The Stacks project

Proof. An element of $(h_ U)_ p$ is given by a triple $(V, y, f)$, where $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $y\in u(V)$ and $f \in h_ U(V) = \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U)$. Two such $(V, y, f)$, $(V', y', f')$ determine the same object if there exists a morphism $\phi : V \to V'$ such that $u(\phi )(y) = y'$ and $f' \circ \phi = f$, and in general you have to take chains of identities like this to get the correct equivalence relation. In any case, every $(V, y, f)$ is equivalent to the element $(U, u(f)(y), \text{id}_ U)$. If $\phi $ exists as above, then the triples $(V, y, f)$, $(V', y', f')$ determine the same triple $(U, u(f)(y), \text{id}_ U) = (U, u(f')(y'), \text{id}_ U)$. This proves that the map $u(U) \to (h_ U)_ p$, $x \mapsto \text{class of }(U, x, \text{id}_ U)$ is bijective. $\square$