and
nonabelian homological algebra
An object $P$ of a category $C$ is projective (with respect to epimorphisms) if it has the left lifting property against epimorphisms.
This means that $P$ is projective if for any morphism $f:P \to B$ and any epimorphism $q:A \to B$, $f$ factors through $q$ by some morphism $P\to A$.
Yet another way to say this is that
An object $P$ is projective precisely if the hom-functor $Hom(P,-)$ preserves epimorphisms.
This generalizes the notion of projective modules over a ring.
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.”
The dual notion is that of injective objects.
A category $C$ has enough projectives if for every object $X$ there is an epimorphism $P\to X$ where $P$ is projective.
Equivalently: if every object admits a projective presentation.
This terminology refers to the existence of projective resolutions, prop. 2 below.
Projective objects and injective objects in abelian categories $\mathcal{A}$ 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 $Ch_\bullet(\mathcal{A})$, with respect to one of the two standard model structures on chain complexes.
The following are equivalent
$X \in \mathcal{A}$ is a projective object (in that it has the left lifting property against epimorphisms, def. 2);
The hom-functor $Hom(X,-) : \mathcal{A} \to$ Ab is an exact functor.
For every object $X$, the hom-functor $Hom(X,-)$ is a left exact functor. So the second statement is equivalently that it is also right exact precisely if $X$ is projective.
of prop. 1
Let
be a short exact sequence and consider
Since $Hom(X,-)$ is generally left exact, by the above remark, it preserves kernels and so $Hom(X,p)$ is a monomorphism and $ker( Hom(X,p) ) \simeq im ( Hom(X,i) )$, generally.
Therefore we are reduced to showing that $Hom(X,p)$ is an epimorphism precisely if $X$ is projective. But this is def. 2.
Let $\mathcal{A}$ be an abelian category.
For $N \in \mathcal{A}$ an object, a projective resolution of $N$ is a chain complex $(Q N)_\bullet \in Ch_\bullet(\mathcal{A})$ equipped with a chain map
(with $N$ regarded as a complex concentrated in degree 0) such that
this morphism is a quasi-isomorphism (this is what makes it a resolution), which is equivalent to
being an exact sequence;
all whose entries $(Q N)_n$ are projective objects.
This means precisely that $Q N \to N$ 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.
If $\mathcal{A}$ has enough projectives in the sense of def. 3, then every object has a projective resolution.
The axiom of choice can be phrased as “all objects of Set are projective.”
See also internally projective object and COSHEP.
If $C$ has pullbacks and epimorphisms are preserved by pullback, as is the case in a pretopos, then $P$ is projective iff any epimorphism $Q\to P$ is split.
An object in Ab, an abelian group, is projective precisely if it is a free group.
For $R$ a commutative ring, an object in $R$Mod, an $R$-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.
The projective objects in compact Hausdorff topological spaces are precisely the extremally disconnected profinite sets.
We list examples of classes of categories that have enough projective, according to def. 3.
Assuming the axiom of choice, for $R$ a ring the category $R$Mod of modules over $R$ is has enough projectives.
See at projective module for more.
Assuming the axiom of choice, a free module $N \simeq R^{(S)}$ is projective.
By remark 6 and the free-forgetful adjunction.
More explicitly: if $S \in Set$ and $F(S) = R^{(S)}$ is the free module on $S$, then a module homomorphism $F(S) \to N$ is specified equivalently by a function $f : S \to U(N)$ from $S$ to the underlying set of $N$, which can be thought of as specifying the images of the unit elements in $R^{(S)} \simeq \oplus_{s \in S} R$ of the ${\vert S\vert}$ copies of $R$.
Accordingly then for $\tilde N \to N$ an epimorphism, the underlying function $U(\tilde N) \to U(N)$ is an epimorphism, and by remark 6 the axiom of choice in Set says that we have all lifts $\tilde f$ in
By adjunction these are equivalently lifts of module homomorphisms
of prop. 3
For $N \in R Mod$ and $U(N) \in Set$ its underlying set, consider the $R$-linear map
out of the direct sum of ${\vert U(N)\vert}$ copies of $N$, which sends the unit element $1 \in R_{n}$ of the $n$-labeled copy of $R$ to the corresponding element of $n$ (and is thus fixed on all other elements by $R$-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 $F U(M)$ 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 $Ab(E)$ of abelian group objects in a general Grothendieck topos $E$ (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:
Let $E$ be a W-pretopos that satisfies COSHEP. Then $Ab(E)$ has enough projectives.
The idea of the proof is that under COSHEP, the underlying object of an abelian group $A$ in $E$ admits an epimorphism from a projective object $p \colon X \to U(A)$ in $E$. Then the corresponding $F(X) \to A$ is an epimorphism out of a projective in $Ab(E)$, for this map is a composite of epimorphisms
(the first is epic because left adjoints preserve epis, whereas the second map, the component of the counit $\varepsilon \colon F U \to id$ at $A$, is epic because $U$ is faithful).
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
For instance section 4.3 of