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Cartesian squares

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Cartesian squares

Definition

We recall that a commutative square

(1)

in a category C\mathbf{C} is said to be cartesian, or to be a pullback, if for every object XX of C\mathbf{C} and every pair of maps f:XBf:X\to B and g:XCg:X\to C such that uf=vgu f=v g , there exists a unique map h:XAh:X\to A, such that ph=fp h=f and qh=gq h=g,

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

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

Lemma

If the category C\mathbf{C} has pullbacks, then the category C I\mathbf{C}^{I} admits a factorisation system (,)(\mathcal{L},\mathcal{R}) in which \mathcal{R} is the class of pullback squares. A square f:XYf:X\to Y belongs to the class \mathcal{L} iff the map f 1f_1 is invertible.

Proof

Left to the reader.

Lemma

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.

Proof 1

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.

Proof 2

Let us suppose that the ambiant category C\mathbf{C} has pullbacks. Then the category C I\mathbf{C}^{I} admits a factorisation system (,)(\mathcal{L},\mathcal{R}) in which \mathcal{R} is the class of pullback squares by lemma 1. 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:ABu:A\to B and v:BCv:B\to C in the category C I\mathbf{C}^{I}. We have vv\in \mathcal{R} by assumption. It follows that uu\in \mathcal{R} iff vuvu \in \mathcal{R}, since the the class \mathcal{R} is closed under composition and has the left cancellation property.

Lemma

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.

Proof

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

The faces p:ABp:A\to B and u:BDu:B\to D are cartesian by hypothesis. Hence also their composite up=vqu p=v q by Lemma 2. It follows that the back face q:ACq:A\to C is cartesian by the same lemma, since the right hand face p:CDp:C\to D is cartesian by hypothesis.

Lemma

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

Proof1

The category E=Set I\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 1. But the right class \mathcal{R} of a factorisation system in a category E\mathbf{E} spans a full reflective subcategory of E I\mathbf{E}^I by the proposition here. This proves that the full subcategory of [I×I,Set][I\times I, \mathbf{Set}] spanned by the cartesian squares is reflective, hence closed under limits.

Proof1

The category I×II\times I is a projective cone 0C0\star C based on the category CC with three objects {1,2,3}\{1,2,3\} and two arrows 1321\rightarrow 3\leftarrow 2,

A square S:I×ISetS:I\times I\to \mathbf{Set} is cartesian iff the projective cone that it defines 0CSet0\star C\to \mathbf{Set} is exact. It is a general fact, valid for any category CC, that the full subcategory of [0C,Set][0\star C, \mathbf{Set}] spanned by the exact projective cones is reflective.

Epicartesian squares

Definition

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

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

Lemma

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.

Proof

By diagram chasing.

Lemma

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.

Proof

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

The faces p:ABp:A\to B and u:BDu:B\to D are epicartesian by hypothesis, hence also their composite up=vqu p=v q by Lemma 6. It follows that the back face q:ACq:A\to C is epicartesian by the same lemma, since the right hand face p:CDp:C\to D is cartesian by hypothesis.

Lemma

A retract of an epicartesian square is epicartesian.

Proof

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

is the square of fixed points of an idempotent morphism e:XXe:X\to X. Let us show that XX' is epicartesian when XX is epicartesian. Let xX 01x\in X'_{01} and yX 10y\in X'_{10} be a pair of elements such that q 1(x)=p 1(y)q_1(x)=p_1(y). Then there exists an element zX 00z\in X_{00} such that p 0(z)=xp_0(z)=x and q 0(z)=yq_0(z)=y, since the square XX is epicartesian by assumption. But then e(z)X 00e(z)\in X'_{00}, and we have p 0(e(z))=e(p 0(z))=e(x)=xp_0(e(z))= e(p_0(z))=e(x)=x, q 0(e(z))=e(q 0(z))=e(y)=yq_0(e(z))= e(q_0(z))=e(y)=y. This shows that the square XX' is epicartesian

Lemma

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

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

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

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

Lemma

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

Proof

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

D:[α]SetD:[\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)D(j)D(i,j):D(i)\to D(j) is surjective for every ij<αi\ge j\lt \alpha. Let us show that the restriction map D(α,0):D(α)D(0)D(\alpha,0):D(\alpha)\to D(0) is surjective. Let P=D/[α]P=D/[\alpha] be the poset of elements of the presheaf D:[α] oSetD:[\alpha]^o \to \mathbf{Set}. More precisely, an element of PP is a pair (x,i)(x,i) where i[α]i\in [\alpha] and xD(i)x\in D(i). By definition, we have (x,i)(y,j)(x,i)\leq (y,j) if iji\ge j and D(i,j)(x)=yD(i,j)(x)=y. It follows from the continuity of DD that the poset PP is co-inductive (ie every chain in PP has an infumum). Moreover the minimal elements of PP are of the form (z,α)(z,\alpha) for zD(α)z\in D(\alpha) since the map D(i)D(j)D(i)\to D(j) is surjective for every α>ij\alpha \gt i\le j.
It follows by Zorn lemma that for every xD(0)x\in D(0) there exists an element zD(α)z\in D(\alpha) such that D(α,0)(z)=xD(\alpha,0)(z)=x.

Lemma

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

Exercises

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

Exercise

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

is \mathcal{R}-cartesian. A morphism f:XYf:X\to Y belongs to \mathcal{L}' iff f 1f_1 is invertible and f 0f_0 belongs to \mathcal{L}.

Exercise

The category [I,Set][I,\mathbf{Set}] admits a weak factorisation system (,)(\mathcal{L},\mathcal{R}) in which a morphism f:XYf:X\to Y belongs to \mathcal{R} iff the corresponding square is epicartesian. A morphism f:XYf:X\to Y belongs to \mathcal{L} iff f 1f_1 is bijective and f 0f_0 is monic.

Revised on September 10, 2012 at 06:42:33 by David Roberts