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projective object

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Definition

General

Definition

An object PP of a category CC is projective (with respect to epimorphisms) if it has the left lifting property against epimorphisms.

This means that PP is projective if for any morphism f:PBf:P \to B and any epimorphism q:ABq:A \to B, ff factors through qq by some morphism PAP\to A.

A q P f B. \array{ && A \\ &{}^{\mathllap{\exists}}\nearrow& \downarrow^{\mathrlap{q}} \\ P &\stackrel{f}{\to}& B } \,.

Yet another way to say this is that

Definition

An object PP is projective precisely if the hom-functor Hom(P,)Hom(P,-) preserves epimorphisms.

Remark

This generalizes the notion of projective modules over a ring.

Remark

There are variations of the definition where “epimorphism” is replaced by some other type of morphism, such as a regular epimorphism or strong epimorphism or the left class in some orthogonal factorization system. In this case one may speak of regular projectives and so on. In a regular category “projective” almost always means “regular projective.”

Remark

The dual notion is that of injective objects.

Definition

A category CC has enough projectives if for every object XX there is an epimorphism PXP\to X where PP is projective.

Equivalently: if every object admits a projective presentation.

This terminology refers to the existence of projective resolutions, prop. 2 below.

In abelian categories

Projective objects and injective objects in abelian categories 𝒜\mathcal{A} are of central interest in homological algebra. Here they appear as parts of cofibrant resolutions and fibrant resolutions, respectively, in the category of chain complexes Ch (𝒜)Ch_\bullet(\mathcal{A}), with respect to one of the two standard model structures on chain complexes.

Equivalent characterizations

Proposition

The following are equivalent

  1. X𝒜X \in \mathcal{A} is a projective object (in that it has the left lifting property against epimorphisms, def. 2);

  2. The hom-functor Hom(X,):𝒜Hom(X,-) : \mathcal{A} \to Ab is an exact functor.

Remark

For every object XX, the hom-functor Hom(X,)Hom(X,-) is a left exact functor. So the second statement is equivalently that it is also right exact precisely if XX is projective.

Proof

of prop. 1

Let

0AiBpC0 0 \to A \stackrel{i}{\hookrightarrow} B \stackrel{p}{\to} C \to 0

be a short exact sequence and consider

Hom(X,A)Hom(X,i)Hom(X,B)Hom(X,p)Hom(X,C). Hom(X,A) \stackrel{Hom(X,i)}{\to} Hom(X,B) \stackrel{Hom(X,p)}{\to} Hom(X,C) \,.

Since Hom(X,)Hom(X,-) is generally left exact, by the above remark, it preserves kernels and so Hom(X,p)Hom(X,p) is a monomorphism and ker(Hom(X,p))im(Hom(X,i))ker( Hom(X,p) ) \simeq im ( Hom(X,i) ), generally.

Therefore we are reduced to showing that Hom(X,p)Hom(X,p) is an epimorphism precisely if XX is projective. But this is def. 2.

Projective resolutions

Let 𝒜\mathcal{A} be an abelian category.

Definition

For N𝒜N \in \mathcal{A} an object, a projective resolution of NN is a chain complex (QN) Ch (𝒜)(Q N)_\bullet \in Ch_\bullet(\mathcal{A}) equipped with a chain map

QNN Q N \to N

(with NN regarded as a complex concentrated in degree 0) such that

  1. this morphism is a quasi-isomorphism (this is what makes it a resolution), which is equivalent to

    (QN) 1(QN) 0N \cdots \to (Q N)_1 \to (Q N)_0 \to N

    being an exact sequence;

  2. all whose entries (QN) n(Q N)_n are projective objects.

Remark

This means precisely that QNNQ N \to N is an cofibrant resolution with respect to the standard model structure on chain complexes (see here) for which the fibrations are the positive-degreewise epimorphisms. Notice that in this model structure every object is fibrant, so that cofibrant resolutions are the only resolutions that need to be considered.

Proposition

If 𝒜\mathcal{A} has enough projectives in the sense of def. 3, then every object has a projective resolution.

Examples

Remark

The axiom of choice can be phrased as “all objects of Set are projective.”

See also internally projective object and COSHEP.

Example

If CC has pullbacks and epimorphisms are preserved by pullback, as is the case in a pretopos, then PP is projective iff any epimorphism QPQ\to P is split.

Example

An object in Ab, an abelian group, is projective precisely if it is a free group.

Example

For RR a commutative ring, an object in RRMod, an RR-module, is projective (a projective module, see there for more details) precisely if it is a direct summand of a free module. See at projective module for more on this.

Example

The projective objects in compact Hausdorff topological spaces are precisely the extremally disconnected profinite sets.

Properties

Existence of enough projectives

We list examples of classes of categories that have enough projective, according to def. 3.

Proposition

Assuming the axiom of choice, for RR a ring the category RRMod of modules over RR is has enough projectives.

See at projective module for more.

Lemma

Assuming the axiom of choice, a free module NR (S)N \simeq R^{(S)} is projective.

Proof

By remark 6 and the free-forgetful adjunction.

More explicitly: if SSetS \in Set and F(S)=R (S)F(S) = R^{(S)} is the free module on SS, then a module homomorphism F(S)NF(S) \to N is specified equivalently by a function f:SU(N)f : S \to U(N) from SS to the underlying set of NN, which can be thought of as specifying the images of the unit elements in R (S) sSRR^{(S)} \simeq \oplus_{s \in S} R of the |S|{\vert S\vert} copies of RR.

Accordingly then for N˜N\tilde N \to N an epimorphism, the underlying function U(N˜)U(N)U(\tilde N) \to U(N) is an epimorphism, and by remark 6 the axiom of choice in Set says that we have all lifts f˜\tilde f in

U(N˜) f˜ S f U(N). \array{ && U(\tilde N) \\ & {}^{\tilde f} \nearrow & \downarrow \\ S &\stackrel{f}{\to}& U(N) } \,.

By adjunction these are equivalently lifts of module homomorphisms

N˜ R (S) N. \array{ && \tilde N \\ & \nearrow & \downarrow \\ R^{(S)} &\stackrel{}{\to}& N } \,.
Proof

of prop. 3

For NRModN \in R Mod and U(N)SetU(N) \in Set its underlying set, consider the RR-linear map

( nU(n)R)N \left( \oplus_{n \in U(n)} R \right) \to N

out of the direct sum of |U(N)|{\vert U(N)\vert} copies of NN, which sends the unit element 1R n1 \in R_{n} of the nn-labeled copy of RR to the corresponding element of nn (and is thus fixed on all other elements by RR-linearity).

This is clearly a surjection and by lemma 1 it is a surjection out of a projective object.

A slightly subtle point is that there is no guarantee that the free module FU(M)F U(M) is actually projective, unless one assumes some form of the axiom of choice. Since the axiom of choice is not available in all toposes, one cannot use this procedure in general to construct, say, projective resolutions of abelian sheaves, hence in the abelian category Ab(E)Ab(E) of abelian group objects in a general Grothendieck topos EE (even though one can construct free resolutions), such as needed in general in abelian sheaf cohomology.

There are however weak forms of the axiom of choice that hold in many toposes, such as the presentation axiom, aka COSHEP. We have the following result:

Proposition

Let EE be a W-pretopos that satisfies COSHEP. Then Ab(E)Ab(E) has enough projectives.

The idea of the proof is that under COSHEP, the underlying object of an abelian group AA in EE admits an epimorphism from a projective object p:XU(A)p \colon X \to U(A) in EE. Then the corresponding F(X)AF(X) \to A is an epimorphism out of a projective in Ab(E)Ab(E), for this map is a composite of epimorphisms

F(X)F(p)FU(A)ε AAF(X) \stackrel{F(p)}{\to} F U(A) \stackrel{\varepsilon_A}{\to} A

(the first is epic because left adjoints preserve epis, whereas the second map, the component of the counit ε:FUid\varepsilon \colon F U \to id at AA, is epic because UU is faithful).

References

For instance section 4.3 of

Revised on November 21, 2013 09:45:59 by Urs Schreiber (188.200.54.65)