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Model structures

Definitions

Definition

We say that a class 𝒲 of maps in a category E has the three-for-two property if for any commutative triangle

in which two of the three maps belong to 𝒲, then so is the third.

Remark

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.

Definition

We shall say that a triple (π’ž,𝒲,β„±) of classes of maps in finitely bicomplete category E is a Quillen model structure, or just a model structure, if the following conditions are satisfied:

  • the class 𝒲 has the three-for-two property;
  • The pairs (π’žβˆ©π’²,β„±) and (π’ž,β„±βˆ©π’²) are weak factorisation systems.

A Quillen model category, or just a model category, is a category E equipped with a model structure.

Remark

The definition above is equivalent to the notion of closed model structure introduced by Quillen. The proof of the equivalence depends on Tierney’s lemma 1 below.

A map in β„± is called a fibration, a map in π’ž a cofibration and a map in 𝒲 a weak equivalence. A map in 𝒲 is also said to be acyclic. It follows from the axioms that every map f:Aβ†’B admits a factorisation f=pu:Aβ†’Eβ†’B with u an acyclic cofibration and p a fibration, and also a factorisation f=qv:Aβ†’Fβ†’B with v a cofibration and q an acyclic fibration.

An object X∈E is said to be fibrant if the map Xβ†’βŠ€ is a fibration, where ⊀ is the terminal object of E. Dually, an object A∈E is said to be cofibrant if the map βŠ₯β†’A is a cofibration, where βŠ₯ 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.

Example

We shall say that a model structure (π’ž,𝒲,β„±) on a category E is trivial if 𝒲 is the class of isomorphisms, in which case π’ž=E=β„±.

Example

We shall say that a model structure (π’ž,𝒲,β„±) on a category E is coarse if 𝒲 is the class of all maps, in which case the pair (π’ž,β„±) is just an arbitrary weak factorisation system in the category E.

For less trivial examples, see 1.

Duality

If (π’ž,𝒲,β„±) is a model structure on a category E, then the triple (β„± o,β„³ o,π’ž o) is a model structure on the opposite category E o.

If B is an object of a category E and β„³ is a class of maps in E, we shall denote by β„³/B the class of maps in E/B whose underlying map in E belongs to β„³. Dually, we shall denote by B\β„³ the class of maps in B\E whose underlying map belongs to β„³.

Slice and coslice

If (π’ž,𝒲,β„±) is a model structure on a category E, then the triple (π’ž/B,𝒲/B,β„±/B) is a model structure in the slice category E/B for any object B∈E. Dually, the triple (B\π’ž,B\𝒲,B\β„±) is a model structure in the coslice category B\E.

Proof

This follows from the analogous properties of weak factorisation systems here.

Proposition

The classes π’ž, π’žβˆ©π’², β„± and β„±βˆ©π’² are closed under composition and retracts. The classes β„± and β„±βˆ©π’² are closed under base changes and products. The classes π’ž and π’žβˆ©π’² are closed under cobase changes and coproducts. The intersection π’žβˆ©π’²βˆ©β„± is the class of isomorphisms.

Proof

This follows directly from the general properties of the classes of a weak factorisation system here.

Corollary

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.

Proof

If an object X is a retract of an object Y, then the map Xβ†’βŠ₯ is a retract of the map Yβ†’βŠ₯. If an object X is the product of a family of object (X i∣i∈I), then the map Xβ†’bot is the product of the family of maps (X iβ†’βŠ₯∣i∈I).

Lemma

(Myles Tierney) The class 𝒲 is closed under retracts.

Proof

Notice first that the class β„±βˆ©π’² is closed under retracts by Proposition 1. Suppose now that a map f:Aβ†’B is a retract of a map g:Xβ†’Y in 𝒲. We want to show fβˆˆπ’². We have a commutative diagram

with ts=1 A and vu=1 B. We suppose first that f is a fibration. In this case, factor g as g=qj:X→Z→Y with j an acyclic cofibration and q a fibration. The map q is acyclic by three-for-two, since q and j are acyclic. The square

has a diagonal filler d:Z→A, since f is a fibration. We get a commutative diagram

The map f is a retract of q, since d(js)=ts=1 A. Thus, f is acyclic, since q is an acyclic fibration. In the general case, factor f as f=pi:A→E→B with i an acyclic cofibration and p a fibration. By taking a pushout we obtain a commutative diagram

where kin 2=g and rin 1=1 E. The map in 2 is a cobase change of i, so in 2 is an acyclic cofibration, since i is. Thus, k is acyclic by three-for-two, since g=kin 2 is acyclic by assumption. So p is acyclic by the first part, since p is a fibration. Finally, f=pi is acyclic, since i is acyclic.

The homotopy category of a model category E is defined to be the category of fractions

Ho(E)=𝒲 βˆ’1E.Ho(\mathbf{ E})=\mathcal{W}^{-1}\mathbf{E}.

The canonical functor H:Eβ†’Ho(E) is a localisation with respects to the maps in 𝒲. Recall that a functor F:Eβ†’K is said to invert a map u:Aβ†’B if the morphism F(u):FAβ†’FB is invertible. The functor H inverts the maps in 𝒲 and for any functor F:Eβ†’K which inverts the maps in 𝒲, there is a unique functor Fβ€²:Ho(E)β†’K such that Fβ€²H=F. We shall see in theorem 2 that a map u:Aβ†’B in the category E is acyclic if and only if it is inverted by the functor H. We shall see in corollary 1 that the category Ho(E) is locally small if the category E is locally small.

We denote by E f (resp. E c, E fc) the full subcategory of E spanned by the fibrant (resp. cofibrant, fibrant-cofibrant) objects of E. If β„³ is a class of maps in E, we shall put

β„³ f=β„³βˆ©E fβ„³ c=β„³βˆ©E candβ„³ fc=β„³βˆ©E fc.\mathcal{M}_f=\mathcal{M} \,\cap\, \mathbf{ E}_f \quad \quad \mathcal{M}_c=\mathcal{M}\,\cap\, \mathbf{ E}_c \quad \mathrm{and} \quad \mathcal{M}_{f c}= \mathcal{M} \,\cap\, \mathbf{ E}_{f c}.

Let us put

Ho(E f)=𝒲 f βˆ’1E f,Ho(E c)=𝒲 c βˆ’1E c,andHo(E fc)=𝒲 fc βˆ’1E fc.Ho(\mathbf{ E}_f)=\mathcal{W}_f^{-1}\mathbf{E}_f, \quad \quad Ho(\mathbf{ E}_c)=\mathcal{W}_c^{-1}\mathbf{E}_c, \quad \mathrm{and}\quad Ho(\mathbf{ E}_{f c})=\mathcal{W}_{f c}^{-1}\mathbf{E}_{f c}.

Then the square of inclusions,

induces a commutative square of categories and canonical functors,

(1)

We shall see in theorem 2 that the four functors in the square are equivalences of categories.

The following result is easy to prove but technically useful:

Proposition

The pair (π’ž fβˆ©π’² f,β„± f) and the pair (π’ž f,β„± fβˆ©π’² f) are weak factorisation systems in the category E f. The pair (π’ž cβˆ©π’² c,β„± c) and the pair (π’ž c,β„± cβˆ©π’² c) are weak factorisation systems in the category E c. The pair (π’ž cfβˆ©π’² cf,β„± cf) and the pair (π’ž cf,β„± cfβˆ©π’² cf) are weak factorisation systems in the category E cf.

Proof

Let us how that the pair (π’ž fβˆ©π’² f,β„± f) is a weak factorisation system in E f. We shall use the characterisation of a weak factorisation system here. Obviously, we have uβ‹”p for every uβˆˆπ’ž fβˆ©π’² f and pβˆˆβ„± f. If f:Xβ†’Y is a map between fibrant objects, let us choose a factoriation f=pu:Xβ†’Eβ†’Y with u an acyclic cofibration and p a fibration. The object E is fibrant, since p is a fibration and Y is fibrant. This shows that uβˆˆπ’ž fβˆ©π’² f and pβˆˆβ„± f. Finally, the class π’ž fβˆ©π’² f is closed under codomain retracts, since a retract of a fibrant object is fibrant. Similarly, the class β„± f is closed under domain retracts.

Cylinders and left homotopies

Lemma

The inclusion in 1:Aβ†’AβŠ”B is a cofibration if B is cofibrant, and the inclusion in 2:Bβ†’AβŠ”B is a cofibration if A is cofibrant.

Proof

The inclusion in 1:Aβ†’AβŠ”B is a cobase change of the map βŠ₯β†’B, since the square

is a pushout. Hence the map in 1 is a cofibration if B is cofibrant (since the class of cofibrations is closed under cobase changes by Proposition 1)

A cylinder for an object A is a quadruple (IA,d 1,d 0,s) obtained by factoring the codiagonal βˆ‡ A=(1 A,1 A):AβŠ”Aβ†’A as a cofibration (d 1,d 0):AβŠ”Aβ†’IA followed by a weak equivalence s:IAβ†’A.

Remark

The notation (IA,d 1,d 0,s) introduced by Quillen suggests that a cylinder represents the first two terms of a cosimplicial object

I *A:Ξ”β†’AI^* A:\Delta \to \mathbf{A}

with I 0A=A and I 1A=IA. This is the notion of a cosimplicial framing of an object A. Beware that if d 0 and d 1 are the maps [0]→[1] in the category Δ, then d 0(0)=1 and d 1(1)=0. We may think of a cylinder (IA,d 1,d 0,s) has an oriented object with two faces, with the face d 1:A→IA representing the source of the cylinder and the face d 0:A→IA representing the target.

The transpose of a cylinder (IA,d 1,d 0,s) is the cylinder (IA,d 0,d 1,s).

Lemma

The maps d 1:A→IA and d 0:A→IA are acyclic, and they are acyclic cofibrations when A is cofibrant.

Proof

The map d 1 and d 0 are acyclic by three-for-two, since we have sd 1=1 A=sd 0 and s is acyclic. If A is cofibrant, then the inclusions in 1:Aβ†’AβŠ”A and in 2:Aβ†’AβŠ”A are cofibrations by Lemma 2. Hence also the composite d 1=(d 1,d 0)in 1 and d 0=(d 1,d 0)in 2 (since the class of cofibrations is closed under composition by Proposition 1).

A mapping cylinder of a map f:Aβ†’B is obtained by factoring the map (f,1 B):AβŠ”Bβ†’B as a cofibration (i A,i B):AβŠ”Bβ†’C(f) followed by a weak equivalence q B:C(f)β†’B. We then have f=q Bi A and q Bi B=1 B. The factorisation

f=q Bi A:A→C(f)→Bf=q_B i_A:A\to C(f)\to B

is called the mapping cylinder factorisation of the map f. The map i B is acyclic by three-for-two, since q B is acyclic and we have q Bi B=1 B.

Lemma

The maps i A:A→C(f) is a cofibration when B is cofibrant, and the map i B:B→C(f) is a cofibration when A is cofibrant.

Proof

The inclusion in 1:Aβ†’AβŠ”B is a cofibration when B is cofibrant by Lemma 2. Hence also the composite i A=(i A,i B)in 1 in this case. The inclusion in 2:Bβ†’AβŠ”B is a cofibration when A is cofibrant by Lemma 2. Hence also the composite i B=(i A,i B)in 2 in this case.

If A is cofibrant, then a mapping cylinder for a map f:A→B can be constructed from a cylinder (IA,d 1,d 0,s) by the following diagram with a pushout square

(2)

We have q Bi A=f and q Bi B=1 B by construction. The map (i A,i B) is a cofibration by cobase change, since the map (d 1,d 0) is a cofibration. Let us show that q B is acyclic. For this, it suffices to show that i B is acyclic by three-for-two, since q Bi B=1 B. The two squares of the following diagram are pushout,

(3)

hence also their composite,

(4)

by the lemma here. This shows that the map i B is a cobase change of the map i 1. But i 1 is an acyclic cofibration by Lemma 3, since A is cofibrant. It follows that i B is an acyclic cofibration (since the class of acyclic cofibrations is closed under cobase changes by Proposition 1).

Lemma

(Ken Brown 1) Let E be a model category and let F:E cβ†’C be a functor defined on the sub-category of cofibrant objects and taking its values in a category C equipped with class 𝒲 of weak equivalences containing the units and satisfying three-for-two. If the functor F takes an acyclic cofibration to a weak equivalence, then it takes an acyclic map to a weak equivalence.

Proof

If f:Aβ†’B is an acyclic map between cofibrant objects, let us choose a mapping cylinder factorisation f=q Bi A:Aβ†’C(f)β†’B. The maps i A and i B are cofibrations by Lemma 4, since A and B are cofibrant. The map i B is acyclic by three-for-two, since q Bi B=1 B and q B is acyclic. Thus, F(i B) is a weak equivalence. Hence also the map F(q B) by three-for-two since we have F(q B)F(i B)=F(q Bi B)=F(1 B)=1 FB and 𝒲 contains the units. The map i A is acyclic by three-for-two, since f=q Bi A and f and q B are acyclic. Thus, F(i A) is a weak equivalence, and it follows by three-for-two that F(f) is a weak equivalence since F(f)=F(q Bi A)=F(q B)F(i A).

Recall that a functor is said to invert a morphism in its domain if it takes this morphism to an isomorphism.

Lemma

(Ken Brown 2)

  • If a functor F:E cβ†’C inverts acyclic cofibrations, then it inverts weak equivalences.

  • If a functor F:E fβ†’C inverts acyclic fibrations, then it inverts weak equivalences.

Proof

This follows from Lemma 5, if 𝒲 is the class of isomorphisms in C.

Remark

Ken Brown’s lemma implies that the inclusion π’ž cβˆ©π’² cβŠ†π’² c induces an isomorphism of categories,

(π’ž cβˆ©π’² c) βˆ’1E cβ†’Ho(E c).(\mathcal{C}_c\cap \mathcal{W}_c)^{-1}\mathbf{ E}_c\to Ho(\mathbf{ E}_c).

If (IA,d 1,d 0,s) is a cyclinder for A, then a left homotopy h:f→ lg between two maps f,g:A→X is a map h:IA→X such that and f=hd 1 and g=hd 0. We shall say that hd 1 is the source of the homotopy h and that hd 0 is the target,

The reverse of an homotopy h:f→ lg is the homotopy g→ lf defined by the same map h:IA→X but on the transpose cylinder (IA,d 0,d 1,s). The homotopy unit f= lf of a map f:A→X is defined by the map fp:IA→A→X.

Two maps f,g:Aβ†’X are left homotopic, f∼ lg, if there exists a left homotopy h:fβ†’ lg with domain some cylinder object for A.

Lemma

The left homotopy relation between the maps A→X can be defined on a fixed cylinder for A, when X is fibrant.

Proof

Let us show that if two maps f,g:A→X are homotopic by virtue of a homotopy defined on a cylinder (IA,i 1,i 0,r), then then they are homotopic by virtue of a homotopy defined on any another cylinder (JA,j 1,j 0,s). By assumption, we have h(i 1,i 0)=(f,g) for a map h:IA→X. Let us choose a factorisation r=r′u:IA→I′A→A with u an acyclic cofibration and r′ a fibration. The map r′ is acyclic by three-for-two, since the maps p and u are. Hence the square

has a diagonal filler k:JA→I′A since the map (j 1,j 0) is a cofibration. But the square

has also a diagonal filler d:I′A→X, since u is an acyclic cofibration and X is fibrant. The composite h′=dk:JA→X is a left homotopy f→ lg.

Lemma

If a functor F:E→K inverts weak equivalences, then the implication

f∼ lgβ‡’F(f)=F(g)f\sim_l g \quad \Rightarrow \quad F(f)=F(g)

is true for any pair of maps f,g:A→B in E. The same result is true for a functor defined on E c or on E fc.

Proof

If (IA,d 1,d 0,s) is a cylinder for A, then the map F(s) is invertible by the assumption of F since s is acyclic. Hence we have F(d 1)=F(d 0), since we have

F(s)F(d 1)=F(sd 1)=F(1 A)=F(sd 0)=F(s)F(d 0).F(s) F(d_1)=F(s d_1)=F(1_A)=F(s d_0)=F(s)F(d_0).

If h:IA→X is a homotopy between two map f,g:A→X, then

F(f)=F(hd 1)=F(h)F(d 1)=F(h)F(d 0)=F(hd 0)=F(g).F(f)=F(h d_1)=F(h)F(d_1)=F(h)F(d_0)=F(h d_0)=F(g).

Let us now consider the case where the domain of the functor F is the category E c. Observe that if A is cofibrant, then so is the object IA in a cylinder (IA,d 1,d 0,s), since the map (d 1,d 0):AβŠ”Aβ†’IA is a cofibration and the object AβŠ”A is cofibrant (since a coproduct of cofibrant objects is cofibrant by Corollary 1). Hence the cylinder (IA,d 1,d 0,s) belongs to the category E c and the proof above can be repeated in this case. Let us now consider the case where the domain of the functor F is the category E fc. In this case the left homotopy relation between the maps Aβ†’B can defined on a fixed cylinder for A by Lemma 7, since B is fibrant. A cylinder for A can be constructed by factoring the map βˆ‡ A:AβŠ”Aβ†’A as an acyclic cofibration (d 1,d 0):AβŠ”Aβ†’IA followed by a fibration s:IAβ†’A. The object IA is cofibrant, since A is cofibrant. But IA is also fibrant, since s is a fibration and A is fibrant. Hence the cylinder (IA,d 1,d 0,p) belongs to the category E fc and the proof above can be repeated.

The left homotopy relation on the set of maps A→X is reflexive and symmetric. We shall denote by π l(A,X) the quotient of the set Hom(A,X) by the equivalence relation generated by the left homotopy relation. The relation is compatible with composition on the left: the implication

f∼ lgβ‡’pf∼ lpgf\sim_l g \quad \Rightarrow \quad p f\sim_l p g

is true for every maps f,g:A→X and p:X→X′. This defines a functor

Ο€ l(A,βˆ’):Eβ†’Set.\pi^l(A,-):\mathbf{E}\to \mathbf{Set}.
Lemma

The left homotopy relation between the maps A→X is an equivalence when A is cofibrant.

Proof

Cylinders for A can be composed as cospan. More precisely, the composite of a cylinder (IA,i 1,i 0,r)withacylinder(J A,j_1,j_0,s)isthecylinder(K A,k_1,k_0,t)$ defined by the following diagram with a pushout square,

The map t:KA→A is defined by the condition tin 1=r and tin 2=s. Let us show that t is acyclic. The map j 1:A→JA is an acyclic cofibration by Lemma 3, since A is cofibrant. It follows that in 1 is an acyclic cofibration, since it is a cobase change of j 1. Hence the map t is acyclic by three-for-two, since we have tin 1=s and the maps in 1 and s are acyclic. It remains to show that the map (k 0,k 1) is a cofibration. For this, we can use the following diagram with a pushout square,

The map k in this diagram is a cobase change of the map (i 1,i 0)βŠ”(j 1,j 0). But the map (i 1,i 0)βŠ”(j 1,j 0) is a cofibration, since the maps (i 1,i 0) and (j 1,j 0) are cofibrations. This proves that the map k is a cofibration. It follows that the composite (k 1,k 0)=k(in 1βŠ”in 3) is a cofibration, since the map in 1βŠ”in 3 is a cofibration by Lemma 2. We have proved that (KA,k 1,k 0,r) is a cylinder for A. We can now prove that the left homotopy relation on the set of maps Aβ†’X is transitive. Let f 1,f 2 and f 3 be three maps Aβ†’X and suppose that h 1:IAβ†’X is a left homotopy f 1β†’ lf 2, and h 2:JAβ†’X is a left homotopy f 2β†’ lf 3. There is then a unique map h 3:KAβ†’X such that h 3in 1=h 1 and h 3in 2=h 2, since h 1i 0=f 2=h 2j 1. This defines a homotopy h 3:f 1β†’ lf 3, since h 3k 1=h 3in 1i 1=h 1i 1=f 1 and h 3k 0=h 3in 2j 0=h 2j 0=f 3.

Lemma

(Covering homotopy theorem) Let A be cofibrant, let f:X→Y be a fibration, let a:A→X, and let h:IA→Y be a left homotopy with source fa:A→Y. Then there exists a left homotopy H:IA→X with source a such that fH=h.

Proof

The square

has a diagonal filler H:IA→X, since d 1 is an acyclic cofibration by Lemma 3 and f is a fibration.

Lemma

(Homotopy lifting lemma) Let f:X→Y be an acyclic fibration, let a:A→X and b:A→X, and let h:IA→Y be a left homotopy fa→ lfb. Then there exists a map H:IA→X defining a left homotopy a→ lb such that fH=h.

Proof

The square

has a diagonal filler H:IA→X, since (d 1,d 0) is a cofibration and f is an acylic fibration.

Lemma

If A is cofibrant, then the functor Ο€ l(A,βˆ’):Eβ†’Set inverts acyclic maps between fibrant objects.

Proof

Let us first show that the functor Ο€ l(A,βˆ’) inverts acyclic fibrations. If f:Xβ†’Y is an acyclic fibration and y:Aβ†’Y, then the square

has a diagonal filler, since A is cofibrant and f is an acyclic fibration. Hence there exists a map x:Aβ†’X such that fx=y. This shows that the map Ο€(A,f) is surjective. Let us show that it is injective. If a,b:Aβ†’X and fa∼ lfb, then a∼ lb by the homotopy lifting lemma 11. We have proved that the map Ο€ l(A,f) is bijective. It then follows from Ken Brown’s lemma 6 that that the functor Ο€ l(A,βˆ’) inverts acyclic maps between fibrant objects.

Lemma

A map which is left homotopic to an acyclic map is acyclic.

Proof

Let h:IA→B be a left homotopy between two maps hd 1=u and hd 0=v. If v is acyclic, then so is h by three-for-two, since d 0 is acyclic. Hence the composite u=hd 1 is acyclic by three-for-two, since d 1 is acyclic.

Path objects and right homotopies

Lemma*

(Dual to Lemma 2) The projection pr 1:X×Y→X is a fibration if Y is fibrant, and the projection pr 2:X×Y→Y is a fibration if X is fibrant

A path object for an object X is a quadruple (PX,βˆ‚ 1,βˆ‚ 0,Οƒ) obtained by factoring the diagonal Ξ” X=(1 X,1 X):Xβ†’XΓ—X as a weak equivalence Οƒ:Xβ†’PX followed by a fibration (βˆ‚ 1,βˆ‚ 0):PXβ†’XΓ—X.

Remark

The notation (PX,βˆ‚ 1,βˆ‚ 0,Οƒ) introduced by Quillen suggests that a path object represents the first two terms of a simplicial object

P *X:Ξ”β†’AP_* X:\Delta \to \mathbf{A}

with P 0X=X and P 1X=PX. This is the notion of simplicial framing of an object X. Beware that the source of a 1-simplex f in a simplicial set S is the vertex βˆ‚ 1(f)∈S 0, and that its target is the vertex βˆ‚ 0(f)∈S 0.

The transpose of a path object (PX,βˆ‚ 1,βˆ‚ 0,Οƒ) is the path object (PX,βˆ‚ 0,βˆ‚ 1,Οƒ).

Lemma*

(Dual to Lemma 3) The maps βˆ‚ 1:PXβ†’X and βˆ‚ 0:PXβ†’X are acyclic, and they are are acyclic fibrations when X is fibrant.

A mapping path object of a map f:X→Y is obtained by factoring the map
(1 X,f):X→X×Y as a weak equivalence i X:X→P(f) followed by a fibration (p X,p Y):P(f)→X×Y. By construction, we have f=p Yi X and p Xi X=1 X. The factorisation

f=p Yi X:X→P(f)→Yf=p_Y i_X:X\to P(f)\to Y

is called the mapping path factorisation of the map f. The map p X is acyclic by three-for-two, since i X is acyclic and p Xi X=1 X.

Lemma*

(Dual to Lemma 4) The maps p X:PX→X and p Y:PY→Y are fibrations when X and Y are fibrant.

If Y is fibrant, then a mapping path object for a map f:Xβ†’Y can be constructed from a path object (PY,βˆ‚ 1,βˆ‚ 0,Οƒ) for Y by the following diagram with a pullback square,

(5)

We have p Xi X=1 X and p Yi X=f by construction. The map (p X,p Y) is a fibration by base change, since the map (βˆ‚ 1,βˆ‚ 0) is. Let us show that the map i X is acyclic. For this, it suffices to show that p X is acyclic by three-for-two, since p Xi X=1 X. The two squares of the following diagram are cartesian,

(6)

hence also their composite,

(7)

by the lemma here. Hence the map p X is a base change of the map βˆ‚ 0. But βˆ‚ 1 is an acyclic fibration by Lemma 15, since Y is fibrant. This shows that the map p X is acyclic (since the base change of an acyclic fibration is an acyclic fibration by Proposition 1).

If (PX,βˆ‚ 1,βˆ‚ 0,Οƒ) is a path object for X, then a right homotopy h:fβ†’ rg between two maps f,g:Aβ†’X is defined to be a map h:Aβ†’PX such that f=βˆ‚ 1h and g=βˆ‚ 0h. We shall say that βˆ‚ 1h is the source of the homotopy h and that βˆ‚ 0h is its target .

The reverse of h is the homotopy th:gβ†’ rh defined by the same map h:Aβ†’PX but on the transpose path object (PX,βˆ‚ 0,βˆ‚ 1,Οƒ). The unit homotopy f= rf of a map f:Aβ†’X is the map Οƒf:Aβ†’Xβ†’PX.

Two maps f,g:Aβ†’X are right homotopic, f∼ rg, if there exists a right homotopy h:fβ†’ rg with codomain a path object for X.

Lemma*

(Dual to Lemma 7) The right homotopy relation between the maps A→X can be defined on a fixed path object for X when A is cofibrant.

We shall denote by Ο€ r(A,X) the quotient of Hom(A,X) by the equivalence relation generated by the right homotopy relation. The right homotopy relation is compatible with composition on the right:

f∼ rgβ‡’fu∼ rguf\sim_r g \quad \Rightarrow \quad f u\sim_r g u

for every map u:Aβ€²β†’A. We thus obtain a functor

Ο€ r(βˆ’,X):E oβ†’Set.\pi^r(-,X):\mathbf{E}^o\to \mathbf{Set}.
Lemma*

(Dual to Lemma 9) The right homotopy relation between the maps A→X is an equivalence when X is fibrant.

Lemma*

(Homotopy extension theorem, dual to Lemma 10). Let X be fibrant, let u:A→B be a cofibration, let b:B→X, and let h:A→PX be a right homotopy with source bu:A→X. Then there exists a right homotopy H:B→PX with source b such that Hu=h.

Lemma*

(Homotopy prolongation lemma, dual to Lemma 11) Let X be fibrant, let u:A→B be an acyclic cofibration, let a:B→X and b:B→X, and let h:A→PX be a right homotopy au→ rbu. Then there exists a map H:B→PX defining a right homotopy a→ rb such that Hu=h.

Lemma*

(Dual to Lemma 12) If X is fibrant, then the functor Ο€ r(βˆ’,X):E oβ†’Set inverts acyclic maps between cofibrant objects.

Lemma*

(Dual to Lemma 13) A map which is right homotopic to an acyclic map is acyclic

Double homotopies

If (IA,d 1,d 0,s) is a cylinder object for A and (PX,βˆ‚ 1,βˆ‚ 0,Οƒ) is a path object for X, then a map H:IAβ†’PX is double homotopy between four maps Aβ†’X,

The four corners of the square are representing maps A→X, the horizontal sides are representing left homotopies, and the vertical sides are representing right homotopies.

Lemma

If X is fibrant, then every open box of three homotopies, opened at the top, between four maps f ij:A→X,

can be filled by a double homotopy H:IAβ†’PX (ie βˆ‚ 0H=h 0, Hd 1=v 1 and Hd 0=v 0).

Proof

The square

has a diagonal filler H:IAβ†’PX, since (d 1,d 0) is a cofibration and βˆ‚ 0 is an acyclic fibration by Lemma 15.

Lemma

If X is fibrant, then the right homotopy relation on the set of maps A→X implies the left homotopy relation. Dually, if A is cofibrant, then the left homotopy relation implies the right homotopy relation. Hence the two relations coincide when A is cofibrant and X is fibrant.

Proof

Let h:f→ rg be a right homotopy between two maps A→X. By Lemma 23, the open box of homotopies

can be filled by a double homotopy H:IAβ†’PX, since X is fibrant. This yields a left homotopy βˆ‚ 1H:fβ†’ lg.

When A is cofibrant and X is fibrant, then two maps f,g:Aβ†’X are said to be homotopic if they are left (or right) homotopic; we shall denote this relation by f∼g. We shall denote by Ο€(A,X) the quotient of the set Hom(A,X) by the homotopy relation:

Ο€(A,X)=Hom(A,X)/∼.\pi(A,X)=Hom(A,X)/\sim.

By definition, Ο€(A,X)=Ο€ r(A,X)=Ο€ l(A,X). This defines a functor

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π:E c o×E f→Set,\pi:\mathbf{E}_c^o\,\times\, \mathbf{E}_f\to \mathbf{Set},

that is, a distributor π:E c⇒E f.

The homotopy category

The homotopy relation ∼ is compatible with the composition law

Hom(Y,Z)Γ—Hom(X,Y)β†’Hom(X,Z)Hom(Y,Z)\,\times \, Hom(X,Y) \to Hom(X,Z)

if X,Y and Z are fibrant-cofibrant objects. It thus induces a composition law

Ο€(Y,Z)Γ—Ο€(X,Y)β†’Ο€(X,Z)\pi(Y,Z)\,\times \, \pi(X,Y)\to \pi(X,Z)

and this defines a category Ο€E fc if we put

(Ο€E fc)(X,Y)=Ο€(X,Y)(\pi\mathbf{E}_{f c})(X,Y)=\pi(X,Y)

for X,Y∈E fc.

Definition

We say that a map in E cf is a homotopy equivalence if it is invertible in the category Ο€E fc.

It is obvious from the definition that the class of homotopy equivalences has the three-for-two property.

A map f:Xβ†’Y in E fc is a homotopy equivalence iff there exists a map g:Yβ†’X such that gf∼1 X and fg∼1 Y.

Let us denote by H:E→Ho(E), H′:E fc→Ho(E fc) and P:E fc→π(E fc), the canonical functors. The functor H′ takes homotopic maps u,v:X→Y to the same morphism by Lemma 8. It follows that there is a unique functor U:πE fc→Ho(E fc) such that the following triangle commutes,

We shall prove in Theorem 1 below that the functor U is an isomorphism of categories. We will need a lemma:

Lemma

If Ξ£ is a set of morphisms in a category A, then the localisation functor H:Aβ†’Ξ£ βˆ’1A is an epimorphism of category. Moreover, for any category M, the functor

H *:[Ξ£ βˆ’1A,M]β†’[A,M]H^*:\mathbf{[}\Sigma^{-1}\mathbf{A},\mathbf{M}\mathbf{]} \to \mathbf{[}\mathbf{A},\mathbf{M}\mathbf{]}

induced by H is fully faithful and it induces an isomorphism between the category [Ξ£ βˆ’1A,M] and the full subcategory of [A,M] spanned by the functors Aβ†’M inverting the elements of Ξ£. In particular, two functors Q 0,Q 1:Aβ†’M are isomorphic iff the functors Q 0H and Q 1H are isomorphic.

Proof

Left to the reader.

Theorem

The functor U:πE fc→Ho(E fc) defined above is an isomorphism of categories. A map in E fc is acyclic iff it is a homotopy equivalence.

Proof

(Mark Hovey) Let us first show that if a map f:X→Y in the category E fc is acyclic, then it is a homotopy equivalence. The map π(A,f):π(A,X)→π(A,Y) is bijective for every cofibrant object A by Lemma 12, since f is an acyclic map between fibrant object. It follows that f is inverted by the Yoneda functor

πE fc→[πE fc o,Sets].\pi\mathbf{E}_{f c}\to [\pi\mathbf{E}_{f c}^o, \mathbf{Sets}].

But the Yoneda functor is conservative, since it is fully faithful by Yoneda. It follows that f is invertible in the category πE fc. This shows that f is a homotopy equivalence. Let us now prove that the functor U is an isomorphism of categories. The canonical functor P:E fc→π(E fc) inverts acyclic maps by what we just proved. Hence there is a unique functor T:Ho(E fc)→πE fc such that the following triangle commutes,

Let us show that the functors U and T are mutually inverses. Let us observe that the functor P is an epimorphism, since it is surjective on objects and full. But we have TUP=THβ€²=P. It follows that we have TU=Id, since P is an epimorphism. On the other hand, the functor Hβ€² is an epimorphism by Lemma 25, since it is a localisation. It follows that we have UT=Id, since we have UTHβ€²=UP=Hβ€². We have proved that the functor U and T are mutually inverse. Let us now show that a homotopy equivalence f:Xβ†’Y is acyclic. We shall first consider the case where f is a fibration. There exists a map g:Xβ†’Y such that fg∼1 Y and gf∼1 X, since f is a homotopy equivalence by assumption. There is then a left homotopy h:fgβ†’ l1 Y defined on a cylinder IY. I then follows from the covering homotopy theorem 10 that there exists a left homotopy H:IYβ†’X such that Hi 0=g, since f is a fibration. Let us put q=Hi 1. Then fq=1 Y and q∼g. Thus, qf∼gf∼1 X, since the homotopy relation is a congruence. Hence the map qf is acyclic by Lemma 1, since 1 X is acyclic. But the map f:Xβ†’Y is retract of the map qf:Xβ†’X, since the following diagram commutes and we have fq=1 Y,

It then follows by Lemma 1 that f is a weak equivalence. The implication (⇐) is proved in the case where f is a fibration. In the general case, let us choose a factorisation f=pu:Xβ†’Eβ†’Y with u an acyclic cofibration and p a fibration. The map u is a homotopy equivalence by the first part of the proof, since it is acyclic. Thus, p is a homotopy equivalence by three-for-two for homotopy equivalences, since f is a homotopy equivalence by assumption. Thus, p is acyclic since it is a fibration. Hence the composite f=pu is acyclic by three-for-two.

Definition

A fibrant replacement of an object X is a fibrant object X′ together with an acyclic cofibration X→X′. A cofibrant replacement of an object X is a cofibrant object X′ together with an acyclic fibration X′→X.

A fibrant replacement of X is obtained by factoring the map Xβ†’βŠ€ as an acyclic cofibration i X:Xβ†’RX followed by a fibration RXβ†’βŠ€. If X is fibrant, we can take RX=X and i X=1 X. Similarly, a cofibrant replacement of X is obtained by factoring the map βŠ₯β†’X as a cofibration βŠ₯β†’QX followed by an acyclic fibration q X:QXβ†’X. If X is cofibrant, we can take QX=X and q X=1 X.

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 q Xi X:QX→RX is acyclic. It can thus be factored as an acyclic cofibration j X:QX→WX followed by an acyclic fibration p X:WX→RX,

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The object WX is a fibrant-cofibrant replacement of the object X. If X is cofibrant, we can take WX=RX, and if X is fibrant, we can take WX=QX (in which case we have WX=X, when X is fibrant-cofibrant).

Lemma

For every map u:X→Y, there exits two maps R(u):RX→RY and W(u):WX→WY fitting in the following commutative diagram,

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The map R(u) is unique up to a right homotopy, and the map W(u) unique up to homotopy. The maps R(u) and W(u) are acyclic if u is acyclic. If u:Xβ†’Y and v:Yβ†’Z, then R(vu)∼ rR(v)R(u) and W(vu)∼W(v)W(u).

Proof

The map R(u):RX→RY exists because RY is fibrant and the map i X is an acyclic cofibration. After choosing R(u), we can choose the map W(u):WX→WY, since the object WX is cofibrant and the map p Y is an acyclic fibration. The map R(u) is unique up to a right homotopy by Lemma 20, since RY is fibrant. Hence also the composite R(u)p X:WX→RY, since the right homotopy relation is compatible with composition on the right. But this composite is also unique up to a left homotopy by Lemma 24, since the object RY is fibrant. It follows by Lemma 11 that the map W(u) is unique up to a left homotopy, since p Y is an acyclic fibration. The first statement is proved. Let us prove the second statement. If u is a weak equivalence, then so are the maps R(u) and W(u) by three-for-two, since the vertical sides of the diagram (9) are acyclic. Let us prove the third statement. If we compose horizontally the following diagram,

we obtain the following diagram,

which should be compared with the diagram,

It then follows from the homotopy uniqueness that we have R(vu)∼ rR(v)R(u) and W(vu)∼W(v)W(u).

Theorem

The four canonical functors in the following square are equivalences of categories,

A map u:A→B is acyclic iff it is inverted by the canonical functor H:E→Ho(E).

Proof

Let us first show that the diagonal functor F:Ho(E fc)β†’Ho(E) is an equivalence of categories. We shall use the following commutative square of functors,

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where the top functor is the inclusion and the vertical functor are the canonical functors. We have F(Hβ€²(u))=H(u) for every map u∈E fc. For every map u:Xβ†’Y in E, the map W(u):WXβ†’WY is well defined up to homotopy by Lemma 26. Hence the morphism Hβ€²W(u):WXβ†’WY is independant of the choice of W(u) by Lemma 8. If u:Xβ†’Y and v:Yβ†’Z, then W(vu)∼W(v)W(u) by Lemma 26. Thus, Hβ€²W(vu)=Hβ€²W(v)Hβ€²W(u). This defines a functor Hβ€²W:Eβ†’Ho(E fc). The functor Hβ€²W takes a weak equivalence to an isomorphism, since W takes a weak equivalence to a weak equivalence by Lemma 26. It follows that there is a unique functor G:Ho(E)β†’Ho(E fc) such that the following square commutes,

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By construction, have GH(u)=Hβ€²W(u) for every map u:Xβ†’Y. Let us show that the functors F and G are quasi-inverses. Observe first that we have W(u)∼u for a map u in E fc (we could even suppose that W(u)=u); thus GFHβ€²(u)=GH(u)=Hβ€²W(u)=Hβ€²(u) in this case. This shows that GFHβ€²=Hβ€² and it follows that GF=Id, since the functor Hβ€² is epic by Lemma 25. Let us now show that the composite FG is isomorphic to the identity of the category Ho(E). For this, it suffices to show that the functors FGH and H are isomorphic by Lemma 25, since the functor H is a localisation. We have FGH(X)=WX for every object X. Moreover, for every map u:Xβ†’Y, the map FGH(u):FGH(X)β†’FGH(Y) is equal to the map HW(u):WXβ†’WY, since FGH(u)=FHβ€²W(u)=HW(u). The maps i Xβ†’RX and p X:WXβ†’RX are invertible in the category Ho(E) since they are acyclic. Let us put ΞΈ X=H(p X) βˆ’1H(i X):Xβ†’WX in the category Ho(E). This defines a natural isomorphim ΞΈ:Hβ†’FGH since the image by H of the commutative diagram (9) is a commutative diagram

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It then follows from Lemma 25 that there is a natural isomorphism Id→FG. We have proved that the functor F is an equivalence of categories. The proof that the canonical functor Ho(E f)→Ho(E) is an equivalence of categories is similar except that it uses the fibrant replacement R instead of the fibrant-cofibrant replacement W. It then follows by duality that the canonical functor Ho(E c)→Ho(E) is an equivalence of categories. Let us now prove that a map u:X→Y is acyclic iff it is inverted by the canonical functor H:E→Ho(E). The implication (⇒) is clear by definition of the functor H. Conversely, if H(u) is invertible, let us show that u is acyclic. The square (eq:77) shows that the map HW(u) is also invertible since the vertical sides of the square are invertible. But we have HW(u)=FH′W(u). Hence the map H′W(u) is invertible in the category πE cf, since the functor F is an equivalence. This shows that Wu is a homotopy equivalence. Thus, Wu is acyclic by Theorem 1. It then follows by three-for-two that u is acyclic, since the vertical maps in the diagram (9) are acyclic.

Corollary

If the category E is locally small, then so is the category Ho(E)

Proof

The category E fc is locally small since it is a subcategory of E. Hence the category Ο€E fc is locally small, since it is a quotient of the category E fc by a congruence relation. It follows by Theorem 1 that the category Ho(E fc) is locally small. It then follows by Theorem 2 that the category Ho(E) is locally small.

Corollary

A map between cofibrant objects u:A→B is acyclic iff the map π(u,X):π(B,X)→π(A,X) is bijective for every fibrant-cofibrant object X.

Proof

The implication (⇒) was proved in Lemma 21. Conversely, if the map π(u,X):π(B,X)→π(A,X) is bijective for every fibrant-cofibrant object X, let us show that u is acyclic. For this, let us choose fibrant replacements i A:A→RA and i B:B→RB of the objects A and B, together with a map R(u):RA→RB fitting in a commutative square,

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The horizontal maps of the following square are bijective by Lemma 21, since the maps i A and i B are acyclic,

It follows that the map Ο€(R(u),X) is bijective, since the map Ο€(u,X) is bijective. It then follows by using the Yoneda embedding that the map R(u) is invertible in the category Ο€E fc. This shows that the map R(u) is a acyclic by Theorem 1. It then follows by three-for-two applied to the square (12) that the map u is acyclic.

Determination

Proposition

A model structure (π’ž,𝒲,β„±) in a category E is determined by any two of its three classes. A map f:Xβ†’Y is a weak equivalence iff it admits a factorisation f=pu:Xβ†’Eβ†’Y with u an acyclic cofibration and p an acyclic fibration.

Proof

The class π’žβˆ©π’² determines the class β„±, since the pair (π’žβˆ©π’²,β„±) is a weak factorisation system. Hence the pair (π’ž,𝒲) determines the class β„±. Dually, the pair (β„±,𝒲) determines the class π’ž. Let us show that the pair (π’ž,β„±) determines the class 𝒲. Obviously, the pair (π’ž,β„±) determines the pair (π’žβˆ©π’²,β„±βˆ©π’²) since π’žβˆ©π’²= β‹”β„± and β„±βˆ©π’²=π’ž β‹”. Hence it suffices to show that a map f:Xβ†’Y is a weak equivalence iff it admits a factorisation f=pu:Xβ†’Eβ†’Y with u an acyclic cofibration and p an acyclic fibration. The implication (⇐) is clear, since the class 𝒲 is closed under composition. Conversely, if f:Xβ†’Y is a weak equivalence, let us choose a factorisation f=pu with u an acyclic cofibration and p a fibration. The map p is acyclic by three-for-two, since the maps f and u are acyclic. Thus, pβˆˆβ„±βˆ©π’².

Let us denote by M f (resp. M c, M fc) the class of fibrant (resp. cofibrant, fibrant-cofibrant) objects of a model structure M=(π’ž,𝒲,β„±) in a category E.

Proposition

A model structure M=(π’ž,𝒲,β„±) in a category E is determined by its class of cofibrations and its class M f of fibrant objects (resp. its class M fc of fibrant-cofibrant objects). More precisely, if Mβ€²=(π’ž,𝒲′,β„±β€²) is another model structure with the same cofibrations, then the four conditions

π’²βŠ†π’²β€²,Mβ€² fβŠ†M f,Mβ€² fcβŠ†M fc,𝒲 cβŠ†π’²β€² c,\mathcal{W}\subseteq \mathcal{W}', \quad \quad \quad M'_{f}\subseteq M_{f}, \quad \quad \quad M'_{f c}\subseteq M_{f c}, \quad \quad \quad \mathcal{W}_c\subseteq \mathcal{W}'_c ,

are equivalent.

Proof:

We shall use the equality

β„±βˆ©π’²=π’ž β‹”=β„±β€²βˆ©π’²β€²\mathcal{F}\,\cap\, \mathcal{W}=\mathcal{C}^\pitchfork=\mathcal{F}'\,\cap\, \mathcal{W}'

and the inclusion π’ž β‹”βŠ†π’²βˆ©π’²β€². Let us prove the implication (1)β‡’(2). If π’²βŠ†π’²β€², then π’žβˆ©π’²βŠ†π’žβˆ©π’²β€². Thus, β„±β€²βŠ†β„±, since β„±=(π’žβˆ©π’²) β‹” and β„±β€²=(π’žβˆ©π’²β€²) β‹”. It follows that Mβ€² fβŠ†M f. The implication (1)β‡’(2) is proved. The implication (2)β‡’(3) is obvious, since M c=Mβ€² c. Let us prove the implication (3)β‡’(4). If A is cofibrant and XβˆˆβŠ†M fc (resp. XβˆˆβŠ†Mβ€² fc) let us denote by Ο€(A,X) (resp. Ο€β€²(A,X)) the set of homotopy classes of maps Aβ†’X with respect to the model structure M (resp. Mβ€²). We claim that if X∈Mβ€² fc, then the left homotopy relation between the maps Aβ†’X only depends on the weak factorisation system (π’ž,π’ž β‹”). To see this, observe that a cylinder object of A can be constructed by factoring the map βˆ‡ A:AβŠ”Aβ†’A as a cofibration (d 1,d 0):AβŠ”Aβ†’IA followed by a map s:IAβ†’A in π’ž β‹”. The left homotopy relation between the maps Aβ†’X can be defined by using this cylinder alone by Lemma 7, since the object X is fibrant in both model structures under the assumption that Mβ€² fcβŠ†M fc. This shows that the left homotopy relation between the maps Aβ†’X only depends on the system (π’ž,π’ž β‹”) if X∈Mβ€² fc. It follows that we have Ο€β€²(A,X)=Ο€(A,X) if A is cofibrant and X∈Mβ€² fc. We can now prove that 𝒲 cβŠ†π’²β€² c. If a map u:Aβ†’B belongs to 𝒲 c, then the map Ο€(u,X):Ο€(B,X)β†’Ο€(A,X) is bijective for every object X∈M fc by Corollary 2, hence also the map Ο€β€²(u,X):Ο€β€²(B,X)β†’Ο€β€²(A,X) for every object X∈Mβ€² fc. This shows that uβˆˆπ’²β€² c by the same corollary. The inclusion 𝒲 cβŠ†π’²β€² c is proved. Let us now prove the implication (4)β‡’(1). Let u:Aβ†’B be a map in 𝒲. Let us factor the map βŠ₯β†’A as a cofibration βŠ₯β†’Aβ€² followed by a map q A:Aβ€²β†’A in π’ž β‹”, and then factor the composite uq A:Aβ€²β†’B as a cofibration uβ€²:Aβ€²β†’Bβ€² followed by a map q B:Bβ€²β†’B in π’ž β‹”. We obtain a commutative square

with q A,q Bβˆˆπ’²βˆ©π’²β€², since π’ž β‹”βŠ†π’²βˆ©π’²β€². We have uβ€²βˆˆπ’² c by three-for-two, since uβˆˆπ’². Thus, uβ€²βˆˆπ’²β€² c, since 𝒲 cβŠ†π’²β€² c by assumption. It follows by three-for-two that uβˆˆπ’²β€². The implication (4)β‡’(1) is proved.

Corollary

A model structure (π’ž,𝒲,β„±) in a category E is determined by its subcategory of fibrant objects E f together with one of the classes π’ž or β„±βˆ©π’². Dually, a model structure (π’ž,𝒲,β„±) is determined by its subcategory of cofibrant objects E c together with one of the classes β„± or π’žβˆ©π’².

Proof:

The first statement follows from Proposition 4 and by using the fact that π’ž= β‹”(β„±βˆ©π’²). The rest follows by duality.

Corollary

A model structure (π’ž,𝒲,β„±) in a category E is determined by its subcategory of fibrant-cofibrant objects E fc together with one of the classes

π’ž,π’žβˆ©π’²,β„±,β„±βˆ©π’².\mathcal{C}, \quad \mathcal{C}\,\cap\, \mathcal{W}, \quad \mathcal{F}, \quad \mathcal{F}\, \cap\, \mathcal{W}.
Proof:

This follows from Proposition 4 in the case of the class π’ž, and also in the case of the class β„±βˆ©π’², since we have π’ž= β‹”(β„±βˆ©π’²). The rest follows by duality.

Derived functors

Definitions

Let Ξ£ be a set of arrows in a category A, and let P:Aβ†’Ξ£ βˆ’1A be the localisation functor. We saw in Lemma 25 that for any category M, the functor

P *:[Ξ£ βˆ’1A,M]β†’[A,M]P^*:[\Sigma^{-1}\mathbf{A},\mathbf{M}] \to [\mathbf{A},\mathbf{M}]

induced by P is fully faithful, and that it induces an isomorphism between the category [Ξ£ βˆ’1A,M] and the full subcategory of [A,M] spanned by the functors Aβ†’M which inverts the elements of Ξ£. We can thus identify these categories, by using the same notation for a functor G:Ξ£ βˆ’1Aβ†’M and the functor GP:Aβ†’B. In which case the category [Ξ£ βˆ’1A,M] becomes a full subcategory of the category [A,M]. The left Kan extension of a functor F:Aβ†’M along the functor P is a functor Fβ€²:Ξ£ βˆ’1Aβ†’M equipped with a natural transformation Ξ·:Fβ†’Fβ€² which reflects F into the full subcategory [Ξ£ βˆ’1A,M]. More precisely, for any functor G:Ξ£ βˆ’1Aβ†’M and any natural transformation Ξ±:Fβ†’G, there exists a unique natural transformation Ξ±β€²:Fβ€²β†’G such that Ξ±β€²Ξ·=Ξ±,

The functor Fβ€² can thus be regarded as the best right approximation of the functor F by a functor inverting the elements in Ξ£.

Dually, the right Kan extension of a functor F:Aβ†’M along the functor P is a functor Fβ€²:Ξ£ βˆ’1Aβ†’M equipped with a natural transformation Ο΅:Fβ€²β†’F which coreflects F into the full subcategory [Ξ£ βˆ’1A,M]. More precisely, for any functor G:Ξ£ βˆ’1Aβ†’M and any natural transformation Ξ²:Gβ†’F, there exists a unique natural transformation Ξ±β€²:Gβ†’Fβ€² such that ϡα′=Ξ±,

The functor Fβ€² can thus be regarded as the best left approximation of the functor F by a functor inverting the elements in Ξ£.

Definition

Let E be a category equipped with a class 𝒲 of weak equivalences. The right derived functor of a functor F:Eβ†’M is defined to be the best dexter approximation Ξ·:Fβ†’F R of the functor F by a functor F R inverting weak equivalences. Dually, the left derived functor of F is defined to be the best sinister approximation Ο΅:F Lβ†’F of the functor F by a functor F L inverting weak equivalences.

We shall say a right derived functor η:F→F R is absolute if it stays a right derived functor after postcomposing it with any functor U:M→N, that is, if the natural transformation U(η):UF→UF R exibits the functor UF R as the the right derived functor (UF) R of UF. Dually, We shall say a left derived functor ϡ:F L→F is absolute if it stays a left derived functor after postcomposing it with any functor U:M→N, that is, if the natural transformation U(ϡ):UF L→UF exibits the functor UF L as the left derived functor (UF) L of UF.

Proposition

Let E be a model category, and F:E→M be a functor which inverts weak equivalences between fibrant objects. Then the right derived functor η:F→F R exists and it is absolute. Moreover, the map η X:FX→F RX is an isomorphism when X is fibrant.

Proof:

For each object X∈E, let us choose a fibrant replacement i X:Xβ†’RX; we shall take i X=1 X:Xβ†’X when X is fibrant. By Lemma 26, for every map u:Xβ†’Y, there exits a map R(u):RXβ†’RY fitting in the commutative square,

(13)

The map R(u) is unique up to a right homotopy by Lemma 26. Hence the map FR(u):FRX→FRY does not depends on the choice of R(u) by Lemma 8. Moreover, if u:X→Y and v:Y→Z, then we have FR(vu)=FR(v)FR(u) by the same lemma. We thus obtain a functor F R=FR:E→M.

The functor F R inverts weak equivalences, since R(u) is a weak equivalence between fibrant objects when u is a weak equivalence by Lemma 26. Thus, F R:Ho(E)β†’M. Notice that F RX=FX when X is fibrant. The image by F of the square (13) is a commutative square,

It shows that we can define a natural transformation η:F→F R by putting η X=F(i X):FX→FRX for every object X. Notice that we have η X=1 FX when X is fibrant. Let us show that the natural transformation η:F→F R is reflecting the functor F into the subcategory [Ho(E),M] of the category [E,M]. For this we have to prove that for any functor G:Ho(E)→M and any natural transformation α:F→G there exists a unique natural transformation α′:F R→G such that α′η=α,

The map G(i X):GX→GRX in the following square

(14)

is invertible by the assumption on G, since i X is a weak equivalence. We can thus defines a map Ξ±β€² X:FRXβ†’GX by putting Ξ± X=G(i X) βˆ’1Ξ± RX. Let us verify that this defines a natural transformation Ξ±β€²:F Rβ†’G. The right hand square of the following diagram commutes for any map u:Xβ†’Y by the functoriality of G, since the square (13) commutes,

And the left hand square commutes by the naturality of Ξ±. The naturality of Ξ±β€² is proved. Observe that we have G(i X) βˆ’1Ξ± RXF(i X)=Ξ± X for every object X, since the square (14) commutes. This proves that Ξ±β€²Ξ·=Ξ±. It remains to prove the uniqueness of Ξ±β€². If Ξ²:FRβ†’G is a natural transformation such that Ξ²Ξ·=Ξ±, let us show that Ξ²=Ξ±β€². Notice that Ξ² X=Ξ± X=Ξ±β€² X when X is fibrant, since we have Ξ· X=1 FX in this case. In general, let us choose a weak equivalence w:Xβ†’Y with codomain a fibrant object Y (for example, w=i X:Xβ†’RX). The bottom maps in the following two squares are equal since Y is fibrant,

Hence also the top maps, since G(w) is invertible by the assumption on G. Thus, β X=α′ X. We have proved that F R is the right derived functor of F. Moreover, the map η X is invertible when X is fibrant, since we have η X=1 FX in this case. For any functor U:M→N we have (UF) R=(UF)R=U(FR)=UF R and U(η X)=UF(i X). This shows that the right derived functor F R is absolute.

Quillen functors

Definition

We shall say that a functor F:U→V between two model categories is a left Quillen functor if it takes a cofibration to a cofibration and an acyclic cofibration to an acyclic cofibration. Dually, we shall say that F is a right Quillen functor if it takes a fibration to a fibration and an acyclic fibration to an acyclic fibration.

Remark

Most left Quillen functors that we shall consider are cocontinuous and have a right adjoint. Dually, most right Quillen functors have a left adjoint.

Lemma

A left Quillen functor takes an acyclic map between cofibrant objects to an acyclic map. A right Quillen functor takes an acyclic map between fibrant objects to an acyclic map.

Proof:

This follows from Ken Brown’s lemma 5.

Lemma

Let F:U↔V:G be an adjunction F⊒G between two model categories. Then the left adjoint F is a left Quillen functor iff the right adjoint G is a right Quillen functor.

Proof:

Let us denote by (π’ž,𝒲,β„±) the model structure on U and by (π’žβ€²,𝒲′,β„±β€²) the model structure on V. Then the conditions

F(π’ž)βŠ†π’žβ€²andG(β„±β€²βˆ©π’²β€²)βŠ†β„±βˆ©π’²F(\mathcal{C})\subseteq \mathcal{C}' \quad \mathrm{and} \quad G(\mathcal{F}'\,\cap\, \mathcal{W}')\subseteq \mathcal{F}\cap \mathcal{W}

are equivalent by the proposition here. Similarly, the conditions

F(π’žβˆ©π’²)βŠ†π’žβ€²βˆ©π’²β€²andG(β„±β€²)βŠ†β„±F(\mathcal{C}\,\cap\, \mathcal{W})\subseteq \mathcal{C}'\,\cap\, \mathcal{W}' \quad \mathrm{and} \quad G(\mathcal{F}')\subseteq \mathcal{F}

are equivalent.

Definition

We shall say that an adjunction F⊒G between two model categories is a Quillen adjunction if the left adjoint F is a left Quillen functor, or equivalently, if the right adjoint G is a right Quillen functor.

Examples of model categories

Example

The category of simplicial sets SSet admits a model structure, called the Kan model structure, in which the fibrant objects are the Kan complexes and the cofibrations are the monomorphisms. The weak equivalences are the weak homotopy equivalences and the fibrations are the Kan fibrations. The model structure is cartesian closed and proper.

Example

The category Cat admits a model structure, called the natural model structure in which the cofibrations are the functors monic on objects, the weak equivalences are the equivalences of categories and the fibrations are the isofibrations. The model structure is cartesian closed and proper.

Example

The category of simplicial sets SSet admits a model structure, called the model structure for quasi-categories, in which the fibrant objects are the quasi-categories and the cofibrations are the monomorphisms. A weak equivalence is called a categorical equivalence and a fibration is called an isofibration. The model structure is cartesian closed and left proper.

Exercises

Exercise

Gives a direct proof (without using the duality) to the lemmata 14 15 16 17 18 19 20 21 22.

Exercise

Recall (from after Definition 4) that a fibrant-cofibrant replacement of an object X is obtained by factoring the composite q Xi X:QX→RX as an acyclic cofibration j X:QX→WX followed by an acyclic fibration p X:WX→RX. Show that for every map u:X→Y, there are maps Qu, Ru and Wu fitting in a commutative cube:

Exercise

Show that the distributor π:E⇒E f defined in (7) induces a distributor π′:Ho(E c)⇒Ho(E f). Show that the distributor π′ is representable by an equivalence of categories Ho(E c)→Ho(E f).

Exercise

Show that the natural model structure on Cat is characterised by each of the following four groups of conditions:

  • Group 1:

    • the acyclic maps are the equivalences of categories

    • an acyclic map is a fibration iff it is a split epimorphism

  • Group 2:

    • the acyclic maps are the equivalences of categories

    • an acyclic map is a cofibration iff it is a split monomorphism

  • Group 3:

    • every object is fibrant.

    • an acyclic map is a fibration iff it is a surjective equivalence

  • Group 4:

    • every object is cofibrant.

    • an acyclic map is a cofibration iff it is a monic equivalence

References

Papers:

Lecture Notes and Textbooks:

Revised on October 24, 2012 11:38:47 by Raeder