nLab
discrete fibration

Definition

A functor $F:C\to B$ is a discrete fibration if for every object $c$ in $C$, and every morphism of the form $g:b\to F(c)$ in $B$ there is a unique morphism $h:d\to c$ in $C$ such that $F(h)=g$.

A functor $F:C\to B$ is a discrete opfibration if $F^{\mathrm{op}}:C^{\mathrm{op}}\to B^{\mathrm{op}}$ is a discrete fibration.

A discrete fibration is a special case of a Grothendieck fibration.

Given a cartesian category $E$, internal categories $C,B$ in $E$, an internal functor $F:C\to B$ is a discrete fibration of internal categories if the square

$$\begin{array}{ccc}{C}_1& \stackrel{F_1}{\to }& B_1\\ d_0\downarrow & & \downarrow d_0\\ C_0& \stackrel{F_0}{\to }& B_0\end{array}$$\begin{matrix} C_1 &\stackrel{F_1}\to& B_1\\ d_0\downarrow && \downarrow d_0\\ C_0 &\stackrel{F_0}\to& B_0 \end{matrix}

is cartesian.

Discussion via category of elements

Given a discrete fibration $F:C\to B$, define a functor $F^{*}:B^{\mathrm{op}}\to \mathrm{Set}$ as follows:

• For $x$ an object of $B$, let $F^{*}(x)$ be the set of objects $y$ of $C$ such that $F(y)=x$.
• For $g:x\to y$ a morphism of $B$, let $F^{*}(g):F^{*}(y)\to F^{*}(x)$ be the function that maps each element of $F^{*}(y)$ to the unique morphism $h$ determined by the definition of discrete fibration above.

There is a size issue here, is $F^{*}(x)$ in fact small? We say that the fibration has small fibres if so; else we must pass to a larger universe when we define Set.

Conversely, give a functor $F^{*}:B^{\mathrm{op}}\to \mathrm{Set}$, define a category $C$ and a discrete fibration $F:C\to B$ as follows:

• Let $C$ be the category of elements of the functor $F$; that is:

  • an object of $C$ is a pair consisting of an object $x$ of $B$ and an element of $F^{*}(x)$,
  • a morphism from $(x,a)$ to $(y,b)$ in $C$ is a morphism $g:x\to y$ in $B$ such that $F^{*}(g)$ assigns $a$ to $b$.

• The functor from $C$ to $B$ is the obvious forgetful functor.

If you start from $F^{*}$, construct $C$ and $F$, and then construct a new $F^{*}$, it will be equal to the original $F^{*}$. Conversely, if you start with $C$ and $F$, construct $F^{*}$, and then construct a new $C\prime$ and $F\prime$, then there will be an isomorphism of categories between $C$ and $C\prime$, relative to which $F$ and $F\prime$ are equal.

Invariance under equivalence

Note that the definition of fibration refers to equality of morphisms without previously assuming that the sources match, while the construction of $F^{*}$ from $F$ refers to equality of objects. This is also why we get equality of functors and isomorphism of categories in the immediately preceding paragraph. So the only non-evil thing on this page is the idea of a functor to Set. That is the fundamental invariant notion; a discrete fibration is just a convenient way of talking about it.

Generalization for spans internal to a category

Let $E$ be a cartesian category. A span of internal categories $A\stackrel{p}{\leftarrow }C\stackrel{q}{\to }B$ in $\mathrm{Cat}(E)$ is called a discrete fibration from $A$ to $B$ if in the diagram

$$\begin{array}{ccccc}{A}_0& \leftarrow & C_l& & \\ \downarrow & & \downarrow i_l& & \\ A& \stackrel{p}{\leftarrow }& C& \stackrel{i_r}{\leftarrow }& C_r\\ & & q\downarrow & & \downarrow \\ & & B& \leftarrow & B_0\end{array}$$\begin{matrix} A_0 & \leftarrow & C_l && \\ \downarrow &&\downarrow i_l &&\\ A &\stackrel{p}\leftarrow & C &\stackrel{i_r}\leftarrow & C_r\\ &&q \downarrow && \downarrow \\ && B &\leftarrow & B_0 \end{matrix}

in which the two squares are the cartesian satisfies the following 3 properties:

• $p\circ i_r:C_1\to A$ is a discrete fibration

• $q\circ i_l:C_l\to B$ is a discrete opfibration

• Let $X$ be defined as the pullback

  $$\begin{array}{ccc}X& \to & (C_r{)}_1\\ \downarrow & & \downarrow \\ (C_l{)}_1& \to & C_0\end{array}$$\begin{matrix} X & \to & (C_r)_1 \\ \downarrow &&\downarrow \\ (C_l)_1 &\to & C_0 \end{matrix}

  and $j:X\hookrightarrow C_1{\times }_{C_0}{C}_1$ the canonical inclusion. Then the morphism $c\circ j:X\to C_1$, where $c:C_1{\times }_{C_0}{C}_1\to C_1$ is the composition morphism of internal category $C$, is invertible.

Example. Given internal functors $a:A\to D$ and $b:B\to D$ in $E$, the obvious span $A\leftarrow a\downarrow b\to B$ is a discrete fibration from $A$ to $B$.