Joyal's CatLab
Grothendieck fibrations

Please, feel free to leave a comment.

Uday Reddy, 2012-05-18. There is a typo in Example 6 (Mod/Ring). A morphism of modules is said to be from (A,M)(B,M)(A,M) \to (B,M). One of these should be NN. Which one?

**Category theory** ##Contents * Contributors * References * Introduction * Basic category theory * Weak factorisation systems * Factorisation systems * Distributors and barrels * Model structures on Cat * Homotopy factorisation systems in Cat * Accessible categories * Locally presentable categories * Algebraic theories and varieties of algebras

Edit this sidebar

Contents

Cartesian morphisms

Definition

We shall say that a morphism f:abf:a\to b in a category XX is cartesian with respect to a functor p:XYp:X\to Y, or that ff is pp-cartesian, if for every object xXx\in X, and every pair of morphisms u:xbu:x\to b and v:p(x)p(a)v:p(x)\to p(a) such that p(u)=p(f)vp(u)=p(f)v, there exists a unique morphism w:xaw:x\to a such that fw=uf w=u and p(w)=vp(w)=v,

Dually, we shall say that a morphism f:abf:a\to b is cocartesian with respect to a functor p:XYp:X\to Y, or that ff is pp-cocartesian, if the opposite morphism f o:b oa of^o:b^o\to a^o in X oX^o is cartesian with respect to the functor p o:X oY op^o:X^o\to Y^o.

Let us spell out the definition of a cocartesian morphism. A morphism f:abf:a\to b in a category XX is cocartesian with respect to a functor p:XYp:X\to Y iff for every object yXy\in X, and every pair of morphisms u:ayu:a\to y and v:p(b)p(y)v:p(b)\to p(y) such that p(u)=vp(f)p(u)=v p(f), there exists a unique morphism w:byw:b\to y such that wf=uw f=u and p(w)=vp(w)=v.

Example

A morphism in a category XX is invertible iff it is cartesian with respect to the functor X1X\to 1 iff it is cartesian with respect to every functor XYX\to Y.

Example

Let C [1]\mathbf{C}^{[1]} be the arrow category of a category C\mathbf{C}. A morphism f:XYf:X\to Y in C [1]\mathbf{C}^{[1]} is a commutative square of maps in C\mathbf{C},

(1)

If d 0:C [1]Cd_0:\mathbf{C}^{[1]}\to \mathbf{C} is the target functor (which associates to a map x:X 0X 1x:X_0\to X_1 its target X 1X_1), then a morphism f:XYf:X\to Y in C [1]\mathbf{C}^{[1]} is cartesian with respect to d 1d_1 iff the square (1) is cartesian. If d 1:C [1]Cd_1:\mathbf{C}^{[1]}\to \mathbf{C} is the source functor (which associates to a map x:X 0X 1x:X_0\to X_1 its source X 0X_0), then a morphism f:XYf:X\to Y in C [1]\mathbf{C}^{[1]} is cocartesian with respect to d 1d_1 iff the square (1) is cocartesian.

Example

A morphism f:XYf:X\to Y in C [1]\mathbf{C}^{[1]} is cocartesian with respect to the target functor d 0:C [1]Cd_0:\mathbf{C}^{[1]}\to \mathbf{C} iff the map f 0f_0 is invertible, and ff is cartesian with respect to the source functor d 1:C [1]Cd_1:\mathbf{C}^{[1]}\to \mathbf{C} iff the map f 1f_1 is invertible.

Proposition

A morphism f:abf:a\to b in a category XX is cartesian with respect to a functor p:XYp:X\to Y iff it satisfies the following equivalent conditions:

(1) for every object xXx\in X the following square is cartesian in the category of sets,

where the vertical maps are induced by the functor pp.

(2) the following square of categories is cartesian in the category of categories,

where the vertical functors are induced by the functor pp, and where the functor f !f_! and p(f) !p(f)_! are respectively defined by composing with ff and p(f)p(f).

Proof

Left as an exercise to the reader.

Proposition

Suppose that a morphism g:bcg:b\to c in a category XX is cartesian with respect to a functor p:XYp:X\to Y. Then a morphism f:abf:a\to b in XX is pp-cartesian iff the composite morphism gf:acg f:a\to c is pp-cartesian.

Proof

The hypothesis implies that the right hand square of the following commutative diagram is cartesian.

It then follows from the lemma here that the left hand square is cartesian iff the composite square is cartesian.

Proposition

Let f:abf:a\to b be a morphism in XX, and let p:XYp:X\to Y and q:YZq:Y\to Z be two functors. If the morphism p(f):p(a)p(b)p(f):p(a)\to p(b) is qq-cartesian then ff is pp-cartesian iff it is qpq p-cartesian.

Proof

The hypothesis implies that the bottom square of the following commutative diagram is cartesian,

It then follows from the lemma here that the top square is cartesian iff the composite square is cartesian.

Grothendieck fibrations

Definition

We shall say that a functor p:XYp:X\to Y is a Grothendieck fibration if for every object bXb\in X and every morphism gYg\in Y with target p(b)p(b) there exists a pp-cartesian morphism fXf\in X with target bb such that p(f)=gp(f)=g,

Dually, we shall say that a functor p:XYp:X\to Y is a Grothendieck opfibration if the opposite functor p o:X oY op^o:X^o\to Y^o is a Grothendieck fibration. We shall say that a functor p:XYp:X\to Y is a Grothendieck bifibration if it is both a Grothendieck fibration and a Grothendieck opfibration.

Let us spell out the definition of a Grothendieck opfibration. A functor p:XYp:X\to Y is a Grothendieck opfibration iff for every object aXa\in X and every morphism gYg\in Y with source p(a)p(a) there exists a cocartesian morphism fXf\in X with source aa such that p(f)=gp(f)=g,

Example

Recall that a functor p:EBp:E\to B is said to be a discrete fibration if for every object bEb\in E and every morphism gBg\in B with target p(b)p(b) there exists a unique morphism fEf\in E with target bb such that p(f)=gp(f)=g. Every discrete fibration is a Grothendieck fibration.

Example

Let C [1]\mathbf{C}^{[1]} be the arrow category of a category C\mathbf{C}. Then the target functor d 0:C [1]Cd_0:\mathbf{C}^{[1]}\to \mathbf{C} is a Grothendieck opfibration, and it is a fibration iff the category C\mathbf{C} has pullbacks. Dually the source functor d 1:C [1]Cd_1:\mathbf{C}^{[1]}\to \mathbf{C} is a Grothendieck fibration, and it is an opfibration iff the category C\mathbf{C} has pushouts.

Example

Let Mod/Ring\mathbf{Mod/Ring} be the category whose objects are the pairs (A,M)(A,M), where AA is a ring and MM is a left AA-module, and whose morphisms (A,M)(B,M)(A,M)\to (B,M) are the pairs (f,u)(f,u), where f:ABf:A\to B is a ring homorphism and u:MNu:M\to N is a map of left AA-modules. If Ring\mathbf{Ring} is the category of rings, then the obvious forgetful functor U:Mod/RingRingU:\mathbf{Mod/Ring}\to \mathbf{Ring} is a Grothendieck bifibration. A morphism (f,u):(A,M)(B,M)(f,u):(A,M)\to (B,M) is UU-cartesian iff uu is invertible, and it is UU-cocartesian iff the induce map of BB-modules B AMNB\otimes_A M\to N is invertible.

Proposition

The class of Grothendieck fibrations in the category CAT\mathbf{CAT} is closed under composition and base changes.

Proof

Let us show that the composite of two Grothendieck fibrations p:XYp:X\to Y and q:YZq:Y\to Z is a Grothendieck fibration.

Proof

Let us show that for every object bXb\in X and every morphism hZh\in Z with target qp(b) q p(b), there exists a qpq p-cartesian morphism fXf\in X with target bb and such that qp(f)=hq p(f)=h. By hypothesis, there exists a qq-cartesian morphism gYg\in Y with target p(b)p(b) such that q(g)=hq(g)=h, since qq is a Grothendieck fibration. But then, there exists a pp-cartesian morphism fXf\in X with target bb such that q(f)=gq(f)=g, since pp is a Grothendieck fibration,

The morphism ff is qpq p-cartesian by Proposition 6 and we have qp(f)=hq p(f)=h. This proves that qpq p is a Grothendieck fibration. Let us now prove that the base change of a Gothendieck fibration q:XYq:X\to Y along a functor v:YYv:Y'\to Y is a Grothendieck fibration q:XYq':X'\to Y',

(2)

For this we have to show that for every object bXb'\in X' and every morphism gYg'\in Y' with target p(b)p'(b'), there exists a pp'-cartesian morphism fXf'\in X' with target bb' such that p(f)=gp'(f')=g'. Let us put b=u(b)Xb=u(b')\in X and g=v(g)Yg=v(g')\in Y. The target of gg is equal to p(b)p(b),

Hence there exists a pp-cartesian morphism fXf\in X with target bb such that p(f)=gp(f)=g, since the functor pp is a Grothendieck fibration by hypothesis,

There is then a unique morphism fEf'\in E such that p(f)=gp'(f')=g and u(f)=fu(f')=f, since the square (2) is cartesian. If f:abf':a'\to b', then f:abf:a\to b where a=u(a)a=u(a'),

It remains to show that ff' is pp'-cartesian. For this it suffices to show that the left hand face of the following cube is cartesian by Proposition 2,

But the back and front faces of the cube are cartesian, since the square (2) is cartesian. The right hand face is also cartesian by Proposition 2, since the morphism ff is pp-cartesian. It then follows from the lemma here that the left hand face is cartesian.

Exercises

Exercise

If the image of a morphism f:abf:a\to b by a functor p:EBp:E\to B is invertible, then ff is pp-cartesian iff it is invertible.

Exercise

The notion of a cartesian morphism with respect to a functor p:XYp:X\to Y can be analysed by using the collage category C=X pY=X D(p)YC=X\star_p Y=X\star_{D(p)} Y of the distribuor D(p):XYD(p)\colon X ⇸Y defined by putting D(p)(x,y)=Y(p(x),y)D(p)(x,y)=Y(p(x),y) for xXx\in X and yYy\in Y. The natural inclusion YX pYY\subseteq X\star_p Y turns the category YY into a full reflexive subcategory of X pYX\star_p Y. If xXx\in X, then the element θ(x)D(x,p(x))\theta(x)\in D(x,p(x)), defined by the identity of p(x)p(x) is reflecting the object xx into the subcategory YY. For every morphism f:abf:a\to b in XX there is a naturality square in X pYX\star_p Y,

Show that the morphism ff is pp-cartesian iff the naturality square is cartesian in the category X pYX\star_p Y.

Revised on September 12, 2012 at 23:28:47 by Jacques Carette