We say that a class of maps in a category has the three-for-two property if for any commutative triangle
in which two of the three maps belong to , then so is the third.
We call this property three-for-two rather than two-for-three because this is like getting three apples for the price of two in a food store.
We shall say that a triple of classes of maps in finitely bicomplete category is a Quillen model structure, or just a model structure, if the following conditions are satisfied: * the class has the three-for-two property; * The pairs and are weak factorisation systems.
A Quillen model category, or just a model category, is a category equipped with a model structure.
The definition above is equivalent to the notion of closed model structure introduced by Quillen. The proof of the equivalence depends on Tierneyβs lemma below.
A map in is called a fibration, a map in a cofibration and a map in a weak equivalence. A map in is also said to be acyclic. It follows from the axioms that every map admits a factorisation with an acyclic cofibration and a fibration, and also a factorisation with a cofibration and an acyclic fibration.
An object is said to be fibrant if the map is a fibration, where is the terminal object of . Dually, an object is said to be cofibrant if the map is a cofibration, where is the initial object. We shall say that an object is fibrant-cofibrant if it is both fibrant and cofibrant.
A model structure is said to be right proper if the base change of a weak equivalence along a fibration is a weak equivalence. Dually, a model structure is said to be left proper if the cobase change of a weak equivalence along a cofibration is a weak equivalence. A model structure is said to be proper if it is both left and right proper.
We shall say that a model structure on a category is trivial if is the class of isomorphisms, in which case .
We shall say that a model structure on a category is coarse if is the class of all maps, in which case the pair is just an arbitrary weak factorisation system in the category .
For less trivial examples, see .
If is a model structure on a category , then the triple is a model structure on the opposite category .
If is an object of a category and is a class of maps in , we shall denote by the class of maps in whose underlying map in belongs to . Dually, we shall denote by the class of maps in whose underlying map belongs to .
If is a model structure on a category , then the triple is a model structure in the slice category for any object . Dually, the triple is a model structure in the coslice category .
This follows from the analogous properties of weak factorisation systems here.
The classes , , and are closed under composition and retracts. The classes and are closed under base changes and products. The classes and are closed under cobase changes and coproducts. The intersection is the class of isomorphisms.
This follows directly from the general properties of the classes of a weak factorisation system here.
A retract of a fibrant object is fibrant. A product of a family of fibrant objects is fibrant. Dually, a retract of a cofibrant object is cofibrant. A coproduct of a family of cofibrant objects is cofibrant.
If an object is a retract of an object , then the map is a retract of the map . If an object is the product of a family of object , then the map is the product of the family of maps .
(Myles Tierney) The class is closed under retracts.
Notice first that the class is closed under retracts by Proposition . Suppose now that a map is a retract of a map in . We want to show . We have a commutative diagram
with and . We suppose first that is a fibration. In this case, factor as with an acyclic cofibration and a fibration. The map is acyclic by three-for-two, since and are acyclic. The square
has a diagonal filler , since is a fibration. We get a commutative diagram
The map is a retract of , since . Thus, is acyclic, since is an acyclic fibration. In the general case, factor as with an acyclic cofibration and a fibration. By taking a pushout we obtain a commutative diagram
where and . The map is a cobase change of , so is an acyclic cofibration, since is. Thus, is acyclic by three-for-two, since is acyclic by assumption. So is acyclic by the first part, since is a fibration. Finally, is acyclic, since is acyclic.
The homotopy category of a model category is defined to be the category of fractions
The canonical functor is a localisation with respects to the maps in . Recall that a functor is said to invert a map if the morphism is invertible. The functor inverts the maps in and for any functor which inverts the maps in , there is a unique functor such that . We shall see in theorem that a map in the category is acyclic if and only if it is inverted by the functor . We shall see in corollary that the category is locally small if the category is locally small.
We denote by (resp. , ) the full subcategory of spanned by the fibrant (resp. cofibrant, fibrant-cofibrant) objects of . If is a class of maps in , we shall put
Let us put
Then the square of inclusions,
induces a commutative square of categories and canonical functors,
We shall see in theorem that the four functors in the square are equivalences of categories.
The following result is easy to prove but technically useful:
The pair and the pair are weak factorisation systems in the category . The pair and the pair are weak factorisation systems in the category . The pair and the pair are weak factorisation systems in the category .
Let us how that the pair is a weak factorisation system in . We shall use the characterisation of a weak factorisation system here. Obviously, we have for every and . If is a map between fibrant objects, let us choose a factoriation with an acyclic cofibration and a fibration. The object is fibrant, since is a fibration and is fibrant. This shows that and . Finally, the class is closed under codomain retracts, since a retract of a fibrant object is fibrant. Similarly, the class is closed under domain retracts.
The inclusion is a cofibration if is cofibrant, and the inclusion is a cofibration if is cofibrant.
The inclusion is a cobase change of the map , since the square
is a pushout. Hence the map is a cofibration if is cofibrant (since the class of cofibrations is closed under cobase changes by Proposition )
A cylinder for an object is a quadruple obtained by factoring the codiagonal as a cofibration followed by a weak equivalence .
The notation introduced by Quillen suggests that a cylinder represents the first two terms of a cosimplicial object
with and . This is the notion of a cosimplicial framing? of an object . Beware that if and are the maps in the category , then and . We may think of a cylinder has an oriented object with two faces, with the face representing the source of the cylinder and the face representing the target.
The transpose of a cylinder is the cylinder .
The maps and are acyclic, and they are acyclic cofibrations when is cofibrant.
The map and are acyclic by three-for-two, since we have and is acyclic. If is cofibrant, then the inclusions and are cofibrations by Lemma . Hence also the composite and (since the class of cofibrations is closed under composition by Proposition ).
A mapping cylinder of a map is obtained by factoring the map as a cofibration followed by a weak equivalence . We then have and . The factorisation
is called the mapping cylinder factorisation of the map . The map is acyclic by three-for-two, since is acyclic and we have .
The maps is a cofibration when is cofibrant, and the map is a cofibration when is cofibrant.
The inclusion is a cofibration when is cofibrant by Lemma . Hence also the composite in this case. The inclusion is a cofibration when is cofibrant by Lemma . Hence also the composite in this case.
If is cofibrant, then a mapping cylinder for a map can be constructed from a cylinder by the following diagram with a pushout square
We have and by construction. The map is a cofibration by cobase change, since the map is a cofibration. Let us show that is acyclic. For this, it suffices to show that is acyclic by three-for-two, since . The two squares of the following diagram are pushout,
hence also their composite,
by the lemma here. This shows that the map is a cobase change of the map . But is an acyclic cofibration by Lemma , since is cofibrant. It follows that is an acyclic cofibration (since the class of acyclic cofibrations is closed under cobase changes by Proposition ).
(Ken Brown 1) Let be a model category and let be a functor defined on the sub-category of cofibrant objects and taking its values in a category equipped with class of weak equivalences containing the units and satisfying three-for-two. If the functor takes an acyclic cofibration to a weak equivalence, then it takes an acyclic map to a weak equivalence.
If is an acyclic map between cofibrant objects, let us choose a mapping cylinder factorisation . The maps and are cofibrations by Lemma , since and are cofibrant. The map is acyclic by three-for-two, since and is acyclic. Thus, is a weak equivalence. Hence also the map by three-for-two since we have and contains the units. The map is acyclic by three-for-two, since and and are acyclic. Thus, is a weak equivalence, and it follows by three-for-two that is a weak equivalence since .
Recall that a functor is said to invert a morphism in its domain if it takes this morphism to an isomorphism.
(Ken Brown 2)
If a functor inverts acyclic cofibrations, then it inverts weak equivalences.
If a functor inverts acyclic fibrations, then it inverts weak equivalences.
Ken Brownβs lemma implies that the inclusion induces an isomorphism of categories,
If is a cyclinder for , then a left homotopy between two maps is a map such that and and . We shall say that is the source of the homotopy and that is the target,
The reverse of an homotopy is the homotopy defined by the same map but on the transpose cylinder . The homotopy unit of a map is defined by the map .
Two maps are left homotopic, , if there exists a left homotopy with domain some cylinder object for .
The left homotopy relation between the maps can be defined on a fixed cylinder for , when is fibrant.
Let us show that if two maps are homotopic by virtue of a homotopy defined on a cylinder , then then they are homotopic by virtue of a homotopy defined on any another cylinder . By assumption, we have for a map . Let us choose a factorisation with an acyclic cofibration and a fibration. The map is acyclic by three-for-two, since the maps and are. Hence the square
has a diagonal filler since the map is a cofibration. But the square
has also a diagonal filler , since is an acyclic cofibration and is fibrant. The composite is a left homotopy .
If a functor inverts weak equivalences, then the implication
is true for any pair of maps in . The same result is true for a functor defined on or on .
If is a cylinder for , then the map is invertible by the assumption of since is acyclic. Hence we have , since we have
If is a homotopy between two map , then
Let us now consider the case where the domain of the functor is the category . Observe that if is cofibrant, then so is the object in a cylinder , since the map is a cofibration and the object is cofibrant (since a coproduct of cofibrant objects is cofibrant by Corollary ). Hence the cylinder belongs to the category and the proof above can be repeated in this case. Let us now consider the case where the domain of the functor is the category . In this case the left homotopy relation between the maps can defined on a fixed cylinder for by Lemma , since is fibrant. A cylinder for can be constructed by factoring the map as an acyclic cofibration followed by a fibration . The object is cofibrant, since is cofibrant. But is also fibrant, since is a fibration and is fibrant. Hence the cylinder belongs to the category and the proof above can be repeated.
The left homotopy relation on the set of maps is reflexive and symmetric. We shall denote by the quotient of the set by the equivalence relation generated by the left homotopy relation. The relation is compatible with composition on the left: the implication
is true for every maps and . This defines a functor
The left homotopy relation between the maps is an equivalence when is cofibrant.
Cylinders for can be composed as cospan. More precisely, the composite of a cylinder with a cylinder is the cylinder defined by the following diagram with a pushout square,
The map is defined by the condition and . Let us show that is acyclic. The map is an acyclic cofibration by Lemma , since is cofibrant. It follows that is an acyclic cofibration, since it is a cobase change of . Hence the map is acyclic by three-for-two, since we have and the maps and are acyclic. It remains to show that the map is a cofibration. For this, we can use the following diagram with a pushout square,
The map in this diagram is a cobase change of the map . But the map is a cofibration, since the maps and are cofibrations. This proves that the map is a cofibration. It follows that the composite is a cofibration, since the map is a cofibration by Lemma . We have proved that is a cylinder for . We can now prove that the left homotopy relation on the set of maps is transitive. Let and be three maps and suppose that is a left homotopy , and is a left homotopy . There is then a unique map such that and , since . This defines a homotopy , since and .
(Covering homotopy theorem) Let be cofibrant, let be a fibration, let , and let be a left homotopy with source . Then there exists a left homotopy with source such that .
The square
has a diagonal filler , since is an acyclic cofibration by Lemma and is a fibration.
(Homotopy lifting lemma) Let be an acyclic fibration, let and , and let be a left homotopy . Then there exists a map defining a left homotopy such that .
The square
has a diagonal filler , since is a cofibration and is an acylic fibration.
If is cofibrant, then the functor inverts acyclic maps between fibrant objects.
Let us first show that the functor inverts acyclic fibrations. If is an acyclic fibration and , then the square
has a diagonal filler, since is cofibrant and is an acyclic fibration. Hence there exists a map such that . This shows that the map is surjective. Let us show that it is injective. If and , then by the homotopy lifting lemma . We have proved that the map is bijective. It then follows from Ken Brownβs lemma that that the functor inverts acyclic maps between fibrant objects.
A map which is left homotopic to an acyclic map is acyclic.
Let be a left homotopy between two maps and . If is acyclic, then so is by three-for-two, since is acyclic. Hence the composite is acyclic by three-for-two, since is acyclic.
(Dual to Lemma ) The projection is a fibration if is fibrant, and the projection is a fibration if is fibrant
A path object for an object is a quadruple obtained by factoring the diagonal as a weak equivalence followed by a fibration .
The notation introduced by Quillen suggests that a path object represents the first two terms of a simplicial object
with and . This is the notion of simplicial framing? of an object . Beware that the source of a 1-simplex in a simplicial set is the vertex , and that its target is the vertex .
The transpose of a path object is the path object .
(Dual to Lemma ) The maps and are acyclic, and they are are acyclic fibrations when is fibrant.
A mapping path object of a map is obtained by factoring the map
as a weak equivalence followed by a fibration . By construction, we have and . The factorisation
is called the mapping path factorisation of the map . The map is acyclic by three-for-two, since is acyclic and .
If is fibrant, then a mapping path object for a map can be constructed from a path object for by the following diagram with a pullback square,
We have and by construction. The map is a fibration by base change, since the map is. Let us show that the map is acyclic. For this, it suffices to show that is acyclic by three-for-two, since . The two squares of the following diagram are cartesian,
hence also their composite,
by the lemma here. Hence the map is a base change of the map . But is an acyclic fibration by Lemma , since is fibrant. This shows that the map is acyclic (since the base change of an acyclic fibration is an acyclic fibration by Proposition ).
If is a path object for , then a right homotopy between two maps is defined to be a map such that and . We shall say that is the source of the homotopy and that is its target .
The reverse of is the homotopy defined by the same map but on the transpose path object . The unit homotopy of a map is the map .
Two maps are right homotopic, , if there exists a right homotopy with codomain a path object for .
(Dual to Lemma ) The right homotopy relation between the maps can be defined on a fixed path object for when is cofibrant.
We shall denote by the quotient of by the equivalence relation generated by the right homotopy relation. The right homotopy relation is compatible with composition on the right:
for every map . We thus obtain a functor
(Dual to Lemma ) The right homotopy relation between the maps is an equivalence when is fibrant.
(Homotopy extension theorem, dual to Lemma ). Let be fibrant, let be a cofibration, let , and let be a right homotopy with source . Then there exists a right homotopy with source such that .
(Homotopy prolongation lemma, dual to Lemma ) Let be fibrant, let be an acyclic cofibration, let and , and let be a right homotopy . Then there exists a map defining a right homotopy such that .
(Dual to Lemma ) If is fibrant, then the functor inverts acyclic maps between cofibrant objects.
If is a cylinder object for and is a path object for , then a map is double homotopy between four maps ,
The four corners of the square are representing maps , the horizontal sides are representing left homotopies, and the vertical sides are representing right homotopies.
If is fibrant, then every open box of three homotopies, opened at the top, between four maps ,
can be filled by a double homotopy (ie , and ).
The square
has a diagonal filler , since is a cofibration and is an acyclic fibration by Lemma .
If is fibrant, then the right homotopy relation on the set of maps implies the left homotopy relation. Dually, if is cofibrant, then the left homotopy relation implies the right homotopy relation. Hence the two relations coincide when is cofibrant and is fibrant.
Let be a right homotopy between two maps . By Lemma , the open box of homotopies
can be filled by a double homotopy , since is fibrant. This yields a left homotopy .
When is cofibrant and is fibrant, then two maps are said to be homotopic if they are left (or right) homotopic; we shall denote this relation by . We shall denote by the quotient of the set by the homotopy relation:
By definition, This defines a functor
that is, a distributor .
The homotopy relation is compatible with the composition law
if and are fibrant-cofibrant objects. It thus induces a composition law
and this defines a category if we put
for .
We say that a map in is a homotopy equivalence if it is invertible in the category .
It is obvious from the definition that the class of homotopy equivalences has the three-for-two property.
A map in is a homotopy equivalence iff there exists a map such that and .
Let us denote by , and , the canonical functors. The functor takes homotopic maps to the same morphism by Lemma . It follows that there is a unique functor such that the following triangle commutes,
We shall prove in Theorem below that the functor is an isomorphism of categories. We will need a lemma:
If is a set of morphisms in a category , then the localisation functor is an epimorphism of category. Moreover, for any category , the functor
induced by is fully faithful and it induces an isomorphism between the category and the full subcategory of spanned by the functors inverting the elements of . In particular, two functors are isomorphic iff the functors and are isomorphic.
Left to the reader.
The functor defined above is an isomorphism of categories. A map in is acyclic iff it is a homotopy equivalence.
(Mark Hovey) Let us first show that if a map in the category is acyclic, then it is a homotopy equivalence. The map is bijective for every cofibrant object by Lemma , since is an acyclic map between fibrant object. It follows that is inverted by the Yoneda functor
But the Yoneda functor is conservative, since it is fully faithful by Yoneda?. It follows that is invertible in the category . This shows that is a homotopy equivalence. Let us now prove that the functor is an isomorphism of categories. The canonical functor inverts acyclic maps by what we just proved. Hence there is a unique functor such that the following triangle commutes,
Let us show that the functors and are mutually inverses. Let us observe that the functor is an epimorphism, since it is surjective on objects and full. But we have . It follows that we have , since is an epimorphism. On the other hand, the functor is an epimorphism by Lemma , since it is a localisation. It follows that we have , since we have . We have proved that the functor and are mutually inverse. Let us now show that a homotopy equivalence is acyclic. We shall first consider the case where is a fibration. There exists a map such that and , since is a homotopy equivalence by assumption. There is then a left homotopy defined on a cylinder . I then follows from the covering homotopy theorem that there exists a left homotopy such that , since is a fibration. Let us put . Then and . Thus, , since the homotopy relation is a congruence. Hence the map is acyclic by Lemma , since is acyclic. But the map is retract of the map , since the following diagram commutes and we have ,
It then follows by Lemma that is a weak equivalence. The implication () is proved in the case where is a fibration. In the general case, let us choose a factorisation with an acyclic cofibration and a fibration. The map is a homotopy equivalence by the first part of the proof, since it is acyclic. Thus, is a homotopy equivalence by three-for-two for homotopy equivalences, since is a homotopy equivalence by assumption. Thus, is acyclic since it is a fibration. Hence the composite is acyclic by three-for-two.
A fibrant replacement of an object is a fibrant object together with an acyclic cofibration . A cofibrant replacement of an object is a cofibrant object together with an acyclic fibration .
A fibrant replacement of is obtained by factoring the map as an acyclic cofibration followed by a fibration . If is fibrant, we can take and . Similarly, a cofibrant replacement of is obtained by factoring the map as a cofibration followed by an acyclic fibration . If is cofibrant, we can take and .
A fibrant replacement of a cofibrant object is cofibrant; it is thus fibrant-cofibrant. Dually a cofibrant replacement of a fibrant object is fibrant-cofibrant.
The composite is acyclic. It can thus be factored as an acyclic cofibration followed by an acyclic fibration ,
The object is a fibrant-cofibrant replacement of the object . If is cofibrant, we can take , and if is fibrant, we can take (in which case we have , when is fibrant-cofibrant).
For every map , there exits two maps and fitting in the following commutative diagram,