Proof

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

Proof

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

Proof

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 $(X_i\to \bot| i\in I)$.

Proof

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.

Proof

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.

Proof

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 )

Proof

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 ).

Proof

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.

Proof

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 $F(f)=F(q_B i_A)=F(q_B)F(i_A)$.

Proof

This follows from Lemma , if $\mathcal{W}$ is the class of isomorphisms in $\mathbf{C}$.

Proof

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 $f\to_l g$.

Proof

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.

Proof

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 $h_3 k_0 =h_3 in_2 j_0= h_2 j_0 =f_3$.

Proof

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.

Proof

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.

Proof

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 Brown’s lemma that that the functor $\pi^l(A,-)$ inverts acyclic maps between fibrant objects.

Proof

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.