**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 exists $W \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, morphisms $W \to U$, $W \to V$ and $h_ W \to F$ such that

\[ \xymatrix{ h_ W \ar[d] \ar[r] & h_ U \times h_ V \ar[d]^{\xi \times \xi '} \\ F \ar[r] & F \times F } \]

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

\[ h_{W'} = h_ W \times _{h_ U \times h_ U} h_ U = (h_ U \times _{\xi , F, \xi '} h_ U) \times _{h_ U \times h_ U} h_ U = F \times _{F \times F} h_ U. \]

Thus $\Delta $ is representable and this finishes the proof.
$\square$

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