# Kerodon

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Remark 9.10.6.3. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a functor of categories, let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$, and suppose we are given objects $\widetilde{X}, \widetilde{Y} \in \operatorname{Ob}(\operatorname{\mathcal{D}})$ with $U( \widetilde{X} ) = X$ and $U( \widetilde{Y} ) = Y$. Then:

• Suppose that $U$ is a fibration in sets. Then the function $f^{\ast }$ satisfies $f^{\ast }( \widetilde{Y} ) = \widetilde{X}$ if and only if there exists a morphism $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfying $U( \widetilde{f} ) = f$. If this condition is satisfied, then the morphism $\widetilde{f}$ is unique.

• Suppose that $U$ is an opfibration in sets. Then the function $f_!$ satisfies $f_{!}( \widetilde{X} ) = \widetilde{Y}$ if and only if there exists a morphism $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfying $U( \widetilde{f} ) = f$. If this condition is satisfied, then the morphism $\widetilde{f}$ is unique.