Lemma 4.33.11. Let \mathcal{C} be a category. Let U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}). If p : \mathcal{S} \to \mathcal{C} is a fibred category and p factors through p' : \mathcal{S} \to \mathcal{C}/U then p' : \mathcal{S} \to \mathcal{C}/U is a fibred category.
Proof. Suppose that \varphi : x' \to x is strongly cartesian with respect to p. We claim that \varphi is strongly cartesian with respect to p' also. Set g = p'(\varphi ), so that g : V'/U \to V/U for some morphisms f : V \to U and f' : V' \to U. Let z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}). Set p'(z) = (W \to U). To show that \varphi is strongly cartesian for p' we have to show
given by \psi ' \longmapsto (\varphi \circ \psi ', p'(\psi ')) is bijective. Suppose given an element (\psi , h) of the right hand side, then in particular g \circ h = p(\psi ), and by the condition that \varphi is strongly cartesian we get a unique morphism \psi ' : z \to x' with \psi = \varphi \circ \psi ' and p(\psi ') = h. OK, and now p'(\psi ') : W/U \to V/U is a morphism whose corresponding map W \to V is h, hence equal to h as a morphism in \mathcal{C}/U. Thus \psi ' is a unique morphism z \to x' which maps to the given pair (\psi , h). This proves the claim.
Finally, suppose given g : V'/U \to V/U and x with p'(x) = V/U. Since p : \mathcal{S} \to \mathcal{C} is a fibred category we see there exists a strongly cartesian morphism \varphi : x' \to x with p(\varphi ) = g. By the same argument as above it follows that p'(\varphi ) = g : V'/U \to V/U. And as seen above the morphism \varphi is strongly cartesian. Thus the conditions of Definition 4.33.5 are satisfied and we win. \square
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