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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #7488 by Louis Carlin on
Comment #7634 by Stacks Project on
There are also: