Lemma 4.33.8. Let $\mathcal{C}$ be a category. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$. Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent as categories over $\mathcal{C}$. Then $\mathcal{S}_1$ is fibred over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is fibred over $\mathcal{C}$.

**Proof.**
Denote $p_ i : \mathcal{S}_ i \to \mathcal{C}$ the given functors. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$, $G : \mathcal{S}_2 \to \mathcal{S}_1$ be functors over $\mathcal{C}$, and let $i : F \circ G \to \text{id}_{\mathcal{S}_2}$, $j : G \circ F \to \text{id}_{\mathcal{S}_1}$ be isomorphisms of functors over $\mathcal{C}$. We claim that in this case $F$ maps strongly cartesian morphisms to strongly cartesian morphisms. Namely, suppose that $\varphi : y \to x$ is strongly cartesian in $\mathcal{S}_1$. Set $f : V \to U$ equal to $p_1(\varphi )$. Suppose that $z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_2)$, with $W = p_2(z')$, and we are given $g : W \to V$ and $\psi ' : z' \to F(x)$ such that $p_2(\psi ') = f \circ g$. Then

is a morphism in $\mathcal{S}_1$ with $p_1(\psi ) = f \circ g$. Hence by assumption there exists a unique morphism $\xi : G(z') \to y$ lying over $g$ such that $\psi = \varphi \circ \xi $. This in turn gives a morphism

lying over $g$ with $\psi ' = F(\varphi ) \circ \xi '$. We omit the verification that $\xi '$ is unique. $\square$

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