Lemma 96.5.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent
$f$ is representable, and
for every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ the functor $\mathcal{X}^{opp} \to \textit{Sets}$, $x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(f(x), y)$ is representable.
Proof. According to the discussion in Algebraic Stacks, Section 94.6 we see that $f$ is representable if and only if for every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ lying over $U$ the $2$-fibre product $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ is representable, i.e., of the form $(\mathit{Sch}/V_ y)_{fppf}$ for some scheme $V_ y$ over $U$. Objects in this $2$-fibre products are triples $(h : V \to U, x, \alpha : f(x) \to h^*y)$ where $\alpha $ lies over $\text{id}_ V$. Dropping the $h$ from the notation we see that this is equivalent to the data of an object $x$ of $\mathcal{X}$ and a morphism $f(x) \to y$. Hence the $2$-fibre product is representable by $V_ y$ and $f(x_ y) \to y$ where $x_ y$ is an object of $\mathcal{X}$ over $V_ y$ if and only if the functor in (2) is representable by $x_ y$ with universal object a map $f(x_ y) \to y$. $\square$
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