Kerodon

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Example 8.4.0.28. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories, and let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{E}}$. Assume that $U(f)$ is an isomorphism in $\operatorname{\mathcal{C}}$. Then the following conditions are equivalent:

  • The morphism $f$ is $U$-cartesian.

  • The morphism $f$ is $U$-cocartesian.

  • The morphism $f$ is an isomorphism in $\operatorname{\mathcal{E}}$.

In particular, every isomorphism in $\operatorname{\mathcal{E}}$ is both $U$-cartesian and $U$-cocartesian.