# Kerodon

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Construction 9.10.6.1 (Contravariant Transport). Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets. Using Example 4.2.3.4 and Remark 4.2.3.6, we see that for each object $X \in \operatorname{\mathcal{C}}$, the fiber $\operatorname{\mathcal{D}}_{X} = \{ X\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ is a discrete category.

Let $f: X \rightarrow Y$ be a morphism in the category $\operatorname{\mathcal{C}}$. For each object $\widetilde{Y} \in \operatorname{\mathcal{D}}_{Y}$, our assumption that $U$ is a fibration in sets guarantees that there exists a unique pair $(\widetilde{X}, \widetilde{f} )$, where $\widetilde{X}$ is an object of the fiber $\operatorname{\mathcal{D}}_{X}$ and $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfies $U( \widetilde{f} ) = f$. Note that the object $\widetilde{X}$ depends only on $f$ and $\widetilde{Y}$. To emphasize this dependence, we will denote $\widetilde{X}$ by $f^{\ast }( \widetilde{Y} )$. The construction $\widetilde{Y} \mapsto f^{\ast }( \widetilde{Y} )$ then determines a function $f^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$.