Lemma 4.2.18. Let $F : \mathcal{A} \to \mathcal{B}$ be a fully faithful functor. Suppose for every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ we are given an object $j(X)$ of $\mathcal{A}$ and an isomorphism $i_ X : X \to F(j(X))$. Then there is a unique functor $j : \mathcal{B} \to \mathcal{A}$ such that $j$ extends the rule on objects, and the isomorphisms $i_ X$ define an isomorphism of functors $\text{id}_\mathcal {B} \to F \circ j$. Moreover, $j$ and $F$ are quasi-inverse equivalences of categories.

**Proof.**
To construct $j : \mathcal{B} \to \mathcal{A}$, there are two steps. Firstly, we define the map $j : \mathop{\mathrm{Ob}}\nolimits (\mathcal{B}) \to \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ that associates $j(X)$ to $X \in \mathcal{B}$. Secondly, if $X,Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and $\phi : X \to Y$, we consider $\phi ' := i_ Y \circ \phi \circ i_ X^{-1}$. There is an unique $\varphi $ verifying $F(\varphi ) = \phi '$, using that $F$ is fully faithful. We define $j(\phi ) = \varphi $. We omit the verification that $j$ is a functor. By construction the diagram

commutes. Hence, as each $i_ X$ is an isomorphism, $\{ i_ X\} _ X$ is an isomorphism of functors $\text{id}_\mathcal {B} \to F\circ j$. To conclude, we have to also prove that $j \circ F$ is isomorphic to $\text{id}_\mathcal {A}$. However, since $F$ is fully faithful, in order to do this it suffices to prove this after post-composing with $F$, i.e., it suffices to show that $F \circ j \circ F$ is isomorphic to $F \circ \text{id}_\mathcal {A}$ (small detail omitted). Since $F \circ j \cong \text{id}_\mathcal {B}$ this is clear. $\square$

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