When is a right-adjoint fully-faithful

Question

Let $F:\mathcal{C}\rightarrow \mathcal{D}$ be a functor and $G$ its right-adjoint. Let $S:=\{f\in \operatorname{Mor}\mathcal{C}:F(f)\in\mathcal{Iso}(\mathcal{D})\}$. I've read that $G$ is fully-faithful if and only if $\bar{F}:\mathcal{C}\lbrack S^{-1}\rbrack\rightarrow \mathcal{D}$ is an equivalence of categories. However, I can't find a proof (by myself or in the literature). Can someone give me a proof of this?

Answer 1

This is Proposition 1.3 in Gabriel and Zisman's Calculus of Fractions and Homotopy Theory.

Answer 2

Quite clearly $F$ sends arrows in $S$ to isomorphisms (that's the definition).

Now let $K: C\to E$ be any functor that also inverts arrows in $S$, and define $K^\# : D\to E$ by $K^\# := K\circ G$. Then we have a natural transformation $K\eta : K\to K\circ G\circ F$

Now $G$ being fully faithful implies that $\eta \in S$ (pointwise) so that $K\eta$ is actually a natural isomorphism. From this it follows that $[D,E]\to^{F^*} [C,E]$ is essentially surjective on the subcategory of $S$-inverting functors.

Now suppose you have a natural transformation $\theta : K\circ F\to L\circ F$ where $K,L: D\to E$ are functors. We wish to find $\delta: K\to L$ such that $\theta = \delta F$. Now let $x\in D$ and consider $\epsilon_x : FGx \to x$. Then if $\delta$ works, we have a naturality square :
$\require{AMScd} \begin{CD} KFGx @>{\delta_{FGx}}>> LFGx\\ @V{K(\epsilon_x)}VV @VV{L(\epsilon_x)}V\\ Kx @>>{\delta_x}> Lx \end{CD}$

which, by $\theta= \delta F$ is actually

$\require{AMScd} \begin{CD} KFGx @>{\theta_{Gx}}>> LFGx\\ @V{K(\epsilon_x)}VV @VV{L(\epsilon_x)}V\\ Kx @>>{\delta_x}> Lx \end{CD}$

Note that $G$ being fully faithful also implies that $\epsilon_x$ is an isomorphism, so that we get $\delta_x = L(\epsilon_x)\circ \theta_{Gx}\circ K(\epsilon_x^{-1})$, thus proving faithfulness of $F^*$, and giving us a guess as to how to prove fullness : starting from $\theta$, put $\delta := L\epsilon \circ \theta G \circ K\epsilon^{-1}$ which is clearly a natural transformation given its definition.

Next, use the naturality square above to show that $\delta FG = \theta G$. Next, look at $\eta_a: a\to GFa$ in $C$ and apply naturality of $\theta$ to it to get :

$\require{AMScd} \begin{CD} KFa @>{KF(\eta_a)}>> KFGFx\\ @V{\theta_a}VV @VV{\theta_{GFa}}V\\ LFa @>>{LF(\eta_a)}> LFGFa \end{CD}$

and naturality of $\delta F$ to get :

$\require{AMScd} \begin{CD} KFa @>{KF(\eta_a)}>> KFGFx\\ @V{\delta_{Fa}}VV @VV{\delta_{FGFa}}V\\ LFa @>>{LF(\eta_a)}> LFGFa \end{CD}$

The horizontal arrows are isomorphisms because $\eta_a\in S$ and the vertical arrows on the right are equal by what we observed above : it follows that the vertical arrows on the left are also equal : $\delta_{Fa} = \theta_a$, i.e. $\delta F= \theta$; thus proving that $F^*$ is full.

But that's the definition of $F:C\to D$ being a localization functor at $S$, so that (by definition) $D=C[S^{-1}]$ (where $=$ means something like "has the same universal property therefore is equivalent to any model of")

So the things we used about $G$ being fully faithful are that : i) $F(\eta_a)$ is an isomorphism for all $a$

ii) $\epsilon_x$ is an isomorphism for all $x$ These are standard properties of adjunctions where the right adjoint is fully faithful, if you don't know them you should prove them (the intuitive idea is that if $G$ is fully faithful then $D$ is a full "subcategory" of $C$, embedded with $G$; and $F$ just takes the best $D$-approximation : but if you're already in $D$, then obviously this best $D$-approximation is yourself)