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 $(A,M) \to (B,M)$. One of these should be $N$. Which one?

**Category theory**
##Contents
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* Weak factorisation systems
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* Model structures on Cat
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We shall say that a morphism $f:a\to b$ in a category $X$ is **cartesian** with respect to a functor $p:X\to Y$, or that $f$ is $p$-**cartesian**, if for every object $x\in X$, and every pair of morphisms $u:x\to b$ and $v:p(x)\to p(a)$ such that $p(u)=p(f)v$, there exists a unique morphism $w:x\to a$ such that $f w=u$ and $p(w)=v$,

Dually, we shall say that a morphism $f:a\to b$ is **cocartesian** with respect to a functor $p:X\to Y$, or that $f$ is $p$-**cocartesian**, if the opposite morphism $f^o:b^o\to a^o$ in $X^o$ is cartesian with respect to the functor $p^o:X^o\to Y^o$.

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

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

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

(1)

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If $d_0:\mathbf{C}^{[1]}\to \mathbf{C}$ is the target functor (which associates to a map $x:X_0\to X_1$ its target $X_1$), then a morphism $f:X\to Y$ in $\mathbf{C}^{[1]}$ is cartesian with respect to $d_1$ iff the square (1) is cartesian. If $d_1:\mathbf{C}^{[1]}\to \mathbf{C}$ is the source functor (which associates to a map $x:X_0\to X_1$ its source $X_0$), then a morphism $f:X\to Y$ in $\mathbf{C}^{[1]}$ is cocartesian with respect to $d_1$ iff the square (1) is cocartesian.

A morphism $f:X\to Y$ in $\mathbf{C}^{[1]}$ is cocartesian with respect to the target functor $d_0:\mathbf{C}^{[1]}\to \mathbf{C}$ iff the map $f_0$ is invertible, and $f$ is cartesian with respect to the source functor $d_1:\mathbf{C}^{[1]}\to \mathbf{C}$ iff the map $f_1$ is invertible.

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

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

where the vertical maps are induced by the functor $p$.

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

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

Left as an exercise to the reader.

Suppose that a morphism $g:b\to c$ in a category $X$ is cartesian with respect to a functor $p:X\to Y$. Then a morphism $f:a\to b$ in $X$ is $p$-cartesian iff the composite morphism $g f:a\to c$ is $p$-cartesian.

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.

Let $f:a\to b$ be a morphism in $X$, and let $p:X\to Y$ and $q:Y\to Z$ be two functors. If the morphism $p(f):p(a)\to p(b)$ is $q$-cartesian then $f$ is $p$-cartesian iff it is $q p$-cartesian.

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.

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

Dually, we shall say that a functor $p:X\to Y$ is a **Grothendieck opfibration** if the opposite functor $p^o:X^o\to Y^o$ is a Grothendieck fibration. We shall say that a functor $p:X\to Y$ 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 $p:X\to Y$ is a Grothendieck opfibration iff for every object $a\in X$ and every morphism $g\in Y$ with source $p(a)$ there exists a cocartesian morphism $f\in X$ with source $a$ such that $p(f)=g$,

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

Let $\mathbf{C}^{[1]}$ be the arrow category of a category $\mathbf{C}$. Then the target functor $d_0:\mathbf{C}^{[1]}\to \mathbf{C}$ is a Grothendieck opfibration, and it is a fibration iff the category $\mathbf{C}$ has pullbacks. Dually the source functor $d_1:\mathbf{C}^{[1]}\to \mathbf{C}$ is a Grothendieck fibration, and it is an opfibration iff the category $\mathbf{C}$ has pushouts.

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

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

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

Let us show that for every object $b\in X$ and every morphism $h\in Z$ with target $q p(b)$, there exists a $q p$-cartesian morphism $f\in X$ with target $b$ and such that $q p(f)=h$. By hypothesis, there exists a $q$-cartesian morphism $g\in Y$ with target $p(b)$ such that $q(g)=h$, since $q$ is a Grothendieck fibration. But then, there exists a $p$-cartesian morphism $f\in X$ with target $b$ such that $q(f)=g$, since $p$ is a Grothendieck fibration,

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

(2)

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

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

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

It remains to show that $f'$ is $p'$-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 $f$ is $p$-cartesian. It then follows from the lemma here that the left hand face is cartesian.

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

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

Show that the morphism $f$ is $p$-cartesian iff the naturality square is cartesian in the category $X\star_p Y$.

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