Lemma 8.2.1. This actually does give a presheaf.

**Proof.**
Let $g : V'/U \to V/U$ be as above and similarly $g' : V''/U \to V'/U$ be morphisms in $\mathcal{C}/U$. So $f' = f \circ g$ and $f'' = f' \circ g' = f \circ g \circ g'$. Let $\phi \in \mathop{Mor}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y)$. Then we have

which is what we want, namely $\phi |_{V''} = (\phi |_{V'})|_{V''}$. The first equality holds because $\alpha _{g', g}$ is a transformation of functors, and hence

commutes. The second equality holds because of property (d) of a pseudo functor since $f' = f \circ g$ (see Categories, Definition 4.29.5). The last equality follows from the fact that $(g')^*$ is a functor. $\square$

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