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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