Proof

Left to the reader.

Proof

($1\Rightarrow 2$) If $f$ is invertible, then the square

has a diagonal filler $f^{-1}y:B\to X$. Thus, $u\,\pitchfork\, f$ for any arrow $u$, and hence $\mathbf{E}\,\pitchfork\, f$. ($2\Rightarrow 3$) If $\mathbf{E}\,\pitchfork\, f$, then $f\,\pitchfork\,f$. ($3\Rightarrow 1$) If $f\,\pitchfork\,f$, then the square

has a diagonal filler $g:Y\to X$ and this shows that $f$ is invertible. The equivalences ($1\Leftrightarrow 2\Leftrightarrow 3$) are proved. The equivalences ($1\Leftrightarrow 4\Leftrightarrow 3$) are proved similarly.

Proof

That class ${}^\pitchfork\mathcal{M}$ contains the isomorphisms by Scholie . Let us show that it is closed under composition. We shall use the properties of epicartesian squares. Let us show that if two morphisms $u:A\to B$ and $v:B\to C$ belongs to ${}^\pitchfork\mathcal{M}$, then so does their composite $v u:A\to C$. For any morphism $f:X\to Y$, the square $Sq(v u,f)$

can be obtained by composing horizontally the squares $Sq(u,f)$ and $Sq(v,f)$,

The squares $Sq(u,f)$ and $Sq(v,f)$ are epicartesian if $f\in \mathcal{M}$; hence their composite is epicartesian by the lemma here. This shows that $v u\in {}^\pitchfork\mathcal{M}$. We have proved that the class ${}^\pitchfork\mathcal{M}$ is closed under composition. Let us now show that the class $\mathcal{M}^\pitchfork$ is closed under retracts. If a map $f:X\to Y$ is a retract of a map $g:U\to V$, then the square $Sq(u,f)$ is a retract of the square $Sq(u,g)$ for any map $u:A\to B$. But a retract of an epicartesian square is epicartesian by the lemma here. It follows that the class $\mathcal{M}^\pitchfork$ is closed under retracts. Let us show that the class $\mathcal{M}^\pitchfork$ is closed under products. If a map $f:X\to Y$ is the product of a family of maps $f_i:X_i\to Y_i$ ($i\in I$), then the square $Sq(u,f)$ is the product of the family of squares $Sq(u,f_i)$. But the product of a family of epicartesian squares is epicartesians by the lemma here. This shows that the class $\mathcal{M}^\pitchfork$ is closed under products. Let us show that the class $\mathcal{M}^\pitchfork$ is closed under base changes. Suppose that we have a pullback square

with $f\in \mathcal{M}^\pitchfork$ and let us prove that $g\in \mathcal{M}^\pitchfork$. It suffices to show that the square $Sq(u,g)$ is epicartesian for every morphism $u:A\to B$ in $\mathcal{M}$. But the square is the back face of the following commutative cube,

The left and the right faces of the cube are cartesian, since the square (2) is cartesian and the functors $hom(A,-)$ and $hom(B,-)$ preserve limits. The front face is epicartesian since we have $u\pitchfork p$. Hence the back face is epicartesian by the cube lemma here.

Proof

This follows from Lemma and Lemma since $\mathcal{R}=\mathcal{L}^\pitchfork$ and $\mathcal{L}={}^\pitchfork\mathcal{R}$.

Proof

The implication ($\Rightarrow$) is clear since the classes of a weak factorisation system are closed under retracts by Proposition . Let us prove the implication ($\Leftarrow$). We have $\mathcal{R} \subseteq \mathcal{L}^\pitchfork$ since we have $\mathcal{L}\,\pitchfork\, \mathcal{R}$ by the hypothesis. Let us show that we have $\mathcal{L}^\pitchfork \subseteq \mathcal{R}$. If a map $f:X\to Y$ belongs to $\mathcal{L}^\pitchfork$, let us choose a $(\mathcal{L},\mathcal{R})$-factorisation $f=p u:X\to E\to Y$. The square

has a diagonal filler $r:E\to X$ since we have $u\,\pitchfork\, f$. Hence we have $f r=p$ and $r u =1_X$ and this shows that $f$ is a domain retract of $p$. Thus, $f\in \mathcal{R}$, since $\mathcal{R}$ is closed under domain retracts by hypothesis.

Proof

The condition $\mathcal{L}\,\pitchfork\,\mathcal{R}$ implies the condition $\mathcal{L}'\,\pitchfork\,\mathcal{R}'$ by Lemma . It is easy to see that $\mathcal{L}'$ is closed under codomain retracts, and that $\mathcal{R}'$ is closed under domain retracts. The result then follows from Theorem .

Proof

Let us show that if $\alpha \gt 0$ is a limit ordinal then every continuous op-chain $C:[\alpha]\to \mathbf{E}$ with a base in $\mathcal{M}^\pitchfork$ has its composite in $\mathcal{M}^\pitchfork$. Let us denote by $c(j,i)$ the transition map $C(j)\to C(i)$ defined for $0\le i\le j\le \alpha$. For any morphism $u:A\to B$, let us denote by $Sq(u,C)$ the contravariant functor $[\alpha]\to \mathbf{Set}^I$ obtained by putting $Sq(u,C)(i)=hom(u,C(i)):hom(B,C(i))\to hom(A,C(i))$ for $i\in [\alpha]$. By definition, the functor $Sq(u,C)$ takes a pair $i\leq j$ to the square $Sq(u,c(j,i))$,

Beware that here the square $Sq(u,c(j,i))$ is defining a morphism from the its top horizontal line to the bottom horizontal line; this means that we are presently using the vertical composition in the category of squares). The (vertical) op-chain $Sq(u,C):[\alpha]\to \mathbf{Set}^I$ is continuous, since $C$ is continuous. If $u\in \mathcal{M}$, then the square $Sq(u,c(j,i))$ is epicartesian for every $i\leq j\lt \alpha$ by the assumption on $C$. It follows that the square $Sq(u,c(\alpha,0))$ is epicartesian by the lemma here. This show that $c(\alpha,0)$ belongs to $\mathcal{M}^\pitchfork$, and hence that $\mathcal{M}^\pitchfork$ is closed under transfinite op-compositions.

Proof

The class ${}^\pitchfork\mathcal{M}$ contains the isomorphisms and it is closed under composition, retracts and cobase changes by . And it is closed under transfinite compositions by Lemma dualised.

Proof

Let $\mathcal{M}$ be a cellular class of maps in a cocomplete category $\mathbf{E}$. We shall say that an object $A\in \mathbf{E}$ cofibrant, if the map $\bot \to A$ belongs to $\mathcal{M}$, where $\bot$ is the initial object of $\mathbf{E}$. We shall first prove that the coproduct of a family of cofibrant objects is cofibrant. Let us first show that the coproduct of a finite family of cofibrant objects is cofibrant. The identity map $\bot \to \bot$ belongs to $\mathcal{M}$, since $\mathcal{M}$ contains the isomorphisms. Hence the object $\bot$ is cofibrant. This show that the coproduct of the empty family of objects is cofibrant. It remains to show that the coproduct of a finite non-empty family of cofibrant objects is cofibrant. For this it suffices to show that the coproduct of two cofibrant objects is cofibrant. If $A$ and $B$ are cofibrant, consider the pushout square

The map $i_B$ is a cobase change of the map $\bot \to A$. Thus, $i_B\in \mathcal{M}$, since $A$ is cofibrant and $\mathcal{M}$ is closed under cobase change. The map $\bot \to B$ also belongs to $\mathcal{M}$, since $B$ is cofibrant. Hence the composite $\bot \to B\to A\sqcup B$ belongs to $\mathcal{M}$, since $\mathcal{M}$ is closed under composition. This shows that $A\sqcup B$ is cofibrant. Let us now show that the coproduct

of an infinite family of cofibrant objects $(A_i:i\in I)$ is cofibrant. We shall argue by induction on the ordinal $\alpha=\mathrm{Card}(I)$. If $j\lt \alpha$, let us put

There is an obvious canonical map $C(j)\to C(k)$ for $j\le k \le \alpha$ and this defines a cocontinuous chain $C:[\alpha] \to \mathbf{E}$. Notice that $C(0)=\bot$ and $C(\alpha)=A$. Hence we can prove that $A$ is cofibrant by showing that the composite of $C$ belongs to $\mathcal{M}$. For this it suffices to show that the base of $C$ belongs to $\mathcal{M}$, since $\mathcal{M}$ is closed under transfinite compositions. But the object

is cofibrant for every $j\le k\lt \alpha$ by the induction hypothesis, since $k\lt \alpha$. And the transition map $C(j)\to C(k)$ is a base change of the map $\bot \to C(j,k)$ since we have a pushout square

This shows that the transition map $C(j)\to C(k)$ belongs to $\mathcal{M}$ for every $j\le k\lt \alpha$. We have proved that the base of $C$ belongs to $\mathcal{M}$ and hence that the object $A$ is cofibrant. Let us now show that the class $\mathcal{M}$ is closed under coproducts. For this, let us show that the coproduct $u:A\to B$ of a family of maps $u_i:A_i\to B_i$ ($i\in I$) in $\mathcal{M}$ belongs to $\mathcal{M}$. For this, let us denote by $\mathcal{M}'$ the class of maps in the category $A\backslash \mathbf{E}$ whose underlying map in $\mathbf{E}$ belongs to $\mathcal{M}$. It is easy to verify that the class $\mathcal{M}'$ satisfies the hypothesis of the proposition. Let us put $E_i=B_i\sqcup_{A_i}A$ for each $i\in I$,

The object $(B,u)$ of $A\backslash \mathbf{E}$ is the coproduct of the family of objects $(E_i,v_i)$ for $i\in I$. The map $v_i:A\to E_i$ belongs to $\mathcal{M}$, since $u_i\in \mathcal{M}$ by assumption, and since the class $\mathcal{M}$ is closed under cobase change. Hence the object $(E_i,v_i)$ of the category $A\backslash \mathbf{E}$ is cofibrant with respect to the class $\mathcal{M}'$. It follows that the object $(B,u)$ is cofibrant by the first part of the proof. This proves that $u\in \mathcal{M}$.

Proof(Part 0)

We first explain the rough idea of proof in the case $\alpha=\omega$. We begin by constructing a functor

together with a natural transformation $\theta:Id\to F$ having the following properties: for every arrow $\sigma:A\to B$ in $\Sigma$ and every map $x:A\to X$, there exists a map $x^\sigma:B\to F(X)$ fitting in a commutative square

The object $R(X)$ is then taken to be the colimit of the infinite sequence,

where $\theta^n=\theta_{F^n(X)}$, and natural transformation $\rho:Id\to R$ is defined by the canonical map $X\to R(X)$. The nice properties of $\rho$ are deduced from the nice properties of $\theta$. Let us show that the object $R(X)$ is $\Sigma$-fibrant. If $v_n:F^n(X)\to R(X)$ denotes the canonical map, then we have a commutative triangle

for every $n\geq 0$. The domain of every map $\sigma:A\to B$ in $\Sigma$ is compact by hypothesis. It follows that for every map $x:A\to R(X)$, there exist an integer $n\ge 0$ together with a map $y:A\to F^n(X)$ such that $x=v_n y$. But there is then a map $y^\sigma:B\to F^{n+1}(X)$ fitting in a commutative square

If $z=v_{n+1}y^\sigma$, then $z\sigma=v_{n+1}y^\sigma \sigma =v_{n+1} \theta^n y= v_{n} y=x.$ This shows that $R(X)$ is $\Sigma$-fibrant. Let us describe the construction of the functor $F$ in the case where $\Sigma$ consists of a single map $\sigma:A\to B$. If $E$ is a set we shall denote by $E\times A$ the coproduct of $E$ copies of $A$. The functor $E\mapsto E\times A$ is left adjoint to the functor $X\mapsto hom(A,X)$. Let $\epsilon(A,X):hom(A,X)\times A\to X$ be the counit of the adjunction. By definition, we have $\epsilon(A,X)i_x=x$ for every $x:A\to X$, where $i_x:A\to hom(A,X)\times A$ is the inclusion indexed by $x$. The object $F(X)$ and the map $\theta_X:X\to F(X)$ are then defined by a pushout square

For every map $x:A\to X$, the composite of the squares

is a square

Proof(Part 1)

We now give a full proof in the case $\alpha=\omega$. For every object $X\in \mathbf{E}$ let us put

where $s(\sigma)$ is the source of the map $\sigma$. This defines a functor $S:\mathbf{E}\to \mathbf{E}.$ The counits

induces a map $\epsilon_X:S(X)\to X$. This defines a natural transformation $\epsilon:S\to Id$, where $Id$ denotes the identity functor. By definition, if $\sigma:A\to B$ is a map in $\Sigma$, then for every map $x:A\to X$ we have $\epsilon_X i_{(\sigma,x)}=x$, where $i_{(\sigma,x)}:A\to S(X)$ is the inclusion indexed by $(\sigma,x)$. For every object $X\in \mathbf{E}$ let us put

where $t(\sigma)$ is the target of the map $\sigma$. This defines a functor $T:\mathbf{E}\to \mathbf{E}.$ The coproduct over $\sigma\in \Sigma$ of the maps

is a map $\phi_X:S(X)\to T(X)$. This defines a natural transformation $\phi:S\to T.$ Let us denote by $F(X)$ the object defined by the pushout square