In this section we work out fibre products in the category of contravariant functors from a category to the category of sets. This will later be superseded during the discussion of sites, presheaves, sheaves. Of some interest is the notion of a “representable morphism” between such functors.
Lemma 4.8.1. Let $\mathcal{C}$ be a category. Let $F, G, H : \mathcal{C}^{opp} \to \textit{Sets}$ be functors. Let $a : F \to G$ and $b : H \to G$ be transformations of functors. Then the fibre product $F \times _{a, G, b} H$ in the category $\textit{PSh}(\mathcal{C})$ exists and is given by the formula for any object $X$ of $\mathcal{C}$.
\[ (F \times _{a, G, b} H)(X) = F(X) \times _{a_ X, G(X), b_ X} H(X) \]
Proof. Omitted. $\square$
As a special case suppose we have a morphism $a : F \to G$, an object $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and an element $\xi \in G(U)$. According to the Yoneda Lemma 4.3.5 this gives a transformation $\xi : h_ U \to G$. The fibre product in this case is described by the rule
If $F$, $G$ are also representable, then this is the functor representing the fibre product, if it exists, see Section 4.6. The analogy with Definition 4.6.4 prompts us to define a notion of representable transformations.
Definition 4.8.2. Let $\mathcal{C}$ be a category. Let $F, G : \mathcal{C}^{opp} \to \textit{Sets}$ be functors. We say a morphism $a : F \to G$ is representable, or that $F$ is relatively representable over $G$, if for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and any $\xi \in G(U)$ the functor $h_ U \times _ G F$ is representable.
Lemma 4.8.3. Let $\mathcal{C}$ be a category. Let $a : F \to G$ be a morphism of contravariant functors from $\mathcal{C}$ to $\textit{Sets}$. If $a$ is representable, and $G$ is a representable functor, then $F$ is representable.
Proof. Omitted. $\square$
Lemma 4.8.4. Let $\mathcal{C}$ be a category. Let $F : \mathcal{C}^{opp} \to \textit{Sets}$ be a functor. Assume $\mathcal{C}$ has products of pairs of objects and fibre products. The following are equivalent:
the diagonal $\Delta : F \to F \times F$ is representable,
for every $U$ in $\mathcal{C}$, and any $\xi \in F(U)$ the map $\xi : h_ U \to F$ is representable,
for every pair $U, V$ in $\mathcal{C}$ and any $\xi \in F(U)$, $\xi ' \in F(V)$ the fibre product $h_ U \times _{\xi , F, \xi '} h_ V$ is representable.
Proof. We will continue to use the Yoneda lemma to identify $F(U)$ with transformations $h_ U \to F$ of functors.
Equivalence of (2) and (3). Let $U, \xi , V, \xi '$ be as in (3). Both (2) and (3) tell us exactly that $h_ U \times _{\xi , F, \xi '} h_ V$ is representable; the only difference is that the statement (3) is symmetric in $U$ and $V$ whereas (2) is not.
Assume condition (1). Let $U, \xi , V, \xi '$ be as in (3). Note that $h_ U \times h_ V = h_{U \times V}$ is representable. Denote $\eta : h_{U \times V} \to F \times F$ the map corresponding to the product $\xi \times \xi ' : h_ U \times h_ V \to F \times F$. Then the fibre product $F \times _{\Delta , F \times F, \eta } h_{U \times V}$ is representable by assumption. This means there exist $W \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, morphisms $W \to U$, $W \to V$ and $h_ W \to F$ such that
is cartesian. Using the explicit description of fibre products in Lemma 4.8.1 the reader sees that this implies that $h_ W = h_ U \times _{\xi , F, \xi '} h_ V$ as desired.
Assume the equivalent conditions (2) and (3). Let $U$ be an object of $\mathcal{C}$ and let $(\xi , \xi ') \in (F \times F)(U)$. By (3) the fibre product $h_ U \times _{\xi , F, \xi '} h_ U$ is representable. Choose an object $W$ and an isomorphism $h_ W \to h_ U \times _{\xi , F, \xi '} h_ U$. The two projections $\text{pr}_ i : h_ U \times _{\xi , F, \xi '} h_ U \to h_ U$ correspond to morphisms $p_ i : W \to U$ by Yoneda. Consider $W' = W \times _{(p_1, p_2), U \times U} U$. It is formal to show that $W'$ represents $F \times _{\Delta , F \times F} h_ U$ because
Thus $\Delta $ is representable and this finishes the proof. $\square$
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