We recall that a commutative square

(1)

```
```

in a category $\mathbf{C}$ is said to be **cartesian**, or to be a **pullback**, if for every object $X$ of $\mathbf{C}$ and every pair of maps $f:X\to B$ and $g:X\to C$ such that $u f=v g$, there exists a unique map $h:X\to A$, such that $p h=f$ and $q h=g$,

In the category of sets, a square (1) is cartesian iff for every pair of elements $(b,c)\in B\times C$ such that $u(a)=v(c)$, there exists a unique element $a\in A$, such that $p(a)=b$ and $q(a)=c$. In general, a square (1) in a category $\mathbf{C}$ is cartesian iff the following square in the category of sets is cartesian for every object $X$ of $\mathbf{C}$,

Let $\mathbf{C}^{I}$ be the arrow category of a category $\mathbf{C}$. A morphism $f:X\to Y$ in the category $\mathbf{C}^{I}$ is a commutative square in the category $\mathbf{C}$,

If the category $\mathbf{C}$ has pullbacks, then the category $\mathbf{C}^{I}$ 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 the class $\mathcal{L}$ iff the map $f_1$ is invertible.

Left to the reader.

Suppose that we have a commutative diagram

(2)

```
```

in which the right hand square is cartesian. Then the left hand square is cartesian iff the composite square is cartesian.

It suffices to prove the result in the case of a diagram in the category of sets, in which case the proof is left to the reader.

Let us suppose that the ambiant category $\mathbf{C}$ has pullbacks. Then the category $\mathbf{C}^{I}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of pullback squares by lemma . Hence the class $\mathcal{R}$ is closed under composition and has the left cancellation property by the proposition here. The diagram can be represented by two maps $u:A\to B$ and $v:B\to C$ in the category $\mathbf{C}^{I}$. We have $v\in \mathcal{R}$ by assumption. It follows that $u\in \mathcal{R}$ iff $vu \in \mathcal{R}$, since the the class $\mathcal{R}$ is closed under composition and has the left cancellation property.

Suppose that we have a commutative cube

(3)

```
```

in which the right face is cartesian. If the left and front faces are cartesian, then so is the back face.

If $\mathbf{C}$ denotes the ambiant category, then the cube (viewed from high above) is a commutative square in the arrow category $\mathbf{C}^{I}$,

The faces $p:A\to B$ and $u:B\to D$ are cartesian by hypothesis. Hence also their composite $u p=v q$ by Lemma . It follows that the back face $q:A\to C$ is cartesian by the same lemma, since the right hand face $p:C\to D$ is cartesian by hypothesis.

The full subcategory of $[I\times I, \mathbf{Set}]$ spanned by the cartesian squares is reflective, hence also closed under arbitrary limits.

The category $\mathbf{E}=\mathbf{Set}^{I}$ admits a factorisation system $(\mathcal{L},\mathcal{R})$ in which $\mathcal{R}$ is the class of cartesian squares by lemma . But the right class $\mathcal{R}$ of a factorisation system in a category $\mathbf{E}$ spans a full reflective subcategory of $\mathbf{E}^I$ by the proposition here. This proves that the full subcategory of $[I\times I, \mathbf{Set}]$ spanned by the cartesian squares is reflective, hence closed under limits.

The category $I\times I$ is a projective cone $0\star C$ based on the category $C$ with three objects $\{1,2,3\}$ and two arrows $1\rightarrow 3\leftarrow 2$,

A square $S:I\times I\to \mathbf{Set}$ is cartesian iff the projective cone that it defines $0\star C\to \mathbf{Set}$ is exact. It is a general fact, valid for any category $C$, that the full subcategory of $[0\star C, \mathbf{Set}]$ spanned by the exact projective cones is reflective.

We shall say that a commutative square in the category of sets,

is **epicartesian** if the induced map $(p,q):A\to B\times_D C$ is surjective. In other words, if for every pair of elements $(b,c)\in B\times D$ such that $u(b)=v(c)$, there exists an element $a\in A$ *not necessarly unique* such that $p(a)=b$ and $q(a)=c$.

Suppose that we have the following commutative diagram in the category of sets,

If the two squares are epicartesian, then so is their composite. Conversely, if the right hand square is cartesian and the composite square is epicartesian, then the left hand square is epicartesian.

By diagram chasing.

Suppose that the right hand face of commutative cube in the category of sets is cartesian,

If the left and front faces are epicartesian, then so is the back face.

The cube viewed from high above is a commutative square in the arrow category ${Set}^{I}$,

The faces $p:A\to B$ and $u:B\to D$ are epicartesian by hypothesis, hence also their composite $u p=v q$ by Lemma . It follows that the back face $q:A\to C$ is epicartesian by the same lemma, since the right hand face $p:C\to D$ is cartesian by hypothesis.

A retract of an epicartesian square is epicartesian.

A retract $S'$ of a set $S$ is the set of fixed points of an idempotent map $e:S\to S$. Similarly, a retract $X'$ of a square $X$,

is the square of fixed points of an idempotent morphism $e:X\to X$. Let us show that $X'$ is epicartesian when $X$ is epicartesian. Let $x\in X'_{01}$ and $y\in X'_{10}$ be a pair of elements such that $q_1(x)=p_1(y)$. Then there exists an element $z\in X_{00}$ such that $p_0(z)=x$ and $q_0(z)=y$, since the square $X$ is epicartesian by assumption. But then $e(z)\in X'_{00}$, and we have $p_0(e(z))= e(p_0(z))=e(x)=x$, $q_0(e(z))= e(q_0(z))=e(y)=y$. This shows that the square $X'$ is epicartesian

The class of epicartesian squares is closed under arbitrary products in the category $[I\times I, \mathbf{Set}]$.

Recall from the theory of weak factorisation systems that if $\mathbf{E}$ is a category and $\alpha$ is an ordinal, then a contravariant functor $C:[\alpha]\to \mathbf{E}$ is called an *opchain*. The opchain is *continuous* if the canonical map

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

is an isomorphism for every non-zero limit ordinal $j\in [\alpha]$. The *composite* of $C$ is defined to be the canonical map $C(\alpha)\to C(0)$. The *base* of $C$ is the restriction of $C$ to $[\alpha)$. We shall say that a subcategory $\mathcal{C}\subseteq \mathbf{E}$ is closed under *transfinite op-compositions* if for any limit ordinal $\alpha\gt 0$, any continuous op-chain $C:\alpha \to \mathbf{E}$ with a base in $\mathcal{C}$ has a composite in $\mathcal{C}$.

In the category of sets, the subcategory of surjections is closed under transfinite op-compositions.

Let $\alpha \gt 0$ be a limit ordinal, and let

$D:[\alpha] \to \mathbf{Set}$

be a continuous opchain with a base in the sub-category of surjections. This last condition means that the “restriction map” $D(i,j):D(i)\to D(j)$ is surjective for every $i\ge j\lt \alpha$. Let us show that the restriction map $D(\alpha,0):D(\alpha)\to D(0)$ is surjective. Let $P=D/[\alpha]$ be the poset of elements of the presheaf $D:[\alpha]^o \to \mathbf{Set}$. More precisely, an element of $P$ is a pair $(x,i)$ where $i\in [\alpha]$ and $x\in D(i)$. By definition, we have $(x,i)\leq (y,j)$ if $i\ge j$ and $D(i,j)(x)=y$. It follows from the continuity of $D$ that the poset $P$ is co-inductive (ie every chain in $P$ has an infumum). Moreover the minimal elements of $P$ are of the form $(z,\alpha)$ for $z\in D(\alpha)$ since the map $D(i)\to D(j)$ is surjective for every $\alpha \gt i\le j$.

It follows by Zorn lemma that for every $x\in D(0)$ there exists an element $z\in D(\alpha)$ such that $D(\alpha,0)(z)=x$.

The class of epicartesian squares is closed under transfinite op-composition in the category $[I,\mathbf{Set}]$.

More generally, if $\mathcal{R}$ is a class of maps in a category pullbacks $\mathbf{C}$, we shall say that the commutative square (1) is $\mathcal{R}$-**cartesian** if the canonical map $(p,q):A\to B\times_D C$ belongs to $\mathcal{R}$. Recall that a commutative square in $\mathbf{C}$

Let $(\mathcal{L},\mathcal{R})$ be a weak factorisation system in a category with pullbacks $\mathbf{C}$. Then the category $[I,\mathbf{C}]$ admits a weak factorisation system $(\mathcal{L}',\mathcal{R}')$ in which a morphism $f:X\to Y$ belongs to $\mathcal{R}'$ iff the corresponding square

is $\mathcal{R}$-cartesian. A morphism $f:X\to Y$ belongs to $\mathcal{L}'$ iff $f_1$ is invertible and $f_0$ belongs to $\mathcal{L}$.

The category $[I,\mathbf{Set}]$ admits a weak factorisation system $(\mathcal{L},\mathcal{R})$ in which a morphism $f:X\to Y$ belongs to $\mathcal{R}$ iff the corresponding square is epicartesian. A morphism $f:X\to Y$ belongs to $\mathcal{L}$ iff $f_1$ is bijective and $f_0$ is monic.

Revised on November 20, 2020 at 21:48:54
by
Dmitri Pavlov