We say that a class $\mathcal{W}$ of maps in a category $\mathbf{E}$ has the three-for-two property if for any commutative triangle
in which two of the three maps belong to $\mathcal{W}$, 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 $(\mathcal{C},\mathcal{W},\mathcal{F})$ of classes of maps in finitely bicomplete category $\mathbf{E}$ is a Quillen model structure, or just a model structure, if the following conditions are satisfied: * the class $\mathcal{W}$ has the three-for-two property; * The pairs $( \mathcal{C}\,\cap \,\mathcal{W},\mathcal{F})$ and $(\mathcal{C},\mathcal{F}\,\cap\, \mathcal{W})$ are weak factorisation systems.
A Quillen model category, or just a model category, is a category $\mathbf{E}$ 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 $\mathcal{F}$ is called a fibration, a map in $\mathcal{C}$ a cofibration and a map in $\mathcal{W}$ a weak equivalence. A map in $\mathcal{W}$ is also said to be acyclic. It follows from the axioms that every map $f:A\to B$ admits a factorisation $f=p u:A\to E\to B$ with $u$ an acyclic cofibration and $p$ a fibration, and also a factorisation $f=q v:A\to F\to B$ with $v$ a cofibration and $q$ an acyclic fibration.
An object $X\in \mathbf{E}$ is said to be fibrant if the map $X\to \top$ is a fibration, where $\top$ is the terminal object of $\mathbf{E}$. Dually, an object $A\in \mathbf{E}$ is said to be cofibrant if the map $\bot \to A$ is a cofibration, where $\bot$ 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 $( \mathcal{C},\mathcal{W},\mathcal{F})$ on a category $\mathbf{E}$ is trivial if $\mathcal{W}$ is the class of isomorphisms, in which case $\mathcal{C}=\mathbf{E}=\mathcal{F}$.
We shall say that a model structure $( \mathcal{C},\mathcal{W},\mathcal{F})$ on a category $\mathbf{E}$ is coarse if $\mathcal{W}$ is the class of all maps, in which case the pair $(\mathcal{C},\mathcal{F})$ is just an arbitrary weak factorisation system in the category $\mathbf{E}$.
For less trivial examples, see .
If $( \mathcal{C},\mathcal{W},\mathcal{F})$ is a model structure on a category $\mathbf{E}$, then the triple $(\mathcal{F}^o,\mathcal{M}^o,\mathcal{C}^o)$ is a model structure on the opposite category $\mathbf{E}^o$.
If $B$ is an object of a category $\mathbf{E}$ and $\mathcal{M}$ is a class of maps in $\mathbf{E}$, we shall denote by $\mathcal{M}/B$ the class of maps in $\mathbf{E}/B$ whose underlying map in $\mathbf{E}$ belongs to $\mathcal{M}$. Dually, we shall denote by $B\backslash \mathcal{M}$ the class of maps in $B\backslash \mathbf{E}$ whose underlying map belongs to $\mathcal{M}$.
If $( \mathcal{C},\mathcal{W},\mathcal{F})$ is a model structure on a category $\mathbf{E}$, then the triple $(\mathcal{C}/B,\mathcal{W}/B, \mathcal{F}/B)$ is a model structure in the slice category $\mathbf{E}/B$ for any object $B\in \mathbf{E}$. Dually, the triple $(B\backslash \mathcal{C},B\backslash \mathcal{W},B\backslash \mathcal{F})$ is a model structure in the coslice category $B\backslash \mathbf{E}$.
This follows from the analogous properties of weak factorisation systems here.
The classes $\mathcal{C}$, $\mathcal{C}\,\cap \,\mathcal{W}$, $\mathcal{F}$ and $\mathcal{F}\,\cap \,\mathcal{W}$ are closed under composition and retracts. The classes $\mathcal{F}$ and $\mathcal{F}\,\cap \,\mathcal{W}$ are closed under base changes and products. The classes $\mathcal{C}$ and $\mathcal{C}\,\cap \,\mathcal{W}$ are closed under cobase changes and coproducts. The intersection $\mathcal{C}\,\cap \,\mathcal{W}\,\cap \, \mathcal{F}$ 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 $X$ is a retract of an object $Y$, then the map $X\to \bot$ is a retract of the map $Y\to \bot$. If an object $X$ is the product of a family of object $(X_i| i\in I)$, then the map $X\to bot$ is the product of the family of maps $(X_i\to \bot| i\in I)$.
(Myles Tierney) The class $\mathcal{W}$ is closed under retracts.
Notice first that the class $\mathcal{F}\,\cap\,\mathcal{W}$ is closed under retracts by Proposition . Suppose now that a map $f: A \to B$ is a retract of a map $g: X \to Y$ in $\mathcal{W}$. We want to show $f \in \mathcal{W}$. We have a commutative diagram
with $t s = 1_A$ and $v u = 1_B$. We suppose first that $f$ is a fibration. In this case, factor $g$ as $g = q j: X \to Z \to Y$ 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 $d: Z \to A$, since $f$ is a fibration. We get a commutative diagram
The map $f$ is a retract of $q$, since $d(j s) = t s =1_A$. Thus, $f$ is acyclic, since $q$ is an acyclic fibration. In the general case, factor $f$ as $f = p i: A \to E \to B$ with $i$ an acyclic cofibration and $p$ a fibration. By taking a pushout we obtain a commutative diagram
where $k in_2 =g$ and $r in_1 = 1_E$. 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 $g = k in_2$ is acyclic by assumption. So $p$ is acyclic by the first part, since $p$ is a fibration. Finally, $f = p i$ is acyclic, since $i$ is acyclic.
The homotopy category of a model category $\mathbf{E}$ is defined to be the category of fractions
The canonical functor $H:\mathbf{ E}\to Ho(\mathbf{ E})$ is a localisation with respects to the maps in $\mathcal{W}$. Recall that a functor $F:\mathbf{E} \to \mathbf{K}$ is said to invert a map $u:A\to B$ if the morphism $F(u):F A\to F B$ is invertible. The functor $H$ inverts the maps in $\mathcal{W}$ and for any functor $F:\mathbf{E} \to \mathbf{K}$ which inverts the maps in $\mathcal{W}$, there is a unique functor $F': Ho(\mathbf{ E})\to \mathbf{K}$ such that $F'H=F$. We shall see in theorem that a map $u:A\to B$ in the category $\mathbf{E}$ is acyclic if and only if it is inverted by the functor $H$. We shall see in corollary that the category $Ho(\mathbf{ E})$ is locally small if the category $\mathbf{ E}$ is locally small.
We denote by $\mathbf{E}_f$ (resp. $\mathbf{E}_c$, $\mathbf{E}_{f c}$) the full subcategory of $\mathbf{ E}$ spanned by the fibrant (resp. cofibrant, fibrant-cofibrant) objects of $\mathbf{ E}$. If $\mathcal{M}$ is a class of maps in $\mathbf{ E}$, 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 $( \mathcal{C}_f\,\cap \,\mathcal{W}_f,\mathcal{F}_f)$ and the pair $(\mathcal{C}_f,\mathcal{F}_f\,\cap\, \mathcal{W}_f)$ are weak factorisation systems in the category $\mathbf{ E}_f$. The pair $( \mathcal{C}_c\,\cap \,\mathcal{W}_c,\mathcal{F}_c)$ and the pair $(\mathcal{C}_c,\mathcal{F}_c\,\cap\, \mathcal{W}_c)$ are weak factorisation systems in the category $\mathbf{ E}_c$. The pair $( \mathcal{C}_{cf}\,\cap \,\mathcal{W}_{cf},\mathcal{F}_{cf})$ and the pair $(\mathcal{C}_{cf},\mathcal{F}_{cf}\,\cap\, \mathcal{W}_{cf})$ are weak factorisation systems in the category $\mathbf{ E}_{cf}$.
Let us how that the pair $( \mathcal{C}_f\,\cap \,\mathcal{W}_f,\mathcal{F}_f)$ is a weak factorisation system in $\mathbf{ E}_f$. We shall use the characterisation of a weak factorisation system here. Obviously, we have $u\,\pitchfork\, p$ for every $u\in \mathcal{C}_f\,\cap \,\mathcal{W}_f$ and $p\in \mathcal{F}_f$. If $f:X\to Y$ is a map between fibrant objects, let us choose a factoriation $f=pu:X\to E\to Y$ 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 $u\in \mathcal{C}_f\,\cap \,\mathcal{W}_f$ and $p\in \mathcal{F}_f$. Finally, the class $\mathcal{C}_f\,\cap \,\mathcal{W}_f$ is closed under codomain retracts, since a retract of a fibrant object is fibrant. Similarly, the class $\mathcal{F}_f$ is closed under domain retracts.
The inclusion $in_1:A\to A\sqcup B$ is a cofibration if $B$ is cofibrant, and the inclusion $in_2:B\to A\sqcup B$ is a cofibration if $A$ is cofibrant.
The inclusion $in_1:A\to A\sqcup B$ is a cobase change of the map $\bot \to B$, 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 )
A cylinder for an object $A$ is a quadruple $(I A,d_1,d_0,s)$ obtained by factoring the codiagonal $\nabla_A=(1_A,1_A):A\sqcup A\to A$ as a cofibration $(d_1,d_0):A\sqcup A \to I A$ followed by a weak equivalence $s:I A\to A$.
The notation $(I A,d_1,d_0,s)$ introduced by Quillen suggests that a cylinder represents the first two terms of a cosimplicial object
with $I^0 A =A$ and $I^1 A =I A$. This is the notion of a cosimplicial framing? of an object $A$. Beware that if $d_0$ and $d_1$ are the maps $[0]\to [1]$ in the category $\Delta$, then $d_0(0)=1$ and $d_1(1)=0$. We may think of a cylinder $(I A,d_1,d_0,s)$ has an oriented object with two faces, with the face $d_1:A\to IA$ representing the source of the cylinder and the face $d_0:A\to IA$ representing the target.
The transpose of a cylinder $(I A,d_1,d_0,s)$ is the cylinder $(I A,d_0,d_1,s)$.
The maps $d_1:A\to I A$ and $d_0:A\to I A$ are acyclic, and they are acyclic cofibrations when $A$ is cofibrant.
The map $d_1$ and $d_0$ are acyclic by three-for-two, since we have $s d_1=1_A= s d_0$ and $s$ is acyclic. If $A$ is cofibrant, then the inclusions $in_1:A\to A\sqcup A$ and $in_2:A\to A\sqcup A$ are cofibrations by Lemma . 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 ).
A mapping cylinder of a map $f:A\to B$ is obtained by factoring the map $(f,1_B):A\sqcup B\to B$ as a cofibration $(i_A,i_B):A\sqcup B\to C(f)$ followed by a weak equivalence $q_B:C(f)\to B$. We then have $f=q_B i_A$ and $q_B i_B=1_B$. The factorisation
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 $q_B i_B=1_B$.
The maps $i_A:A\to C(f)$ is a cofibration when $B$ is cofibrant, and the map $i_B:B\to C(f)$ is a cofibration when $A$ is cofibrant.
The inclusion $in_1:A\to A\sqcup B$ is a cofibration when $B$ is cofibrant by Lemma . Hence also the composite $i_A=(i_A,i_B)in_1$ in this case. The inclusion $in_2: B\to A\sqcup B$ is a cofibration when $A$ is cofibrant by Lemma . 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 $f:A\to B$ can be constructed from a cylinder $(I A,d_1,d_0,s)$ by the following diagram with a pushout square
We have $q_B i_A=f$ and $q_B i_B=1_B$ 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 $q_B i_B=1_B$. The two squares of the following diagram are pushout,
hence also their composite,
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 , 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 ).
(Ken Brown 1) Let $\mathbf{E}$ be a model category and let $F:\mathbf{E}_c\to \mathbf{C}$ be a functor defined on the sub-category of cofibrant objects and taking its values in a category $\mathbf{C}$ equipped with class $\mathcal{W}$ 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.
If $f:A\to B$ is an acyclic map between cofibrant objects, let us choose a mapping cylinder factorisation $f=q_B i_A:A\to C(f)\to B$. The maps $i_A$ and $i_B$ are cofibrations by Lemma , since $A$ and $B$ are cofibrant. The map $i_B$ is acyclic by three-for-two, since $q_B i_B=1_B$ 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 $F(q_B)F(i_B)=F(q_B i_B)=F(1_B)=1_{F B}$ and $\mathcal{W}$ contains the units. The map $i_A$ is acyclic by three-for-two, since $f=q_B i_A$ 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 $F(f)=F(q_B i_A)=F(q_B)F(i_A)$.
Recall that a functor is said to invert a morphism in its domain if it takes this morphism to an isomorphism.
(Ken Brown 2)
If a functor $F:\mathbf{E}_c\to \mathbf{C}$ inverts acyclic cofibrations, then it inverts weak equivalences.
If a functor $F:\mathbf{E}_f\to \mathbf{C}$ inverts acyclic fibrations, then it inverts weak equivalences.
Ken Brownβs lemma implies that the inclusion $\mathcal{C}_c\, \cap\, \mathcal{W}_c \subseteq \mathcal{W}_c$ induces an isomorphism of categories,
If $(I A,d_1,d_0,s)$ is a cyclinder for $A$, then a left homotopy $h:f {\rightarrow}_l g$ between two maps $f,g:A\to X$ is a map $h:I A\to X$ such that and $f= h d_1$ and $g= h d_0$. We shall say that $h d_1$ is the source of the homotopy $h$ and that $h d_0$ is the target,
The reverse of an homotopy $h:f {\rightarrow}_l g$ is the homotopy $g {\rightarrow}_l f$ defined by the same map $h:I A\to X$ but on the transpose cylinder $(I A,d_0,d_1,s)$. The homotopy unit $f=_l f$ of a map $f:A\to X$ is defined by the map $f p:I A\to A\to X$.
Two maps $f,g:A\to X$ are left homotopic, $f\sim_l g$, if there exists a left homotopy $h:f\to_l g$ with domain some cylinder object for $A$.
The left homotopy relation between the maps $A\to X$ can be defined on a fixed cylinder for $A$, when $X$ is fibrant.
Let us show that if two maps $f,g:A\to X$ are homotopic by virtue of a homotopy defined on a cylinder $(I A,i_1,i_0,r)$, then then they are homotopic by virtue of a homotopy defined on any another cylinder $(J A,j_1,j_0,s)$. By assumption, we have $h(i_1,i_0)=(f,g)$ for a map $h: I A\to X$. Let us choose a factorisation $r=r'u:I A\to I' A\to A$ 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 $k:J A\to I' A$ since the map $(j_1,j_0)$ is a cofibration. But the square
has also a diagonal filler $d:I' A\to X$, since $u$ is an acyclic cofibration and $X$ is fibrant. The composite $h'=d k:J A\to X$ is a left homotopy $f\to_l g$.
If a functor $F:\mathbf{E}\to \mathbf{K}$ inverts weak equivalences, then the implication
is true for any pair of maps $f,g:A\to B$ in $\mathbf{E}$. The same result is true for a functor defined on $\mathbf{E}_c$ or on $\mathbf{E}_{f c}$.
If $(I A,d_1,d_0,s)$ 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
If $h:I A\to X$ is a homotopy between two map $f,g:A\to X$, then
Let us now consider the case where the domain of the functor $F$ is the category $\mathbf{E}_c$. Observe that if $A$ is cofibrant, then so is the object $I A$ in a cylinder $(I A,d_1,d_0,s)$, since the map $(d_1,d_0):A\sqcup A \to IA$ is a cofibration and the object $A\sqcup A$ is cofibrant (since a coproduct of cofibrant objects is cofibrant by Corollary ). Hence the cylinder $(I A,d_1,d_0,s)$ belongs to the category $\mathbf{E}_{c}$ 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 $\mathbf{E}_{f c}$. In this case the left homotopy relation between the maps $A\to B$ can defined on a fixed cylinder for $A$ by Lemma , since $B$ is fibrant. A cylinder for $A$ can be constructed by factoring the map $\nabla_A:A\sqcup A\to A$ as an acyclic cofibration $(d_1,d_0):A\sqcup A \to I A$ followed by a fibration $s:I A \to A$. The object $I A$ is cofibrant, since $A$ is cofibrant. But $I A$ is also fibrant, since $s$ is a fibration and $A$ is fibrant. Hence the cylinder $(I A,d_1,d_0,p)$ belongs to the category $\mathbf{E}_{f c}$ and the proof above can be repeated.
The left homotopy relation on the set of maps $A\to X$ is reflexive and symmetric. We shall denote by $\pi^l(A,X)$ 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
is true for every maps $f,g:A\to X$ and $p:X\to X'$. This defines a functor
The left homotopy relation between the maps $A\to X$ is an equivalence when $A$ is cofibrant.
Cylinders for $A$ can be composed as cospan. More precisely, the composite of a cylinder $(I A, i_1,i_0,r)$ with a cylinder $(J A,j_1,j_0,s)$ is the cylinder $(K A,k_1,k_0,t)$ defined by the following diagram with a pushout square,
The map $t:K A\to A$ is defined by the condition $t in_1=r$ and $t in_2 =s$. Let us show that $t$ is acyclic. The map $j_1:A\to J A$ is an acyclic cofibration by Lemma , 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 $t in_1=s$ 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 $(i_1,i_0)\sqcup (j_1,j_0)$. But the map $(i_1,i_0)\sqcup (j_1,j_0)$ 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 $(k_1,k_0)=k(in_1\sqcup in_3)$ is a cofibration, since the map $in_1\sqcup in_3$ is a cofibration by Lemma . We have proved that $(K A,k_1,k_0,r)$ is a cylinder for $A$. We can now prove that the left homotopy relation on the set of maps $A\to X$ is transitive. Let $f_1,f_2$ and $f_3$ be three maps $A\to X$ and suppose that $h_1:I A\to X$ is a left homotopy $f_1\to_l f_2$, and $h_2:J A\to X$ is a left homotopy $f_2\to_l f_3$. There is then a unique map $h_3: K A\to X$ such that $h_3 in_1 =h_1$ and $h_3 in_2 =h_2$, since $h_1 i_0 =f_2= h_2 j_1$. This defines a homotopy $h_3:f_1 \to_l f_3$, since $h_3 k_1= h_3 in_1 i_1 =h_1 i_1=f_1$ and $h_3 k_0 =h_3 in_2 j_0= h_2 j_0 =f_3$.
(Covering homotopy theorem) Let $A$ be cofibrant, let $f:X\to Y$ be a fibration, let $a:A\to X$, and let $h: I A\to Y$ be a left homotopy with source $f a:A\to Y$. Then there exists a left homotopy $H: I A\to X$ with source $a$ such that $f H=h$.
The square
has a diagonal filler $H: I A\to X$, since $d_1$ is an acyclic cofibration by Lemma and $f$ is a fibration.
(Homotopy lifting lemma) Let $f:X\to Y$ be an acyclic fibration, let $a:A\to X$ and $b:A\to X$, and let $h: I A\to Y$ be a left homotopy $f a \to_l f b$. Then there exists a map $H: I A\to X$ defining a left homotopy $a \to_l b$ such that $f H=h$.
The square
has a diagonal filler $H: I A\to X$, since $(d_1,d_0)$ is a cofibration and $f$ is an acylic fibration.
If $A$ is cofibrant, then the functor $\pi^l(A,-):\mathbf{E}\to \mathbf{Set}$ inverts acyclic maps between fibrant objects.
Let us first show that the functor $\pi^l(A,-)$ inverts acyclic fibrations. If $f:X\to Y$ is an acyclic fibration and $y:A\to Y$, then the square
has a diagonal filler, since $A$ is cofibrant and $f$ is an acyclic fibration. Hence there exists a map $x:A\to X$ such that $f x=y$. This shows that the map $\pi(A,f)$ is surjective. Let us show that it is injective. If $a,b:A\to X$ and $f a\sim_l f b$, then $a\sim_l b$ by the homotopy lifting lemma . We have proved that the map $\pi^l(A,f)$ is bijective. It then follows from Ken Brownβs lemma that that the functor $\pi^l(A,-)$ inverts acyclic maps between fibrant objects.
A map which is left homotopic to an acyclic map is acyclic.
Let $h: I A \to B$ be a left homotopy between two maps $h d_1=u$ and $h d_0=v$. If $v$ is acyclic, then so is $h$ by three-for-two, since $d_0$ is acyclic. Hence the composite $u=h d_1$ is acyclic by three-for-two, since $d_1$ is acyclic.
(Dual to Lemma ) The projection $pr_1:X\times Y \to X$ is a fibration if $Y$ is fibrant, and the projection $pr_2:X\times Y \to Y$ is a fibration if $X$ is fibrant
A path object for an object $X$ is a quadruple $(P X,\partial_1,\partial_0,\sigma)$ obtained by factoring the diagonal $\Delta_X=(1_X,1_X):X\to X\times X$ as a weak equivalence $\sigma:X\to P X$ followed by a fibration $(\partial_1,\partial_0):P X\to X\times X$.
The notation $(P X,\partial_1,\partial_0,\sigma)$ introduced by Quillen suggests that a path object represents the first two terms of a simplicial object
with $P_0 X = X$ and $P_1 X= P X$. 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 $\partial_1(f)\in S_0$, and that its target is the vertex $\partial_0(f)\in S_0$.
The transpose of a path object $(P X,\partial_1,\partial_0,\sigma)$ is the path object $(P X,\partial_0,\partial_1,\sigma)$.
(Dual to Lemma ) The maps $\partial_1:P X \to X$ and $\partial_0:P X\to X$ are acyclic, and they are are acyclic fibrations when $X$ is fibrant.
A mapping path object of a map $f:X\to Y$ is obtained by factoring the map
$(1_X,f):X\to X\times Y$ as a weak equivalence $i_X: X\to P(f)$ followed by a fibration $(p_X,p_Y):P(f)\to X\times Y$. By construction, we have $f=p_Y i_X$ and $p_X i_X=1_X$. The factorisation
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 $p_X i_X=1_X$.
(Dual to Lemma ) The maps $p_X:P X\to X$ and $p_Y: P Y\to Y$ are fibrations when $X$ and $Y$ are fibrant.
If $Y$ is fibrant, then a mapping path object for a map $f:X\to Y$ can be constructed from a path object $(P Y, \partial_1,\partial_0,\sigma)$ for $Y$ by the following diagram with a pullback square,
We have $p_X i_X=1_X$ and $p_Y i_X=f$ by construction. The map $(p_X,p_Y)$ is a fibration by base change, since the map $(\partial_1,\partial_0)$ 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 $p_X i_X=1_X$. The two squares of the following diagram are cartesian,
hence also their composite,
by the lemma here. Hence the map $p_X$ is a base change of the map $\partial_0$. But $\partial_1$ is an acyclic fibration by Lemma , 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 ).
If $(P X,\partial_1,\partial_0,\sigma)$ is a path object for $X$, then a right homotopy $h:f {\rightarrow}_r g$ between two maps $f,g:A\to X$ is defined to be a map $h:A\to PX$ such that $f=\partial_1 h$ and $g=\partial_0 h$. We shall say that $\partial_1 h$ is the source of the homotopy $h$ and that $\partial_0 h$ is its target .
The reverse of $h$ is the homotopy ${}^t h:g {\rightarrow}_r h$ defined by the same map $h: A\to P X$ but on the transpose path object $(P X,\partial_0,\partial_1,\sigma)$. The unit homotopy $f=_r f$ of a map $f:A\to X$ is the map $\sigma f : A\to X \to P X$.
Two maps $f,g:A\to X$ are right homotopic, $f\sim_r g$, if there exists a right homotopy $h:f\rightarrow_r g$ with codomain a path object for $X$.
(Dual to Lemma ) The right homotopy relation between the maps $A\to X$ can be defined on a fixed path object for $X$ when $A$ is cofibrant.
We shall denote by $\pi^r(A,X)$ 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:
for every map $u:A'\to A$. We thus obtain a functor
(Dual to Lemma ) The right homotopy relation between the maps $A\to X$ is an equivalence when $X$ is fibrant.
(Homotopy extension theorem, dual to Lemma ). Let $X$ be fibrant, let $u:A\to B$ be a cofibration, let $b:B\to X$, and let $h: A \to P X$ be a right homotopy with source $b u:A\to X$. Then there exists a right homotopy $H: B \to P X$ with source $b$ such that $H u =h$.
(Homotopy prolongation lemma, dual to Lemma ) Let $X$ be fibrant, let $u:A\to B$ be an acyclic cofibration, let $a:B\to X$ and $b:B\to X$, and let $h: A\to P X$ be a right homotopy $a u\to_r b u$. Then there exists a map $H: B \to P X$ defining a right homotopy $a \to_r b$ such that $H u =h$.
(Dual to Lemma ) If $X$ is fibrant, then the functor $\pi^r(-,X):\mathbf{E}^o\to \mathbf{Set}$ inverts acyclic maps between cofibrant objects.
If $(I A,d_1,d_0,s)$ is a cylinder object for $A$ and $(P X, \partial_1,\partial_0,\sigma)$ is a path object for $X$, then a map $H:I A\to P X$ is double homotopy between four maps $A\to X$,
The four corners of the square are representing maps $A\to X$, the horizontal sides are representing left homotopies, and the vertical sides are representing right homotopies.
If $X$ is fibrant, then every open box of three homotopies, opened at the top, between four maps $f_{ij}:A\to X$,
can be filled by a double homotopy $H:I A\to P X$ (ie $\partial_0 H=h_0$, $H d_1=v_1$ and $H d_0=v_0$).
The square
has a diagonal filler $H:I A \to P X$, since $(d_1,d_0)$ is a cofibration and $\partial_0$ is an acyclic fibration by Lemma .
If $X$ is fibrant, then the right homotopy relation on the set of maps $A\to X$ 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.
Let $h:f\to_r g$ be a right homotopy between two maps $A\to X$. By Lemma , the open box of homotopies
can be filled by a double homotopy $H:I A \to P X$, since $X$ is fibrant. This yields a left homotopy $\partial_1 H :f\to_l g$.
When $A$ is cofibrant and $X$ is fibrant, then two maps $f,g:A\to X$ are said to be homotopic if they are left (or right) homotopic; we shall denote this relation by $f\sim g$. We shall denote by $\pi(A,X)$ the quotient of the set $Hom(A,X)$ by the homotopy relation:
By definition, $\pi(A,X)=\pi^r(A,X)=\pi^l(A,X).$ This defines a functor
that is, a distributor $\pi:\mathbf{E}_c\Rightarrow \mathbf{E}_f$.
The homotopy relation $\sim$ is compatible with the composition law
if $X,Y$ and $Z$ are fibrant-cofibrant objects. It thus induces a composition law
and this defines a category $\pi\mathbf{E}_{f c}$ if we put
for $X,Y\in \mathbf{E}_{f c}$.
We say that a map in $\mathbf{E}_{cf}$ is a homotopy equivalence if it is invertible in the category $\pi\mathbf{E}_{f c}$.
It is obvious from the definition that the class of homotopy equivalences has the three-for-two property.
A map $f:X\to Y$ in $\mathbf{E}_{f c}$ is a homotopy equivalence iff there exists a map $g:Y\to X$ such that $g f\sim 1_X$ and $f g\sim 1_Y$.
Let us denote by $H:\mathbf{E}\to \mathrm{Ho}(\mathbf{E})$, $H':\mathbf{E}_{f c}\to \mathrm{Ho}(\mathbf{E}_{f c})$ and $P:\mathbf{E}_{f c}\to \pi(\mathbf{E}_{f c})$, the canonical functors. The functor $H'$ takes homotopic maps $u,v:X\to Y$ to the same morphism by Lemma . It follows that there is a unique functor $U:\pi\mathbf{E}_{f c}\to \mathrm{Ho}(\mathbf{E}_{f c})$ such that the following triangle commutes,
We shall prove in Theorem below that the functor $U$ is an isomorphism of categories. We will need a lemma:
If $\Sigma$ is a set of morphisms in a category $\mathbf{A}$, then the localisation functor $H:\mathbf{A} \to \Sigma^{-1}\mathbf{A}$ is an epimorphism of category. Moreover, for any category $\mathbf{M}$, the functor
induced by $H$ is fully faithful and it induces an isomorphism between the category $\mathbf{[}\Sigma^{-1}\mathbf{A},\mathbf{M}\mathbf{]}$ and the full subcategory of $\mathbf{[}\mathbf{A},\mathbf{M}\mathbf{]}$ spanned by the functors $\mathbf{A}\to \mathbf{M}$ inverting the elements of $\Sigma$. In particular, two functors $Q_0,Q_1:\mathbf{A}\to \mathbf{M}$ are isomorphic iff the functors $Q_0H$ and $Q_1H$ are isomorphic.
Left to the reader.
The functor $U:\pi\mathbf{E}_{f c}\to \mathrm{Ho}(\mathbf{E}_{f c})$ defined above is an isomorphism of categories. A map in $\mathbf{E}_{fc}$ is acyclic iff it is a homotopy equivalence.
(Mark Hovey) Let us first show that if a map $f:X\to Y$ in the category $\mathbf{E}_{f c}$ is acyclic, then it is a homotopy equivalence. The map $\pi(A,f):\pi(A,X)\to \pi(A,Y)$ is bijective for every cofibrant object $A$ by Lemma , since $f$ is an acyclic map between fibrant object. It follows that $f$ is inverted by the Yoneda functor
But the Yoneda functor is conservative, since it is fully faithful by Yoneda?. It follows that $f$ is invertible in the category $\pi\mathbf{E}_{f c}$. This shows that $f$ is a homotopy equivalence. Let us now prove that the functor $U$ is an isomorphism of categories. The canonical functor $P:\mathbf{E}_{f c}\to \pi(\mathbf{E}_{f c})$ inverts acyclic maps by what we just proved. Hence there is a unique functor $T:\mathrm{Ho}(\mathbf{E}_{f c})\to \pi\mathbf{E}_{f c}$ 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 $T U P=T H' =P$. It follows that we have $T U =Id$, since $P$ is an epimorphism. On the other hand, the functor $H'$ is an epimorphism by Lemma , since it is a localisation. It follows that we have $U T=Id$, since we have $U T H'=U P =H'$. We have proved that the functor $U$ and $T$ are mutually inverse. Let us now show that a homotopy equivalence $f:X\to Y$ is acyclic. We shall first consider the case where $f$ is a fibration. There exists a map $g:X\to Y$ such that $f g\sim 1_Y$ and $g f\sim 1_X$, since $f$ is a homotopy equivalence by assumption. There is then a left homotopy $h:f g\to_l 1_Y$ defined on a cylinder $I Y$. I then follows from the covering homotopy theorem that there exists a left homotopy $H:I Y\to X$ such that $H i_0=g$, since $f$ is a fibration. Let us put $q=H i_1$. Then $f q =1_Y$ and $q\sim g$. Thus, $q f\sim g f \sim 1_X$, since the homotopy relation is a congruence. Hence the map $q f$ is acyclic by Lemma , since $1_X$ is acyclic. But the map $f:X\to Y$ is retract of the map $q f:X\to X$, since the following diagram commutes and we have $f q =1_Y$,
It then follows by Lemma that $f$ is a weak equivalence. The implication ($\Leftarrow$) is proved in the case where $f$ is a fibration. In the general case, let us choose a factorisation $f=p u:X\to E\to Y$ 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 $f= p u$ is acyclic by three-for-two.
A fibrant replacement of an object $X$ is a fibrant object $X'$ together with an acyclic cofibration $X\to X'$. A cofibrant replacement of an object $X$ is a cofibrant object $X'$ together with an acyclic fibration $X'\to X$.
A fibrant replacement of $X$ is obtained by factoring the map $X\to \top$ as an acyclic cofibration $i_X:X\to R X$ followed by a fibration $R X\to \top$. If $X$ is fibrant, we can take $R X =X$ and $i_X=1_X$. Similarly, a cofibrant replacement of $X$ is obtained by factoring the map $\bot \to X$ as a cofibration $\bot \to Q X$ followed by an acyclic fibration $q_X:Q X\to X$. If $X$ is cofibrant, we can take $Q X =X$ 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 $q_X i_X:Q X\to R X$ is acyclic. It can thus be factored as an acyclic cofibration $j_X:Q X\to W X$ followed by an acyclic fibration $p_X:W X\to R X$,
The object $W X$ is a fibrant-cofibrant replacement of the object $X$. If $X$ is cofibrant, we can take $W X=R X$, and if $X$ is fibrant, we can take $W X =Q X$ (in which case we have $W X =X$, when $X$ is fibrant-cofibrant).
For every map $u:X\to Y$, there exits two maps $R(u):R X\to R Y$ and $W(u):W X\to W Y$ fitting in the following commutative diagram,