Kerodon

Proof. We construct the category $\operatorname{\mathcal{D}}'$ as follows:

• The objects of $\operatorname{\mathcal{D}}'$ are the objects of $\operatorname{\mathcal{C}}$. To avoid confusion, for each object $X \in \operatorname{\mathcal{C}}$, we write $\overline{X}$ for the corresponding object of $\operatorname{\mathcal{D}}'$.

• For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, we take $\operatorname{Hom}_{ \operatorname{\mathcal{D}}' }( \overline{X}, \overline{Y} )$ to be image of the map $F_{X,Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$. To avoid confusion, if $u: F(X) \rightarrow F(Y)$ is a morphism of $\operatorname{\mathcal{D}}$ which belongs to the image of $F_{X,Y}$, we write $\overline{u}: \overline{X} \rightarrow \overline{Y}$ for the corresponding morphism of $\operatorname{\mathcal{D}}'$.

• For every pair of objects $X,Y, Z \in \operatorname{\mathcal{C}}$, the composition law

  \[ \circ : \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( \overline{Y}, \overline{Z} ) \times \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( \overline{X}, \overline{Y} ) \rightarrow \operatorname{Hom}_{ \operatorname{\mathcal{D}}'}( \overline{X}, \overline{Z} ) \]

  is the restriction of the composition law $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(Y), F(Z) ) \times \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Z) )$ for the category $\operatorname{\mathcal{D}}$: that is, it satisfies the formula $\overline{v} \circ \overline{u} = \overline{v \circ u}$.

Let $F': \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}'$ be the functor which carries each object $X \in \operatorname{\mathcal{C}}$ to the object $\overline{X} \in \operatorname{\mathcal{D}}'$, and each morphism $u: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ to the morphism $\overline{ F(u) }: \overline{X} \rightarrow \overline{Y}$ of $\operatorname{\mathcal{D}}'$. Let $G: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ be the functor which carries each object $\overline{X} \in \operatorname{\mathcal{D}}'$ to the object $F(X) \in \operatorname{\mathcal{D}}$, and each morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ of $\operatorname{\mathcal{D}}'$ to the morphism $u: F(X) \rightarrow F(Y)$ of $\operatorname{\mathcal{D}}$. Then the functor $G$ is faithful, the functor $F'$ is full and essentially surjective, and the composition $G \circ F'$ is equal to $F$. $\square$