom adjunction here reads

.

By hypothesis, is projective, so preserves epimorphisms. It follows that preserves projective objects.

In fact, we are surprisingly close to describing all projective modules. Next we need the following observation. Recall that an object is a retract of an object if there are maps such that (so is a split epimorphism and is a split monomorphism). In an abelian category, this condition just means that is a direct summand of .

**Proposition:** A retract of a projective object is projective.

*Proof.* Let be a projective object, be an object, and be maps satisfying , so that is a retract of . Let be an epimorphism. By hypothesis, every map factors through , and we want to show that every map similarly factors through .

Given a map , by hypothesis the composite

factors through . Since , it follows that the composite

also factors through , as desired.

**Corollary:** A retract / direct summand of a free module is a projective module.

*Example.* Let be the ring of continuous functions on a compact Hausdorff space . Any vector bundle on is a direct summand of a trivial vector bundle, so the -module of continuous sections of a vector bundle is a finitely generated projective -module, and in fact by the Serre-Swan theorem this is an equivalence of categories.

This construction already exhausts all projective modules.

**Theorem:** The projective modules over a ring are precisely the retracts of free modules. If is finitely generated, then the free module can also be taken to be finitely generated.

*Proof.* Let be a projective module and let be a free module equipped with an epimorphism (so that is sent to a set of generators of ). Then fits into a short exact sequence

and by hypothesis this short exact sequence splits, so is a retract of as desired.

on February 18, 2016 at 4:52 pm |ArponYou claim that if a linear functor between abelian categories preserves epis then it’s right exact, but it’s not proven by the implications given in the proof. And it seems to be false in general (e.g. http://math.stackexchange.com/a/406602/5226), though of course it’s equivalent for left-exact functors like Hom.

on February 18, 2016 at 6:35 pm |Qiaochu YuanOops, you’re completely right; I never show that implies anything. Thanks for the correction!

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