Lemma 4.42.6. Let $\mathcal{C}$ be a category. Let $\mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids. Assume $\mathcal{C}$ has products of pairs of objects and fibre products. The following are equivalent:

1. The diagonal $\mathcal{S} \to \mathcal{S} \times \mathcal{S}$ is representable.

2. For every $U$ in $\mathcal{C}$, any $G : \mathcal{C}/U \to \mathcal{S}$ is representable.

Proof. Suppose the diagonal is representable, and let $U, G$ be given. Consider any $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and any $G' : \mathcal{C}/V \to \mathcal{S}$. Note that $\mathcal{C}/U \times \mathcal{C}/V = \mathcal{C}/U \times V$ is representable. Hence the fibre product

$\xymatrix{ (\mathcal{C}/U \times V) \times _{(\mathcal{S} \times \mathcal{S})} \mathcal{S} \ar[r] \ar[d] & \mathcal{S} \ar[d] \\ \mathcal{C}/U \times V \ar[r]^{(G, G')} & \mathcal{S} \times \mathcal{S} }$

is representable by assumption. This means there exists $W \to U \times V$ in $\mathcal{C}$, such that

$\xymatrix{ \mathcal{C}/W \ar[d] \ar[r] & \mathcal{S} \ar[d] \\ \mathcal{C}/U \times \mathcal{C}/V \ar[r] & \mathcal{S} \times \mathcal{S} }$

is cartesian. This implies that $\mathcal{C}/W \cong \mathcal{C}/U \times _\mathcal {S} \mathcal{C}/V$ (see Lemma 4.31.11 and Remark 4.35.8) as desired.

Assume (2) holds. Consider any $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and any $(G, G') : \mathcal{C}/V \to \mathcal{S} \times \mathcal{S}$. We have to show that $\mathcal{C}/V \times _{\mathcal{S} \times \mathcal{S}} \mathcal{S}$ is representable. What we know is that $\mathcal{C}/V \times _{G, \mathcal{S}, G'} \mathcal{C}/V$ is representable, say by $a : W \to V$ in $\mathcal{C}/V$. The equivalence

$\mathcal{C}/W \to \mathcal{C}/V \times _{G, \mathcal{S}, G'} \mathcal{C}/V$

followed by the second projection to $\mathcal{C}/V$ gives a second morphism $a' : W \to V$. Consider $W' = W \times _{(a, a'), V \times V} V$. There exists an equivalence

$\mathcal{C}/W' \cong \mathcal{C}/V \times _{\mathcal{S} \times \mathcal{S}} \mathcal{S}$

namely

\begin{eqnarray*} \mathcal{C}/W' & \cong & \mathcal{C}/W \times _{(\mathcal{C}/V \times \mathcal{C}/V)} \mathcal{C}/V \\ & \cong & \left(\mathcal{C}/V \times _{(G, \mathcal{S}, G')} \mathcal{C}/V\right) \times _{(\mathcal{C}/V \times \mathcal{C}/V)} \mathcal{C}/V \\ & \cong & \mathcal{C}/V \times _{(\mathcal{S} \times \mathcal{S})} \mathcal{S} \end{eqnarray*}

(for the last isomorphism see Lemma 4.31.12 and Remark 4.35.8) which proves the lemma. $\square$

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