We shall say that a pair $(\mathcal{L},\mathcal{R})$ of classes of maps in a category $\mathbf{E}$ is a factorisation system if the following conditions are satisfied:
every morphism $f:A\to B$ admits a factorisation $f=p u:A\to E\to B$ with $u\in \mathcal{L}$ and $p\in \mathcal{R}$ and this factorisation is unique up to unique isomorphism;
The classes $\mathcal{L}$ and $\mathcal{R}$ contain the isomorphisms and are closed under composition.
The class $\mathcal{L}$ is called the left class and the class $\mathcal{R}$ the right class of the factorisation system. 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 morphism $f$. The uniqueness condition in the definition means that for any pair of $(\mathcal{L},\mathcal{R})$-factorisations $f=p u:A\to E\to B$ and $f=q v:A\to F\to B$ of the same morphism, there exists a unique isomorphism $i:E\to F$ such that the following diagram commutes,
If $( \mathcal{L},\mathcal{R})$ is a factorisation system in a category $\mathbf{E}$, then the pair $(\mathcal{R}^o,\mathcal{L}^o)$ is a factorisation system in the opposite category $\mathbf{E}^o$.
If $B$ is an object of a category $\mathbf{E}$ and $\mathcal{M}$ is a class of maps, 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 $\mathcal{M}$.
If $(\mathcal{L},\mathcal{R})$ is a factorisation system in a category $\mathbf{E}$, then the pair $(\mathcal{L}/B,\mathcal{R}/B)$ is a factorisation system in the category $\mathbf{E}/B$ for any object $B$ in $\mathbf{E}$. Dually, the pair $(B\backslash \mathcal{L},B\backslash \mathcal{R})$ is a factorisation system in the category $B\backslash \mathbf{E}$.
Left to the reader.
If $Iso$ is the class of isomorphisms of a category $\mathbf{E}$ and if $Map$ is the class of all maps, then the pairs $(Iso,Map)$ and $(Map,Iso)$ are trivial examples of factorisation systems.
The category of sets $\mathbf{Set}$ admits a factorisation system $(Surj,Mono)$, where $Surj$ the class surjections and $Inj$ is the class of injections.
The category of groups $\mathbf{Grp}$ admits a factorisation system $(Surj,Mono)$, where $Surj$ the class of surjective homomorphisms and $Inj$ is the class of injective homomorphisms. More generally, this is true for the category of models of any algebraic theory?.
Let $\mathbf{CRing}$ be the category of commutative rings. We shall say that a ring homomorphism $u:A\to B$ inverts an element $f\in A$ if $u(f)$ is invertible in $B$. We shall say that the morphism $u:A\to B$ is conservative if every element $f\in A$ which is inverted by $u$ is invertible in $A$. For any subset $S\subseteq A$, there is a commutative ring $S^{-1}A$ together with a ring homomorphism $l:A\to S^{-1}A$ which inverts universally every elements of $S$. The universality means that for any ring homomorphism $u:A\to B$ which inverts every element of $S$ there exists a unique homomorphism $u':S^{-1}A\to B$ such that $u' l=u$.
Every homorphism $u:A\to B$ admits a canonical factorisation $u=u' l:A\to S^{-1}A\to B$, where $S\subseteq A$ is the set of elements inverted by $u$. The homomorphism $u'$ is alaways conservative; we shall say that $u$ is a localisation if $u'$ is an isomorphism. The category $\mathbf{CRing}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ is the class of localisations and $\mathcal{R}$ is the class of conservative homomorphisms.
For more examples of factorisation systems in algebra, see Example
We shall say that a class of maps $\mathcal{M}$ in a category has the left cancellation property if the implication
is true for any pair of maps $u:A\to B$ and $v:B\to C$. Dually, we shall say that $\mathcal{M}$ has the right cancellation property if the implication
is true.
The intersection of the classes of a factorisation system $(\mathcal{L},\mathcal{R})$ in a category $\mathbf{E}$ is the class of isomorphisms. Moreover, the left class $\mathcal{L}$ has the right cancellation property and the right class $\mathcal{R}$ has the left cancellation property.
Let $f:A\to B$ be a map in $\mathcal{L}\cap \mathcal{R}$. The trivial factorisations $f=f 1_A$ and $f=1_B f$ are both $(\mathcal{L},\mathcal{R})$-factorisations, since the classes $\mathcal{L}$ and $\mathcal{R}$ contain the units. It follows by uniqueness that there exists an isomorphism $i:A\to B$ such that $f=i 1_A$ and $f=1_B i$,
This shows that $f$ is invertible. Let us prove that the left class $\mathcal{L}$ has the right cancellation property. Let $u:A\to B$ and $v:B\to C$ be two maps. If $u$ and $v u$ belong to $\mathcal{L}$, let us show that $v$ belongs to $\mathcal{L}$. For this, let us choose a $(\mathcal{L},\mathcal{R})$-factorisation $v=p s:B\to E\to C$. We then have two $(\mathcal{L},\mathcal{R})$-factorisations $w=p(s u)$ and $w=1_C(v u)$ of the composite $w=v u$,
Hence there exists an isomorphism $i:E\to C$ such that $i(s u)=w$ and $1_C i=p$. This shows that $p$ is invertible, and hence $p\in \mathcal{L}$, since every isomorphism belongs to $\mathcal{L}$. It follows that $v=p s\in \mathcal{L}$, since $\mathcal{L}$ is closed under composition.
Let $\mathbf{C}^{I}$ be the arrow category of a category $\mathbf{C}$. Then a morphism $f:X\to Y$ in $\mathbf{C}^{I}$ is a commutative square of maps in $\mathbf{C}$,
If $\mathbf{C}$ has pullbacks, then $\mathbf{C}^{I}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of pullback squares. A square $f:X\to Y$ belongs to $\mathcal{L}$ iff the map $f_1$ is invertible. See Example . Suppose that we have a commutative diagram in $\mathbf{C}$
in which the right hand square is cartesian. Then the left hand square is cartesian iff the composite square is cartesian by Proposition . See the page on cartesian squares.
We shall say that a map $u:A\to B$ in a category $\mathbf{E}$ is left orthogonal to a map $f:X\to Y$, or that $f$ is right orthogonal to $u$, if every commutative square
has a unique diagonal filler $d:B\to X$ ($d u=x$ and $f d=y$),
We shall denote this relation by $u\bot f$. Notice that the condition $u\bot f$ means that the following square $Sq(u,f)$
is cartesian in the category of sets. If $\mathcal{C}$ and $\mathcal{F}$ are two classes of maps in $\mathbf{E}$, we shall write $\mathcal{C} \bot \mathcal{F}$ to indicate that we have $u\bot f$ for every $u\in \mathcal{C}$ and $f\in \mathcal{F}$.
If $\mathcal{M}$ is a class of maps in a category $\mathbf{E}$, we shall denote by ${}^\bot\!\mathcal{M}$ (resp. $\mathcal{M}^\bot$) the class of maps left (resp. right) orthogonal to every map in $\mathcal{M}$. We shall say that ${}^\bot\!\!\mathcal{M}$ is the left orthogonal complement of $\mathcal{M}$, and that $\mathcal{M}^\bot$ is its right orthogonal complement.
If $\mathcal{C}$ and $\mathcal{F}$ are two classes of maps in $\mathbf{E}$, then the conditions
are equivalent. The operations $\mathcal{M}\mapsto \mathcal{M}^\bot$ and $\mathcal{M}\mapsto {}^\bot\mathcal{M}$ on classes of maps are contravariant and mutually adjoint. It follows that the operations $\mathcal{M}\mapsto ({}^\bot\mathcal{M})^\bot$ and $\mathcal{M}\mapsto {}^\bot(\mathcal{M}^\bot)$ are closure operators.
In the category $\mathbf{Cat}$, a functor is fully faithful iff it is right orthogonal to the inclusion $i:\partial I \subset I$, where $\partial I$ denotes the discrete category with two objects $0$ and $1$.
If $J=\pi_1I$ is the groupoid generated by one isomorphism $0\simeq 1$, then a functor is conservative iff it is right orthogonal to the inclusion $I\subset J$.
A functor $p:X\to Y$ is a discrete Conduché fibration? if it is right orthogonal to the inclusion $d_1:[1]\to [2]$. This condition means that for every morphism $f:a\to b$ in $X$ and every factorisation $p(f)=vu:p(a)\to e\to p(b)$ of the morphism $p(f)$, there exists a unique factorisation $f=v'u':a\to e'\to b$ of the morphism $f$ such that $p(v')=v$ and $p(u')=u$. Discrete fibrations and a discrete opfibrations are examples discrete Conduché fibrations.
Let $\mathcal{M}$ be a class of maps in a category $\mathbf{E}$. Then the class $\mathcal{M}^\bot$ is closed under limits, composition, base changes and it has the left cancellation property. Dually, the class ${}^\bot\mathcal{M}$ is closed under colimits, composition, cobase changes and it has the right cancellation property.
Let us show that the class $\mathcal{M}^\bot$ is closed under limits. We shall use the fact that the functor $Sq(u,-):\mathbf{E}^{I}\to [I\times I, \mathbf{Set}]$ preserves limits for any map $u:A\to B$ in $\mathbf{E}$. Let us suppose that a map $f:X\to Y$ in $\mathbf{E}$ is the limit of a diagrams of maps $D:K\to \mathbf{E}^{I}$. Let us put $D(k)=f_k:X_k\to Y_k$ for every object $k\in K$. Then the square $Sq(u,f)$ is the limit of the diagram of squares $Sq(u,f_k)$ for $k\in K$. If $f_k$ belongs to $\mathcal{M}^\bot$ for every $k\in K$, let us show that $f$ belongs to $\mathcal{M}^\bot$. The assumption means that the square $Sq(u,f_k)$ is cartesian for every map $u\in \mathcal{M}$. Hence also the limit square $Sq(u,f)$, since the category of cartesian squares is a full reflexive subcategory of the category of all squares $[I\times I, \mathbf{Set}]$ by here. This proves that $f\in \mathcal{M}^\bot$. Let us now prove that the class ${}^\pitchfork\mathcal{M}$ is closed under composition and that it has the right cancellation property. Let $u:A\to B$ and $v:B\to C$ be two maps in $\mathbf{E}$. If $u$ belongs to ${}^\pitchfork\mathcal{M}$ let us show that $v\in {}^\pitchfork\mathcal{M}\Leftrightarrow v u\in {}^\pitchfork\mathcal{M}$. For any morphism $f:X\to Y$, the square $Sq(v u,f)$
is the composite of the squares $Sq(u,f)$ and $Sq(v,f)$,
The square $Sq(u,f)$ is cartesian for every $f\in \mathcal{M}$, since $u\in {}^\pitchfork\mathcal{M}$. It follows from the lemma here that the square $Sq(v,f)$ is cartesian iff the square $Sq(v u,f)$ is cartesian. Thus, $v\in {}^\pitchfork\mathcal{M}\Leftrightarrow v u\in {}^\pitchfork\mathcal{M}$. The remaining properties can be proved similarly, see the proposition here.
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 of $\mathcal{C}$ 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 $\mathbf{E}^{I}$.
A pair $(\mathcal{L},\mathcal{R})$ of classes of maps in a category $\mathbf{E}$ is a factorisation system iff the following three conditions are satisfied:
every map $f:X\to Y$ admits a $(\mathcal{L},\mathcal{R})$-factorisation $f=p u:X\to E\to Y$;
the classes $\mathcal{L}$ and $\mathcal{R}$ are replete;
$\mathcal{L}\bot\mathcal{R}$.
Moreover, in this case the pair $(\mathcal{L},\mathcal{R})$ is a weak factorisation system and we have
($\Rightarrow$) If $(\mathcal{L},\mathcal{R})$ is a factorisation system, let us prove that we have $\mathcal{L}\bot \mathcal{R}$. If $a:A\to A'$ belongs to $\mathcal{L}$ and $b:B\to B'$ belongs to $\mathcal{R}$, let us show that every commutative square
has a unique diagonal filler. For this, let us choose two $(\mathcal{L}, \mathcal{R})$-factorisations $u=p s:A\to E\to B$ and $u'=p's':A'\to E'\to B'$. Then from the commutative diagram
we obtain two $(\mathcal{L}, \mathcal{R})$-factorisations of the same map $A\to B'$,
It follows that there exists a unique isomorphism $i:E'\to E$ such that $i s'a =s$ and $b p i =p'$. Hence the following diagram commutes,
and the composite $d=p i s':A'\to B$ is a diagonal filler of the square (1). It remains to prove the uniqueness of $d$. If $d':A'\to B$ is another diagonal filler of the same square,
let us choose a $(\mathcal{L}, \mathcal{R})$-factorisation $d'=q t:A'\to F\to B$. From the commutative diagram
we can construct two commutative diagrams,
where the first is representing two $(\mathcal{L}, \mathcal{R})$-factorisations of a map $A'\to B'$ and the second two $(\mathcal{L}, \mathcal{R})$-factorisations of a map $A\to B$. Hence there exists a unique isomorphism $k':E'\to F$ such that $k's'=t$ and $b q k'=p'$ and unique isomorphism $k:F\to E$ such that $k t a=s$ and $p k=q$.
It follows from these relations that the following diagram commutes,
Hence also the diagram
The uniqueness of the isomorphism between two $(\mathcal{L}, \mathcal{R})$-factorisations implies that we have $k k'=i$, where $i$ is the isomorphism in the diagram (2). Thus, $d'=q t =(p k)(k's')=p (k k') s'=p i s'=d$. The relation $\mathcal{L}\bot \mathcal{R}$ is proved. ($\Leftarrow$) If the three conditions are satisfied, let us show that the pair $(\mathcal{L},\mathcal{R})$ is a factorisation system. We shall first prove that it is a weak factorisation system by showing that we have,
We have $\mathcal{R}\subseteq \mathcal{L}^\bot$ since we have $\mathcal{L}\bot\mathcal{R}$ by assumption. Obviously, $\mathcal{L}^\bot\subseteq \mathcal{L}^\pitchfork$. Let us then show that $\mathcal{L}^\pitchfork\subseteq \mathcal{R}$. If $f:X\to Y$ belongs to $\mathcal{L}^\pitchfork$, let us choose a factorisation $f=p u:X\to E\to Y$ with $u\in \mathcal{L}$ and $p\in \mathcal{R}$. Then the square
has a diagonal filler $d:E\to X$, since $u\in \mathcal{L}$ and $f\in \mathcal{L}^\pitchfork$. The relations $f d=p$ and $d u=1_X$ implies that the map $u d:E\to E$ is a diagonal filler of the square
But this square has a unique diagonal filler, since we have $\mathcal{L}\bot \mathcal{R}$. It follows that $u d=1_E$. Thus, $u$ is invertible since $d u=1_X$. It follows $f=p u\in \mathcal{R}$, since the class $\mathcal{R}$ is replete. The eqality $\mathcal{R}=\mathcal{L}^\bot=\mathcal{L}^\pitchfork$ is proved. The equality $\mathcal{L}={}^\bot \mathcal{R}={}^\pitchfork \mathcal{R}$ follows by duality. It follows that the pair $(\mathcal{L},\mathcal{R})$ is a weak factorisation system. Hence the classes $\mathcal{L}$ and $\mathcal{R}$ contain the isomorphisms and they are closed under composition by the proposition here. It remains to prove the uniqueness of the $(\mathcal{L},\mathcal{R})$-factorisation of a map $f:A\to B$. Suppose then that we have two $(\mathcal{L}, \mathcal{R})$-factorisations, $f=p u:A\to E\to B$ and $f=q v:A\to F\to B$. Then each of the following squares
has a unique diagonal filler, respectively $d:F\to E$ and $r:E\to F$. The composite $d r:E\to E$ is then a diagonal filler of the square
It follows that we have $d r=1_E$ by uniqueness of a diagonal filler. Similarly, we have $r d=1_F$. This shows that $d$ is invertible.
A weak factorisation system $(\mathcal{L},\mathcal{R})$ is a factorisation system iff we have $\mathcal{L}\bot\mathcal{R}$.
The implication ($\Rightarrow$) follows from Theorem . Conversely, let $(\mathcal{L},\mathcal{R})$ be a weak factorisation system for which we have $\mathcal{L}\bot\mathcal{R}$. The classes $\mathcal{L}$ and $\mathcal{R}$ are replete, since they are closed under composition and they contain the isomorphisms by the proposition here. It then follows from Theorem Theorem that the pair $(\mathcal{L},\mathcal{R})$ is a factorisation system.
A factorisation system $(\mathcal{L},\mathcal{R})$ is determined by any one of its two classes. The class $\mathcal{R}$ is closed under limits, composition, base changes and it has the left cancellation property. Dually, the class $\mathcal{L}$ is closed under colimits, composition, cobase changes and it has the right cancellation property.
Let $(\mathcal{L},\mathcal{R})$ be a factorisation system in a category $\mathbf{E}$. Then the full subcategory of $\mathbf{E}^{I}=[I,\mathbf{E}]$ spanned by the arrows in $\mathcal{L}$ is coreflective, and the full subcategory spanned by the arrows in $\mathcal{R}$ is reflective. Hence the left class of a factorisation system is closed under colimits in the category $\mathbf{E}^{I}$ and the right class is closed under limits.
Let us denote by $\mathcal{R}'$ the full subcategory of $[I,\mathbf{E}]$ spanned by the arrows in $\mathcal{R}$. Every map $u:A\to B$ admits a $(\mathcal{L},\mathcal{R})$-factorisation $u=p i:A\to E\to B$. The pair $(i,1_B)$ is a morphism $u\to p$ in the category $[I,\mathbf{E}]$,
Let us show that the morphism $(i,1_B)$ is reflecting the arrow $u$ in the subcategory $\mathcal{R}'$. For this, it suffices to show that for every arrow $f:X\to Y$ in $\mathcal{R}$ and every commutative square
there exists a unique arrow $z:E\to X$ such that $z i=x$ and $y p=f z$,
But this is clear, since the square
The diagonal of a map $f:X\to Y$ in a category with pullbacks is defined to be the map $\delta(f):X\to X\times_Y X$ in the commutative diagram,
Dually, the codiagonal of a map $u:A\to B$ in a category with pushouts is defined to be the map $\delta^o(u):B\sqcup_A B\to B$ in the commutative diagram
If the diagonal of a map $f:X\to Y$ in a category $\mathbf{E}$ exists, then the condition $u\bot f$ is equivalent to the conjunction of the conditions $u\,\pitchfork\, f$ and $u\,\pitchfork\, \delta(f)$ for any map $u:A\to B$.
If $u\bot f$, let us show that we have $u\,\pitchfork\, f$ and $u\,\pitchfork\, \delta(f)$. Obviously, it suffices to show that we have $u\,\pitchfork\, \delta(f)$. For this, let us show that every commutative square
has a diagonal filler. We have $f x_1=f x_2$, since the following diagram commutes,
Let us put $y=f x_1=f x_2$. We have $(x_1u,x_2u)=(x_1,x_2)u=\delta(f)a=(a,a)$, since the square (3) commutes. This shows that the maps $x_1,x_2:B\to X$ are both filling the diagonal of the following square,
Thus, $x_1=x_2$ since we have $u\bot f$ by assumption. The map $x=x_1=x_2$ is then filling the diagonal of the square (3). The relation $u\,\pitchfork\, \delta(f)$ is proved. Conversely, if $u\,\pitchfork\, f$ and $u\,\pitchfork\, \delta(f)$, let us show that we have $u\bot f$. For this, it suffices to prove the uniqueness of a diagonal filler of a square (4), since the existence follows from the condition $u\pitchfork f$. Suppose then that we have two maps $x_1,x_2:B\to X$ filling the diagonal of the square (4). The relation $f x_1=y=f x_2$ implies that we can define a map $(x_1,x_2):B\to X\times_Y X$. Moreover, the square (3) commutes, since $(x_1,x_1)u=(x_1 u,x_2 u)=(a,a)=\delta(f)(a)$. The square (3) has then a diagonal filler $x:B\to X$, since we have $u\pitchfork \delta(f)$ by assumption. The relation $\delta(f) x=(x_1,x_2)$ implies that $x_1=x=x_2$, since $\delta(f) x=(x,x)$.
(Dual to Lemma ). If the codiagonal of a map $u:A\to B$ in a category $\mathbf{E}$ exists, then the condition $u\bot f$ is equivalent to the conjunction of the conditions $u\,\pitchfork\, f$ and $\delta^o(u)\,\pitchfork\, f$ for any map $f:X\to Y$.
If the category $\mathbf{E}$ has pullbacks, we shall say that a class of maps $\mathcal{M}$ in $\mathbf{E}$ is closed under diagonals if the implication $f\in \mathcal{M}\Rightarrow \delta(f)\in \mathcal{M}$ is true. Dually, if the category $\mathbf{E}$ has pushouts, we shall say that $\mathcal{M}$ is closed under codiagonals if the implication $f\in \mathcal{M}\Rightarrow \delta^o(f)\in \mathcal{M}$ is true.
In a category with pullbacks, a weak factorisation system $(\mathcal{L},\mathcal{R})$ is a factorisation system iff the class $\mathcal{R}$ is closed under diagonals iff it has the left cancellation property
The implication (1)$\Rightarrow$(3) was proved in Proposition . Let us prove the implication (3)$\Rightarrow$(2). If a map $f:X\to Y$ belongs to $\mathcal{R}$ then so is the first projection $pr_1:X\times_Y X\to X$, since the right class of a weak factorisation system is closed under base change by the proposition here. But we have $pr_1\delta(f)=1_X$, and it follows that $\delta(f)$ belongs to $\mathcal{R}$, since the class $\mathcal{R}$ has the left cancellation property by assumption. Let us prove the implication (2)$\Rightarrow$(1). For this, it suffices to show that we have $\mathcal{L}\bot \mathcal{R}$ by Theorem . But if $u\in \mathcal{L}$ and $f\in \mathcal{R}$, then we have $u\pitchfork f$ and $u \pitchfork \delta(f)$, since the class $\mathcal{R}$ is closed under diagonals by assumption. It then follows by Lemma that we have $u\bot f$.
(Dual to Proposition ). In a category with pushouts, a weak factorisation system $(\mathcal{L},\mathcal{R})$ is a factorisation system iff the class $\mathcal{L}$ it is closed under codiagonals iff it has the right cancellation property.
In a finitely bicomplete category, a weak factorisation system $(\mathcal{L},\mathcal{R})$ is a factorisation system iff the class $\mathcal{R}$ is closed under diagonals iff the class $\mathcal{L}$ is closed under codiagonals.
If $u:A\to B$ is an epimorphism, then the conditions $u \bot f$ and $u \pitchfork f$ are equivalent for any map $f:X\to Y$.
If $u \pitchfork f$, let us show that $u \bot f$. For this we have to show that every commutative square
has a unique diagonal filler. The existence is clear since we have $u \pitchfork f$ by hypothesis. Let us prove the uniquess. But if $d_1,d_2: B\to X$ are two diagonal fillers of the square, then we have $d_1u=a=d_2u$. Thus, $d_1=d_2$, since $u$ is an epimorphism.
Let $\Sigma$ be a set of maps in a category with pushout $\mathbf{E}$. Then $\Sigma^\bot =\bigl(\Sigma \,\cup\, \delta^o\Sigma\bigr)^\pitchfork$.
By Lemma we have $\Sigma^\bot =\Sigma^\pitchfork \, \cap \, (\delta^o\Sigma)^\pitchfork.$ This proves the result, since
Recall from here that class of maps $\mathcal{C}$ in a cocomplete category $\mathbf{E}$ is said to be saturated if it satisfies the following conditions: * $\mathcal{C}$ contains the isomorphisms and is closed under composition and transfinite compositions; * $\mathcal{C}$ is closed under cobase changes; * $\mathcal{C}$ is closed under retracts.
Let $\Sigma$ be a set of maps between small objects in a cocomplete category $\mathbf{E}$. Then the pair
is a factorisation system.
The codiagonal of a map between small objects is a map between small objects. Thus, $\tilde \Sigma=\Sigma\, \cup\, \delta^o\Sigma$ is a set of maps between small objects. It then follows from the theorem here that the pair $(\mathcal{L},\mathcal{R})=(Sat(\tilde \Sigma), \tilde \Sigma^\pitchfork)$ is a weak factorisation system. But we have $\tilde \Sigma^\pitchfork=\Sigma^\bot$ by Lemma . Hence it remains to show that the pair $(\mathcal{L},\mathcal{R})$ is a factorisation system. For this, it suffices to show that we have $\mathcal{L}\bot\mathcal{R}$ by Theorem . But we have
since $\Sigma^\bot=\mathcal{R}$. Thus,
by Lemma . It follows that $Sat(\tilde \Sigma)\subseteq {}^\bot\mathcal{R}$, since the class ${}^\bot\mathcal{R}$ is saturated by Proposition . This proves that $\mathcal{L}\bot\mathcal{R}$, and hence that the pair $(\mathcal{L},\mathcal{R})$ is a factorisation system.
Let $\Sigma$ be a set of maps in a locally presentable category? $\mathbf{E}$. Then the pair
is a factorisation system.
Recall that if $A\to B$ is a monomorphism of commutative ring, then an element $b\in B$ is said to be integral over $A$ if it is the root of a monic polynomial $p\in A[x]$. We shall say that a monomorphism of commutative rings $\phi:A\to B$ is integrally closed if every element $b\in B$ integral over $A$ belongs to $A$. The category of commutative rings $\mathbf{CRing}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of integrally closed monomorphisms. We shall say that an homomorphism in the class $\mathcal{L}$ is an integral homomorphism.
If $R$ is a commutative ring, we shall say that an element $r\in R$ is a simple root of a polynomial $p(x)\in R[x]$ if $p(r)=0$ and $p'(r)$ is invertible. Let us denote by $Z(p,R)$ the set of simple roots in $R$ of a polynomial $p$. We shall say that a ring homomorphism $\phi:A\to B$ is separably closed if it induces a bijection $Z(p,A)\to Z(\phi(p),B)$ for every polynomial $p(x)\in A[x]$. For example, a local ring $A$ with maximal ideal $m$ is henselien iff the quotient map $A\to A/m$ is separably closed. The category of commutative rings $\mathbf{CRing}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of separably closed homomorphisms. We shall say that an homomorphism in the class $\mathcal{L}$ is a separable algebraic extension. We conjecture that a ring homormorphism is a separable algebraic extension iff it is formally etale?.
Let $(\mathcal{L},\mathcal{R})$ be a factorisation system in the category $\mathbf{Cat}$. We shall say that a functor $f:A\to B$ is essentially in $\mathcal{L}$ (resp $\mathcal{R}$) if the functor $p$ (resp. $u$) of an $(\mathcal{L}, \mathcal{R})$-factorisation $f=p u:A\to E\to B$ is an equivalence of categories.
Recall that a functor $f:X\to Y$ is said to be full (resp. faithful, fully faithful) if the map $X(a,b)\to Y(f(a),f(b))$ induced by $f$ is surjective (resp. injective, bijective) for every pair of objects à $a,b\in X$. We shall say that a functor $f:X\to Y$ is monic (resp. surjective, bijective) on objects if the induced map $Ob(X)\to Ob(Y)$ is injective (resp. surjective, bijective). The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ the class of full functors bijective on objects and $\mathcal{R}$ is the class of faithful functors.
The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ the class of functors bijective on objects and $\mathcal{R}$ is the class of fully faithful functors. A $(\mathcal{L},\mathcal{R})$-factorisation of a functor $f:A\to B$ is the Gabriel factorisation $f=p u:A\to E\to B$ constructed as follows: $Ob(E)=Ob(A)$ and $E(a,b)=B(f(a),f(b))$ for every pair $a,b\in Ob(A)$. The composition law is obvious. The functors $u$ and $p$ are induced by $f$. A functor is essentially in $\mathcal{L}$ iff it is essentially surjective.
We shall say that a fully faithful functor $f:X\to Y$ is replete if every object of $Y$ which is isomorphic to to an object in the image of $f$ also belongs to this image. The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ the class of essentialy surjective functors and $\mathcal{R}$ is the class of replete fully faithful functors monic on objects. A functor is essentially in $\mathcal{R}$ iff it is fully faithful.
The category of small categories $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ the class of functors surjective on objects and $\mathcal{R}$ is the class of fully faithful functors monic on objects. A functor is essentially in $\mathcal{L}$ iff it is essentially surjective, and a functor is essentially in $\mathcal{R}$ iff it is fully faithful.
(Street and Walters) Recall that a functor between small categories $p:X\to Y$ is said to be a discrete fibration if for every object $x\in X$ and every arrow $g\in Y$ with target $p(x)$, there exists a unique arrow $f\in X$ with target $x$ such that $p(f)=g$. A functor is a discrete fibration iff it is right orthogonal to the map $d_0:[0]\to [1]$. The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L}, \mathcal{R})$ in which $\mathcal{R}$ is the class of discrete fibrations and $\mathcal{L}$ is the class of final functors. Recall that a functor between small categories $u:A \to B$ is final (but we shall eventually say 0-final) iff the category
defined by the pullback square
is connected for every object $b\in B$.
A functor $p:X\to Y$ is called a discrete op-fibration if the opposite functor $p^o:X^o\to Y^o$ is a discrete fibration. A functor $p:X\to Y$ is a discrete opfibration iff for every object $x\in X$ and every arrow $g\in Y$ with source $p(x)$, there exists a unique arrow $f\in X$ with source $x$ such that $p(f)=g$. A functor is a discrete opfibration iff it is right orthogonal to the map $d_1:[0]\to [1]$. The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L}, \mathcal{R})$ in which $\mathcal{R}$ is the class of discrete opfibrations and $\mathcal{L}$ is the class of initial functors. Recall that a functor between small categories $u:A \to B$ is said to be initial (but we shall eventually say 0-initial) if the opposite functor $u^o:A^o\to B^o$ is final. A functor $u:A \to B$ is initial iff the category $A/b= (B/b) \times_{B} A$ defined by the pullback square
is connected for every object $b\in B$.
A functor between groupoids is a discrete fibration iff it is a discrete opfibration, in which case we shall say that it is a covering. We shall say that a functor between groupoids is connected if it is essentially surjective and full. A functor between groupoids is connected iff it is final iff it is initial. The category $\mathbf{Grpd}$ of small groupoids admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of coverings and $\mathcal{L}$ is the class of connected functors. A functor between groupoids is essentially a covering iff it is faithful.
We shall say that a functor $u:A\to B$ in $cat$ is a discrete bifibration, or a 0-covering, if it is both a discrete fibration and a discrete opfibration. If $\mathbf{Grpd}$ is the category of groupoids, then the inclusion functor $\mathbf{Grpd}\subset \mathbf{Cat}$ admits a left adjoint
which associates to a category $A$ its fundamental groupoid? $\pi_1(A)$. By construction, $\pi_1(A)$ is obtained by inverting universally every arrow in $A$. We shall say that a functor $u:A\to B$ is 0-connected if the functor $\pi_1(u):\pi_1(A)\to \pi_1(B)$ is connected. See Example . The category $\mathbf{Cat}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{L}$ the class of connected functors and $\mathcal{R}$ is the class of 0-coverings.
We shall say that a functor $u:A\to B$ inverts an arrow $f\in A$ if the arrow $u(f)$ is invertible in $B$. A functor $u:A\to B$ is conservative iff every arrow $f\in A$ which is inverted by $u$ is invertible in $A$. For any subset $S$ of arrows in small category $A$, there is a small category $S^{-1}A$ together with a functor $l:A\to S^{-1}A$ which inverts universally every arrow in $S$. The universality means that for any functor $u:A\to B$ which inverts every arrow in $S$ there exists a unique functor $u':S^{-1}A\to B$ such that $u'l=u$.
Every functor $u:A\to B$ admits a canonical factorisation $u=u'l:A\to S^{-1}A\to B$, where $S\subseteq A$ is the set of arrows inverted by $u$. We shall say that $u$ is a localisation if $u'$ is an isomorphism. Beware that the functor $u'$ is not conservative in general. Hence the factorisation $u=u'l$ can be repeated with $u':S^{-1}A\to B$ instead of $u$. Let us put $A_1=S^{-1}A$, $u_1=u'$ and let $S_1$ be the set of arrows in $A_1$ inverted by $u_1:A_1\to B$. We then obtain a factorisation $u_1=u_2l_1:A_1\to S_1^{-1}A_1 \to B$ where $l_1:A_1\to S_1^{-1}A_1$ is the canonical functor. Let us put $A_2=S_1^{-1}A_1$. By iterating this process, we obtain an infinite sequ