# When is a right-adjoint fully-faithful

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?

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

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)