nLab
fibration in a 2-category

Recall the notion of a Grothendieck fibration. It generalizes to the notion of a fibration in a 2-category using Grothendieck fibrations themselves, using generalized elements. We here give an alternative, yet equivalent, 2-categorical definition, trying to explain how it specializes to Grothendieck fibrations. The definition is from

• Ross Street, Fibrations in bicategories. Cahiers de Topologie et Géométrie Différentielle Catégoriques, 21 no. 2 (1980), p. 111–160 (numdam)

and is recalled in Mark Weber’s paper Yoneda Structure from 2-toposes, the main source for this page.

Fix a 2-category $𝒦$. For any two morphisms $f:A\to C$ and $c:B\to C$, let $f/g$ be the corresponding comma object and let $f/=g$ be their pullback.

A morphism $p:A\to B$ in it is a fibration when for all morphism $f:X\to B$, the canonical map $i:f/=p\to f/p$ has a right adjoint in $𝒦/X$.

The purpose of this page is the following result:

When $𝒦$ is Cat, fibrations in this sense are precisely Grothendieck fibrations.

This surely needs some unfolding. First, recall that the 2-category $\mathrm{Cat}/X$ has objects the functors $C\to X$, as morphisms the commuting triangles

$$\begin{array}{ccc}C& \stackrel{h}{\to }& C\prime \\ & f\searrow \swarrow g& \\ & X,& \end{array}$$\array{C & \stackrel{h}{\to} & C' \\ & f \searrow \swarrow g & \\ & X, & }

and as 2-cells the natural transformations $\alpha :h_1\to h_2$ such that $g\alpha ={\mathrm{id}}_f$.

Next, recall that for $f:X\to B$ and $p:A\to B$ the comma object $f/p$ has objects the triples $(x,a,\alpha )$, with $\alpha :f(x)\to p(a)$. We will abbreviate the latter as just $\alpha$.

The pullback $f/=p$ has objects the pairs $(x,a)$ with $f(x)=p(a)$, and the above mentioned canonical morphism $f/=p\to f/p$ is simply the inclusion functor of identity maps $\mathrm{id}:f(x)\to p(a)$.

Somewhat unprecisely, seeing both categories $f/p$ and $f/=p$ as sitting over $X$ means that functors between those should be the identity on the $x$ component, and natural transformations should have the identity as their $x$ component.

Let us concentrate on the case $X=B$. Then the pullback category $f/=p$ has as objects the pairs $(b,a)$ such that $b=p(a)$, i.e., just objects $a$ of $A$. And similarly, its morphisms are just morphisms of $A$.

Assuming a right adjoint $q$ to $i$, $q$ sends a morphism $\alpha :b\to p(a)$ to some object $q(\alpha )$ of $A$, which $i$ then sends to the identity on $\mathrm{pq}(\alpha )$, or more exactly the triple $(\mathrm{pq}(\alpha ),q(\alpha ),{\mathrm{id}}_{\mathrm{pq}(\alpha )})$.

The counit for the corresponding adjunction has to be a morphism in $f/p$ from the latter to $\alpha$ itself, i.e., a pair $({ϵ}_0,{ϵ}_1)$ making the square

$$\begin{array}{ccc}\mathrm{pq}(\alpha )& \stackrel{\mathrm{id}}{\to }& \mathrm{pq}(\alpha )\\ {ϵ}_1\downarrow & & \downarrow p({ϵ}_0)\\ b& \stackrel{\alpha }{\to }& p(a)\end{array}$$\array{pq(\alpha) & \stackrel{id}{\to} & pq(\alpha) \\\epsilon_1 \downarrow & & \downarrow p(\epsilon_0) \\ b & \stackrel{\alpha}{\to} & p(a)}

commute.

But, as a 2-cell in $\mathrm{Cat}/B$, this pair must have ${ϵ}_1=\mathrm{id}$, hence the counit is actually providing an arrow ${ϵ}_0$ in $A$, sent to $\alpha$ by $p$.

Moreover, its universal property tells us that for any other morphism in $f/p$ from some $i(a\prime )$ to our $\alpha$, i.e., for any $a\prime$ and any pair $(h,k)$ making the square

$$\begin{array}{ccc}p(a\prime )& \stackrel{\mathrm{id}}{\to }& p(a\prime )\\ k\downarrow & & \downarrow p(h)\\ b& \stackrel{\alpha }{\to }& p(a)\end{array}$$\array{p(a') & \stackrel{id}{\to} & p(a') \\k \downarrow & & \downarrow p(h) \\ b & \stackrel{\alpha}{\to} & p(a)}

commute, there is a unique map $m:a\prime \to q(\alpha )$ in $A$ such that the above square factors in $f/p$ as

$$\begin{array}{ccc}p(a\prime )& \stackrel{\mathrm{id}}{\to }& p(a\prime )\\ p(m)\downarrow & & \downarrow p(m)\\ \mathrm{pq}(\alpha )& \stackrel{\mathrm{id}}{\to }& \mathrm{pq}(\alpha )\\ \mathrm{id}\downarrow & & \downarrow p({ϵ}_0)\\ b& \stackrel{\alpha }{\to }& a.\end{array}$$\array{p(a') & \stackrel{id}{\to} & p(a') \\ p(m) \downarrow & & \downarrow p(m) \\ pq(\alpha) & \stackrel{id}{\to} & pq(\alpha) \\ id \downarrow & & \downarrow p(\epsilon_0) \\ b & \stackrel{\alpha}{\to} & a.}

In other words, the universal property provides a unique $m$ such that ${ϵ}_{0}m=h$ and $p(m)=k$, which exactly asserts that ${ϵ}_0$ is cartesian.

Hence $p$ is a Grothendieck fibration. The other implication remains to be proved, but not today.