Lemma 8.12.1. In the situation above, if $\mathcal{S}$ is a fibred category over $\mathcal{D}$ then $u^ p\mathcal{S}$ is a fibred category over $\mathcal{C}$.

Proof. Please take a look at the discussion surrounding Categories, Definitions 4.33.1 and 4.33.5 before reading this proof. Let $(a, \beta ) : (U, y) \to (U', y')$ be a morphism of $u^ p\mathcal{S}$. We claim that $(a, \beta )$ is strongly cartesian if and only if $\beta$ is strongly cartesian. First, assume $\beta$ is strongly cartesian. Consider any second morphism $(a_1, \beta _1) : (U_1, y_1) \to (U', y')$ of $u^ p\mathcal{S}$. Then

\begin{align*} & \mathop{Mor}\nolimits _{u^ p\mathcal{S}}((U_1, y_1), (U, y)) \\ & = \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U))} \mathop{Mor}\nolimits _\mathcal {S}(y_1, y) \\ & = \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U))} \mathop{Mor}\nolimits _\mathcal {S}(y_1, y') \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U'))} \mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U)) \\ & = \mathop{Mor}\nolimits _\mathcal {S}(y_1, y') \times _{\mathop{Mor}\nolimits _\mathcal {D}(u(U_1), u(U'))} \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \\ & = \mathop{Mor}\nolimits _{u^ p\mathcal{S}}((U_1, y_1), (U', y')) \times _{\mathop{Mor}\nolimits _\mathcal {C}(U_1, U')} \mathop{Mor}\nolimits _\mathcal {C}(U_1, U) \end{align*}

the second equality as $\beta$ is strongly cartesian. Hence we see that indeed $(a, \beta )$ is strongly cartesian. Conversely, suppose that $(a, \beta )$ is strongly cartesian. Choose a strongly cartesian morphism $\beta ' : y'' \to y'$ in $\mathcal{S}$ with $p(\beta ') = u(a)$. Then bot $(a, \beta ) : (U, y) \to (U, y')$ and $(a, \beta ') : (U, y'') \to (U, y)$ are strongly cartesian and lift $a$. Hence, by the uniqueness of strongly cartesian morphisms (see discussion in Categories, Section 4.33) there exists an isomorphism $\iota : y \to y''$ in $\mathcal{S}_{u(U)}$ such that $\beta = \beta ' \circ \iota$, which implies that $\beta$ is strongly cartesian in $\mathcal{S}$ by Categories, Lemma 4.33.2.

Finally, we have to show that given $(U', y')$ and $U \to U'$ we can find a strongly cartesian morphism $(U, y) \to (U', y')$ in $u^ p\mathcal{S}$ lifting the morphism $U \to U'$. This follows from the above as by assumption we can find a strongly cartesian morphism $y \to y'$ lifting the morphism $u(U) \to u(U')$. $\square$

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