Category theory

# Contents

## Main definitions

###### Definition

We shall say that a map $u:A\to B$ in a category $\mathbf{E}$ has the left lifting property with respect to a map $f:X\to Y$, or that $f$ has the right lifting property with respect to $u$, if every commutative square

has a diagonal filler $d:B\to X$,

We shall denote this relation by $u\,\pitchfork\, f$. Notice that the condition $u\,\pitchfork\, f$ means that the following commutative square $Sq(u,f)$,

(1) 

is epicartesian, or equivalently if the map

$Hom(u,f)': hom(B,X)\to hom(B,Y) \times_{hom(A,Y)}hom(A,X)$

is surjective.

###### Notation

If $\mathcal{M}$ is a class of maps, we shall denote by ${}^\pitchfork\!\mathcal{M}$ (resp. $\mathcal{M}^\pitchfork$) the class of maps having the left (resp. the right) lifting property with respect to every map in $\mathcal{M}$. We shall say that ${}^\pitchfork\!\!\mathcal{M}$ is the left complement of $\mathcal{M}$, and that $\mathcal{M}^\pitchfork$ is its right complement.

###### Example

Recall that a map of simplicial sets is said to be a Kan fibration? if it has the right lifting property with respect to the inclusion $h^k_n: \Lambda^k[n] \subset \Delta[n]$ for every $n\gt 0$ and $0\le k\le n$. A simplicial set $X$ is a Kan complex? iff the map $X\to 1$ is a Kan fibration.

###### Example

Let $J$ be the groupoid generated by one isomorphism $0\simeq 1$. Then a functor in the category $\mathbf{Cat}$ is an isofibration? iff it has the right lifting property with respect to the inclusion $\{0\}\subset J$.

###### Definition

We shall say that a pair $(\mathcal{L},\mathcal{R})$ of classes of maps in a category $\mathbf{E}$ is a weak factorisation system if the following conditions are satisfied:

• every map $f:A\to B$ admits a factorisation $f=p u:A\to E\to B$ with $u\in \mathcal{L}$ and $p\in \mathcal{R}$;

• $\mathcal{L}= {}^\pitchfork\mathcal{R}$ and $\mathcal{R}=\mathcal{L}^\pitchfork$;

We shall say that a factorisation $f=p u:A\to E\to B$ with $u\in \mathcal{L}$ and $p\in \mathcal{R}$ is a $(\mathcal{L},\mathcal{R})$-factorisation of the map $f$. The class $\mathcal{L}$ is called the left class of the system, and the class $\mathcal{R}$ is called the right class .

###### Example

Every factorisation system is a weak factorisation system by the theorem here.

###### Example

The category of sets $\mathbf{Set}$ admits a weak factorisation system $(Inj,Surj)$, where $Inj$ the class of injections and $Surj$ is the class of surjections.

For more examples of weak factorisation systems, go to Example .

###### Proposition (Duality)

If $( \mathcal{L},\mathcal{R})$ is a weak factorisation system in a category $\mathbf{E}$, then the pair $(\mathcal{R}^o,\mathcal{L}^o)$ is a weak factorisation system in the opposite category $\mathbf{E}^o$.

If $\mathcal{M}$ is a class of maps in a category $\mathbf{E}$, then for any object $B\in \mathbf{E}$ we shall denote by $\mathcal{M}/B$ the class of maps in the slice category $\mathbf{E}/B$ whose underlying map in $\mathbf{E}$ belongs to $\mathcal{M}$. Dually, we shall denote by $B\backslash \mathcal{M}$ the class of maps in the coslice category $B\backslash \mathbf{E}$ whose underlying map belongs to $\mathcal{M}$.

###### Proposition (Slice and coslice)

If $(\mathcal{L},\mathcal{R})$ is a weak factorisation system in a category $\mathbf{E}$, then the pair $(\mathcal{L}/B,\mathcal{R}/B)$ is a weak factorisation system in the slice category $\mathbf{E}/B$ for any object $B$ in $\mathbf{E}$. Dually, the pair $(B\backslash \mathcal{L},B\backslash \mathcal{R})$ is a weak factorisation system in the coslice category $B\backslash \mathbf{E}$.

## Closure properties

If $\mathcal{C}$ and $\mathcal{F}$ are two classes of maps in $\mathbf{E}$, we shall write $\mathcal{C}\,\pitchfork\, \mathcal{F}$ to indicate that we have $u\pitchfork f$ for every $u\in \mathcal{C}$ and $f\in \mathcal{F}$. The three conditions

$\mathcal{C}\subseteq {}^\pitchfork \mathcal{F}, \quad \quad \mathcal{C} \,\pitchfork\, \mathcal{F},\quad \quad \mathcal{F} \subseteq \mathcal{C}^\pitchfork$

are equivalent. If $\mathcal{C}=\{u\}$, we shall write $u \,\pitchfork\, \mathcal{F}$ instead of $\{u\}\,\pitchfork\,\mathcal{F}$. Similarly, we shall write $\mathcal{C} \,\pitchfork\, f$ instead of $\mathcal{C} \pitchfork \{f\}$.

The operations $\mathcal{M}\mapsto \mathcal{M}^\pitchfork$ and $\mathcal{M}\mapsto {}^\pitchfork\mathcal{M}$ on classes of maps are contravariant and mutually adjoint. It follows that the operations $\mathcal{M}\mapsto ({}^\pitchfork\mathcal{M})^\pitchfork$ and $\mathcal{M}\mapsto {}^\pitchfork(\mathcal{M}^\pitchfork)$ are closure operators.

###### Lemma

The following conditions on a morphism $f:X\to Y$ in a category $\mathbf{E}$ are equivalent:

$f \mathrm{is}\:\mathrm{invertible}, \quad \mathbf{E}\,\pitchfork\, f, \quad f\,\pitchfork\, f, \quad f\,\pitchfork\, \mathbf{E}.$
###### 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.

Recall that a map $u:A\to B$ in a category $\mathbf{E}$ is said to be a retract of another map $v:C\to D$, if $u$ is a retract of $v$ in the category of arrows $\mathbf{E}^{}$. The condition means that there exists four maps $p,i,q,j$ fitting in a commutative diagram

and such that $p i=1_A$ and $q j=1_B$.

###### Definition

We shall say that a class of maps $\mathcal{M}$ in a category $\mathbf{E}$ is closed under retracts if every retract of a map in $\mathcal{M}$ belongs to $\mathcal{M}$.

We recall that the base change of a map $f:X\to Y$ along a map $v:V\to X$ is the map $g:U \to V$ in a pullback square,

Dually, the cobase change of a map $u:A\to B$ along a map $c:A\to C$ is the map $v:C\to D$ in a pushout square,

###### Definition

We shall say that a class of maps $\mathcal{M}$ in a category $\mathbf{E}$ is closed under base changes if the base change of a map in $\mathcal{M}$ belongs to $\mathcal{M}$, when the base change exists. The notion of a class of maps closed under cobase changes is defined dually.

###### Definition

We shall say that a class of maps $\mathcal{M}$ in a category $\mathbf{E}$ is closed under coproducts if the coproduct

$\sqcup_{i\in I}u_i: \sqcup_{i\in I} A_i\to \sqcup_{i\in I} B_i$

of any family of maps $(u_i:i\in I)$ in $\mathcal{M}$ belongs to $\mathcal{M}$, when this coproduct exists. The notion of a class of maps closed under products is defined similarly.

###### Lemma

Let $\mathcal{M}$ be a class of maps in a category $\mathbf{E}$. Then the class $\mathcal{M}^\bot$ contains the isomorphisms and is closed under composition, retracts, products, and base changes. Dually, the class ${}^\bot\mathcal{M}$ is contains the isomorphisms and is closed under composition, retracts, coproducts, and cobase changes.

###### 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

(2) 

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.

###### Proposition

The two classes of a weak factorisation system $(\mathcal{L},\mathcal{R})$ contain the isomorphisms and they are closed under composition and retracts. The right class $\mathcal{R}$ is closed under base changes and products, and the left class $\mathcal{L}$ under cobase changes and coproducts. The intersection $\mathcal{L}\,\cap \,\mathcal{R}$ is the class of isomorphisms.

###### Proof

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

###### Definition

Recall that a map $u:A\to B$ in a category $\mathbf{E}$ is said to be a domain retract of a map $v:C\to B$, if the object $(A,u)$ of the category $\mathbf{E}/B$ is a retract of the object $(C,v)$. There is a dual notion of codomain retract.

###### Definition

We shall say that a class of maps $\mathcal{M}$ in a category $\mathbf{E}$ is closed under domain retracts if every domain retract of a map in $\mathcal{M}$ belongs to $\mathcal{M}$. The notion of a class closed under codomain retract is defined similarly.

###### Theorem

A pair $(\mathcal{L},\mathcal{R})$ of classes of maps in a category $\mathbf{E}$ is a weak factorisation system iff the following conditions are satisfied:

• every map $f:X\to Y$ admits a $(\mathcal{L},\mathcal{R})$-factorisation $f=p u:X\to E\to Y$;
• $\mathcal{L}\,\pitchfork\,\mathcal{R}$;
• the class $\mathcal{L}$ is closed under codomain retracts and the class $\mathcal{R}$ under domain retracts.
###### 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.

Recall that a class $\mathcal{C}$ of objects in a category $\mathbf{E}$ is said to be replete if every object isomorphic to an object in $\mathcal{C}$ belongs to $\mathcal{C}$. We shall say that a class of maps $\mathcal{M}$ in $\mathbf{E}$ is replete, if it is replete as a class of objects of the category $\mathbf{E}^{}$.

###### Corollary

Suppose that a pair $(\mathcal{L},\mathcal{R})$ of classes of maps in a category $\mathbf{E}$ satisfies the following three conditions:

• every map $f:X\to Y$ admits a $(\mathcal{L},\mathcal{R})$-factorisation $f=p u:X\to E\to Y$;

• $\mathcal{L}\,\pitchfork\,\mathcal{R}$;

• the classes $\mathcal{L}$ and $\mathcal{R}$ are replete.

If $\mathcal{L}'$ denotes the class of maps which are codomain retracts of maps in $\mathcal{L}$ and $\mathcal{R}'$ denotes the class of maps which are domain retracts of maps in $\mathcal{R}$, then the pair $(\mathcal{L}',\mathcal{R}')$ is a weak factorisation system.

###### 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 .

## Existence

### Saturated classes

###### Definition

For any ordinal $\alpha$, let us put $[\alpha]=\{i :0\le i \le \alpha \}$ and $[\alpha)=\{i :0\le i \lt \alpha \}$. Let $\mathbf{E}$ be a cocomplete category. We shall say that a functor $C:[\alpha] \to \mathbf{E}$ is a chain of lentgth $\alpha$, or an $\alpha$-chain. The composite of $C$ is defined to be the canonical map $C(0)\to C( \alpha)$. The base of $C$ is the restriction of $C$ to $[\alpha)$. The chain $C$ is cocontinuous if the canonical map

$\mathrm{colim}_{i\lt j} C(i)\to C(j)$

is an isomorphism for every non-zero limit ordinal $j\in [\alpha]$. We shall say that a subcategory $\mathcal{C}\subseteq \mathbf{E}$ is closed under transfinite compositions if for any limit ordinal $\alpha\gt 0$ any cocontinuous chain $C:\alpha \to \mathbf{E}$ with a base in $\mathcal{C}$ has a composite in $\mathcal{C}$.

Dually, if $\mathbf{E}$ is a complete category, and $\alpha$ is an ordinal, we shall say that a contravariant functor $C:[\alpha]\to \mathbf{E}$ is an opchain. The opchain is continuous if the corresponding chain $C^o:[\alpha]\to \mathbf{E}^o$ is cocontinuous. We shall say that a subcategory $\mathcal{C}\subseteq \mathbf{E}$ is closed under transfinite op-compositions if the opposite subcategory $\mathcal{C}^o\subseteq \mathbf{E}^o$ is closed under transfinite compositions.

###### Lemma

The class $\mathcal{M}^\pitchfork$ is closed under transfinite op-compositions for any class of maps $\mathcal{M}$ in a complete category $\mathbf{E}$.

###### 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.

###### Definition

We shall say that a class of maps $\mathcal{C}$ in a cocomplete category $\mathbf{E}$ is cellular if it satisfies the following conditions: * $\mathcal{C}$ contains the isomorphisms and is closed under composition, * $\mathcal{C}$ is closed under transfinite compositions; * $\mathcal{C}$ is closed under cobase changes.

We shall say that $\mathcal{C}$ is saturated if in addition,

• $\mathcal{C}$ is closed under retracts.

Every class of maps $\Sigma\subseteq \mathbf{E}$ is contained in a smallest cellular class $Cell(\Sigma)$ called the cellular class generated by $\Sigma$. Similarly, $\Sigma$ is contained in a smallest saturated class $Sat(\Sigma)$ called the saturated class generated by $\Sigma$.

###### Example

We shall see in Proposition below that the left class of a weak factorisation system in a cocomplete category is saturated.

###### Example

The class of epimorphisms in any cocomplete category is saturated. Let us say that a map in a cocomplete category is surjective if it is left orthogonal to every monomorphisms; then the class of surjective maps in a cocomplete category is saturated.

###### Example

The class of split monomorphisms in any cocomplete category is saturated. The class of monomorphisms in a Grothendieck topos is saturated.

###### Example

The class of monomorphisms in the category of simplicial sets is generated as a cellular class by the set of inclusions $\partial \Delta[n] \subset \Delta[n]$ ($n\geq 0$).

For more examples of saturated classes of the form $Sat(\Sigma)$, go to Example .

###### Proposition

The class ${}^\pitchfork\mathcal{M}$ is saturated for any class of maps $\mathcal{M}$ in a cocomplete category $\mathbf{E}$. In particular, the left class of a weak factorisation system in a cocomplete category is saturated.

###### 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.

###### Lemma

A cellular class of maps is closed under coproducts.

###### 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

$A= \bigsqcup_{i\in I} A_i$

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

$C(j)=\bigsqcup_{i\lt j} A_i.$

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

$C(j,k)=\bigsqcup_{j\le i\lt k} A_i$

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 $u\in \mathcal{M}$.

###### Definition

If $\alpha$ is a regular cardinal, we shall say that a class of maps $\mathcal{M}$ in a complete category $\mathbf{E}$ is $\alpha$-cellular if it satisfies the following conditions: * $\mathcal{M}$ contains the isomorphisms and is closed under composition; * $\mathcal{M}$ is closed under transfinite compositions of cocontinuous chains of length $\le \alpha$; * $\mathcal{M}$ is closed under cobase changes; * $\mathcal{M}$ is closed under coproducts.

We shall say that an $\alpha$-cellular class $\mathcal{C}$ is $\alpha$-saturated if in addition,

• $\mathcal{C}$ is closed under retracts.

Every class of maps $\Sigma\subseteq \mathbf{E}$ is contained in a smallest $\alpha$-cellular class $Cell^\alpha(\Sigma)$ called the $\alpha$-cellular class generated by $\Sigma$. Similarly, $\Sigma$ is contained in a smallest $\alpha$-saturated class $Sat(\Sigma)$ called the $\alpha$-saturated class generated by $\Sigma$.

###### Example

If $\Sigma$ is the set of inclusions $\partial \Delta[n] \subset \Delta[n]$ ($n\geq 0$) in the category $\mathbf{SSet}$ (of simplicial sets), then $Cell^\omega(\Sigma)=Sat(\Sigma)$ is the class of monomorphisms.

###### Example

If $\Sigma$ is the set of inclusions $h^k_n: \Lambda^k[n] \subset \Delta[n]$ for $n\gt 0$ and $0\le k\le n$, then $Sat^\omega(\Sigma)=Sat(\Sigma)$ is the class of anodyne maps.

### Small object argument

###### Definition

If $u$ is a map in a category $\mathbf{E}$, we shall say that an object $X$ in $\mathbf{E}$ is $u$-fibrant if the map

$hom(u,X):hom(B,X)\to hom(A,X)$

is surjective. More generally, if $\Sigma$ is a class of maps in $\mathbf{E}$, we shall say that an object $X$ is $\Sigma$-fibrant if it is $u$-fibrant for every $u\in \Sigma$. When $\mathbf{E}$ has a terminal object $1$, then an object $X$ is $\Sigma$-fibrant iff the map $X\to 1$ belongs to $\Sigma^\pitchfork$.

Recall that an object $A$ in a cocomplete category $\mathbf{E}$ is said to be compact? if the functor

$hom(A,-): \mathbf{E}\to \mathbf{Set}$

preserves directed colimits. More generally, if $\alpha$ is a regular cardinal, then an object $A$ is said to be $\alpha$-compact? if the functor $hom(A,-)$ preserves $\alpha$-directed colimits. An object $A$ is said to be small? if it is $\alpha$-compact for some regular cardinal $\alpha$.

###### Proposition

(Small object argument) Let $\Sigma$ be a set of maps in a cocomplete category $\mathbf{E}$. If the domains of the maps in $\Sigma$ are $\alpha$-compact, then there exists a functor

$R:\mathbf{E}\to \mathbf{E}$

together with a natural transformation $\rho:Id\to R$ such that: * the object $R(X)$ is $\Sigma$-fibrant for every object $X$; * the map $\rho_X:X\to R(X)$ belongs to $Cell^{\alpha}(\Sigma)$ for every $X\in \mathbf{E}$.

Moreover, the functor $R$ preserves $\alpha$-directed colimits.

###### Proof(Part 0)

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

$F:\mathbf{E}\to \mathbf{E}$

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

$S(X)=\bigsqcup_{\sigma\in \Sigma} hom(s(\sigma),X)\,\times\, s(\sigma)$

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

$\epsilon(s(\sigma),X):hom(s(\sigma), X)\,\times\, s(\sigma) \to X$

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

$T(X)= \bigsqcup_{\sigma\in \Sigma} hom(s({\sigma}),X)\,\times\, t({\sigma}),$

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

$hom(s(\sigma),X)\,\times\, \sigma: hom(s(\sigma),X)\,\times\, s(\sigma) \to hom(s({\sigma}),X)\,\times\, t({\sigma})$

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

This defines a functor $F: \mathbf{E}\to \mathbf{E}$ together with a natural transformation $\theta:Id\to F$. Observe that for every map $\sigma:A\to B$ in $\Sigma$ and every map $x:A\to X$, the composite of the squares

is a square

The colimit $R(X)$ of the infinite sequence

is $\Sigma$-fibrant by the part 0 of the proof, where $\theta^n=\theta_{F^n(X)}$. Let us show that the canonical map $\rho_X:X\to R(X)$ belongs to $Cell^{\omega}(\Sigma)$. For this it suffices to show that the maps $\theta_X$ belong to $Cell^{\omega}(\Sigma)$, since an $\omega$-cellular class is closed under $\omega$-compositions. But $\theta_X$ is a cobase change of $\phi_X$, and $\phi_X$ is a coproduct of maps in $\Sigma$. This shows that $\theta_X$ belongs to $Cell^{\omega}(\Sigma)$ by the closure preperties of this class of maps. It remains to show that the functor $R$ preserves directed colimits. The functor $hom(A,-)$ preserves directed colimits for any compact object $A$. Hence, also the functor $hom(A,-)\,\times\, B$ for any object $B$, since the functor $(-)\times B$ is cocontinuous. The functor $R$ is by construction a colimit of functors of the form $hom(A,-)\,\times\, B$, for compact objects $A$. It follows that $R$ preserves directed colimits. This completes the proof of the proposition in the case where $\alpha=\omega$.

###### Proof (Part 2)

Let us now consider the case where $\alpha\gt \omega$. The sequence

can be extended cocontinuously through all the ordinals $\leq \alpha$ by putting

$F^j(X)=\mathrm{colim}_{i\lt j} F^i(X)$

for every limit ordinal $j \le \alpha$ and by putting $F^{j+1}(X)=F(F^{j}(X))$ and

$\theta^j=\theta_{F^j(X)}:F^j(X)\to F^{j+1}(X)$

for every ordinal $j \lt \alpha$. Let us then put $R(X)=F^\alpha(X)$ and let $v_i:F^i(X)\to R(X)$ be the canonical map for $i \lt \alpha$. This defines a functor $R:\mathbf{E}\to \mathbf{E}$ equipped with a natural transformation $\rho_X=v_0:X\to R(X)$. Let us show that the object $R(X)$ is $\Sigma$-fibrant. For every map $\sigma:A\to B$ in $\Sigma$ and every map $x:A\to X$, there exist an ordinal $i \lt \alpha$ together with a map $y:A\to F^i(X)$ such that $x=v_i y$, since the object $A$ is $\alpha$-compact. But there is then a map $y^\sigma:B\to F^{i+1}(X)$ fitting in a commutative square

If $z=v_{i+1}y^\sigma$, then $z\sigma=v_{i+1}y^\sigma \sigma =v_{i+1} \theta^i y= v_{i} y=x.$ This shows that $R(X)$ is $\Sigma$-fibrant. We leave to the reader the verification that $\rho_X$ belongs to $Cell^{\alpha}(\Sigma)$, and the verification that the functor $R$ preserves $\alpha$-directed colimits.

If $\mathbf{E}$ is a category, then an object of the category $\mathbf{E}^{}$ is a composable pair of maps $A\to B\to C$ in the category $\mathbf{E}.$ There is then a composition functor

$\sigma_1:\mathbf{E}^{}\to \mathbf{E}^{}$

which associates to a composable pair $A\to B\to C$ its composite $A\to B$. We shall say that a section

$F:\mathbf{E}^{}\to \mathbf{E}^{}$

of the functor $\sigma_1$ is a factorisation functor. It associates to a map $f:A\to A'$ a factorisation $f=f_1f_0:A\to F(f)\to A'$, and it takes a commutative square

to a commutative diagram,

Moreover, we have $F(v,v')F(u,u')=F(v u,v'u')$ for any pair of composable squares,

###### Proposition

(Functorial factorisation) Let $\Sigma$ be a set of maps in a cocomplete category $\mathbf{E}$. If the domain and codomain of every map in $\Sigma$ is $\alpha$-compact then there exists a factorisation functor

$F:\mathbf{E}^{}\to \mathbf{E}^{}$

which associates to every morphism $f:A\to B$ a factorisation $f=f_1f_0: A\to F(f)\to B$ with $f_0\in Cell^{\alpha}(\Sigma)$ and $f_1\in \Sigma^\pitchfork$. Moreover, the functor $F$ preserves $\alpha$-directed colimits.

###### Proof

We shall use Proposition . For any map $u:A\to B$ in $\mathbf{E}$, let us denote by $\lambda(u)$ the square

viewed as a morphism $u\to 1_B$ in the category $\mathbf{E}^{}$. If $f:X\to Y$ is a map in $\mathbf{E}$, then the condition $u\,\pitchfork\, f$ exactly means that the map

$Hom(\lambda(u),f): Hom(1_B,f)\to Hom(u,f)$

is surjective and hence that $f$ is $\lambda(u)$-fibrant. Hence a map $f:X\to Y$ belongs to $\Sigma^\pitchfork$ iff it is $\lambda(\Sigma)$-fibrant as an object of the category $\mathbf{E}^{}$. It is easy to verify that domain and codomain of a map in $\lambda(\Sigma)$ are $\alpha$-compact, since this is true of the maps in $\Sigma$. It then follows from Proposition that we can construct a functor

$R:\mathbf{E}^{}\to \mathbf{E}^{}$

together with a natural transformation $\rho:Id\to R$. This yields a commutative square