## 4.38 Categories fibred in sets

Definition 4.38.1. A category is called discrete if the only morphisms are the identity morphisms.

A discrete category has only one interesting piece of information: its set of objects. Thus we sometime confuse discrete categories with sets.

Definition 4.38.2. Let $\mathcal{C}$ be a category. A category fibred in sets, or a category fibred in discrete categories is a category fibred in groupoids all of whose fibre categories are discrete.

We want to clarify the relationship between categories fibred in sets and presheaves (see Definition 4.3.3). To do this it makes sense to first make the following definition.

Definition 4.38.3. Let $\mathcal{C}$ be a category. The $2$-category of categories fibred in sets over $\mathcal{C}$ is the sub $2$-category of the category of categories fibred in groupoids over $\mathcal{C}$ (see Definition 4.35.6) defined as follows:

1. Its objects will be categories $p : \mathcal{S} \to \mathcal{C}$ fibred in sets.

2. Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$ will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that $p' \circ G = p$ (since every morphism is strongly cartesian $G$ automatically preserves them).

3. Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$.

Note that every $2$-morphism is automatically an isomorphism. Hence this $2$-category is actually a $(2, 1)$-category. Here is the obligatory lemma on the existence of $2$-fibre products.

Lemma 4.38.4. Let $\mathcal{C}$ be a category. The 2-category of categories fibred in sets over $\mathcal{C}$ has 2-fibre products. More precisely, the 2-fibre product described in Lemma 4.32.3 returns a category fibred in sets if one starts out with such.

Proof. Omitted. $\square$

Example 4.38.5. This example is the analogue of Examples 4.36.1 and 4.37.1 for presheaves instead of “presheaves of categories”. The output will be a category fibred in sets instead of a fibred category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Sets}$ is a presheaf. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast : F(U) \to F(V)$. We construct a category $\mathcal{S}_ F$ fibred in sets over $\mathcal{C}$ as follows. Define

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F) = \{ (U, x) \mid U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), x \in \mathop{\mathrm{Ob}}\nolimits (F(U))\} .$

For $(U, x), (V, y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F)$ we define

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ F}((V, y), (U, x)) & = \{ f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U) \mid f^*x = y\} \end{align*}

Composition is inherited from composition in $\mathcal{C}$ which works as $g^\ast \circ f^\ast = (f \circ g)^\ast$ for a pair of composable morphisms of $\mathcal{C}$. The functor $p_ F : \mathcal{S}_ F \to \mathcal{C}$ is given by the rule $(U, x) \mapsto U$. As every fibre category $\mathcal{S}_{F, U}$ is discrete with underlying set $F(U)$ and we have already see in Example 4.37.1 that $\mathcal{S}_ F$ is a category fibred in groupoids, we conclude that $\mathcal{S}_ F$ is fibred in sets.

Lemma 4.38.6. Let $\mathcal{C}$ be a category. The only $2$-morphisms between categories fibred in sets are identities. In other words, the $2$-category of categories fibred in sets is a category. Moreover, there is an equivalence of categories

$\left\{ \begin{matrix} \text{the category of presheaves} \\ \text{of sets over }\mathcal{C} \end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{the category of categories} \\ \text{fibred in sets over }\mathcal{C} \end{matrix} \right\}$

The functor from left to right is the construction $F \to \mathcal{S}_ F$ discussed in Example 4.38.5. The functor from right to left assigns to $p : \mathcal{S} \to \mathcal{C}$ the presheaf of objects $U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$.

Proof. The first assertion is clear, as the only morphisms in the fibre categories are identities.

Suppose that $p : \mathcal{S} \to \mathcal{C}$ is fibred in sets. Let $f : V \to U$ be a morphism in $\mathcal{C}$ and let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. Then there is exactly one choice for the object $f^\ast x$. Thus we see that $(f \circ g)^\ast x = g^\ast (f^\ast x)$ for $f, g$ as in Lemma 4.35.2. It follows that we may think of the assignments $U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ and $f \mapsto f^\ast$ as a presheaf on $\mathcal{C}$. $\square$

Here is an important example of a category fibred in sets.

Example 4.38.7. Let $\mathcal{C}$ be a category. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider the representable presheaf $h_ X = \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(-, X)$ (see Example 4.3.4). On the other hand, consider the category $p : \mathcal{C}/X \to \mathcal{C}$ from Example 4.2.13. The fibre category $(\mathcal{C}/X)_ U$ has as objects morphisms $h : U \to X$, and only identities as morphisms. Hence we see that under the correspondence of Lemma 4.38.6 we have

$h_ X \longleftrightarrow \mathcal{C}/X.$

In other words, the category $\mathcal{C}/X$ is canonically equivalent to the category $\mathcal{S}_{h_ X}$ associated to $h_ X$ in Example 4.38.5.

For this reason it is tempting to define a “representable” object in the 2-category of categories fibred in groupoids to be a category fibred in sets whose associated presheaf is representable. However, this is would not be a good definition for use since we prefer to have a notion which is invariant under equivalences. To make this precise we study exactly which categories fibred in groupoids are equivalent to categories fibred in sets.

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