Proof

Left to the reader.

Proof

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.

Proof

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.

Proof

($\Rightarrow$) 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.

Proof

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.

Proof

This follows from Theorem and the Lemma .

Proof

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 has a unique diagonal filler $z:E\to X$ by Theorem .

Proof

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 $\delta(f) x=(x,x)$.

Proof

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 $u\bot f$.

Proof

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.

Proof

By Lemma we have $\Sigma^\bot =\Sigma^\pitchfork \, \cap \, (\delta^o\Sigma)^\pitchfork.$ This proves the result, since

Proof

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