Right adjoint is fully faithful iff the counit is an isomorphism (without Yoneda)

Question

Exercise 2.12(a) from Leinster:

Show that for any adjunction, the right adjoint is full and faithful if and only if the counit is an isomorphism.

Note: The exercise is given before the chapter on representables and the Yoneda lemma, so I wouldn't like to use those. There is a similar question which partly covers what I'm asking, but the answer uses the Yoneda lemma and other stuff from Chapter 4 (whereas this exercise is from Chapter 2).

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Suppose $F:\mathscr A\to \mathscr B,\ G:\mathscr B\to\mathscr A$ are functors and $F\dashv G$. So there is a bijection $$\mathscr B(F(A),B)\cong\mathscr A(A,G(B))$$ denoted by $f\mapsto \bar f$ in either direction. The counit of adjunction is the natural transformation $$\epsilon: FG\to 1_\mathscr B$$ whose component at $B\in\mathscr B$ is $$\epsilon_B=\overline{1_{G(B)}}:FG(B)\to B.$$

The question asks to prove that $G$ is full and faithful iff $\epsilon$ is a natural isomorphism. The latter happens iff $\epsilon_B$ is an isomorphism in $\mathscr B$ for all $B\in\mathscr B$. So it suffices to show that $G$ is full and faithful iff each $\epsilon_B$ is an isomorphism.

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I really don't see how to prove either direction. For example, suppose $G$ is fully faithful. Then there is a bijection $$\mathscr B(B,B')\cong \mathscr A(G(B),G(B'))$$ for all $B,B'\in\mathscr B$. Thus there is also a sequence of bijections

$$\mathscr B(B,B')\cong \mathscr A(G(B),G(B'))\cong \mathscr B(FG(B),B')$$ (the cited question calls these bijections natural isomorphisms, but at this point Leinster doesn't even interpret the adjunction bijection (the second bijection above) as a natural transformation (he only gives some "naturality conditions" which at this point are not interpreted as a natural isomorphism; and I also treat the first $\cong$ above as a mere bijection).

So every arrow $g:B\to B'$ corresponds in a unique way to an arrow $FG(B)\to B'$. But (1) I don't know any explicit formula for this correspondence (the first $\cong$ is just applying $G$; the second is taking bar, but there is no explicit definition of a bar in Leinter's text), and (2) even if I knew the explicit correspondence law, I don't see how it would help me.

The other direction is also not clear.

Addition: the naturality conditions in Leinster's notation:

Answer 1

This result uses the naturalness of the map $$ \Phi \colon \mathrm{Hom}(A,GB) \xrightarrow\sim \mathrm{Hom}(FA,B). $$ In particular, if $f\colon A\to GB$ and $g\colon B\to B’$, then $\Phi(G(g)f)=g\Phi(f)$. Since $\varepsilon_B=\Phi(\mathrm{id}_{GB})$, we see that $\Phi(G(g))=g\varepsilon_B$.

Thus the map you are interested in $$ \mathrm{Hom}(B,B') \to \mathrm{Hom}(GB,GB') \xrightarrow\sim \mathrm{Hom}(FGB,B') $$ coming from applying $G$ and then using adjointness is given explicitly as $g\mapsto g\varepsilon_B$.

The result is now clear: $G$ is fully faithful if and only if this composition is always an isomorphism, which is if and only if $\varepsilon$ is a natural isomorphism.

Answer 2

I think we can do this from scratch as follows:

first, prove a lemma: for any arrows $x, y : b'\to b$ in $\mathscr B$, we have $x\circ \epsilon_b = y\circ \epsilon_{b'} \Leftrightarrow Gx = Gy:$

Since $\epsilon$ is a natural transformation, we have $x\circ \epsilon_{b'}=\epsilon_{b}\circ FGx$ and similarly for $y$. Then, $x\circ \epsilon_b = y\circ \epsilon_b\Leftrightarrow \epsilon_{b}\circ FGx=\epsilon_{b}\circ FGy\Leftrightarrow \overline {Gx}=\overline {Gy}\Leftrightarrow Gx=Gy$, the last equality true because $^-$ is a bijection.

Now, if $\epsilon$ is an isomorphism then in particular, it is epic, so the lemma says $x=y\Leftrightarrow Gx=Gy$; that is, $G$ is faithful. On the other hand, if $G$ is faithful, then the lemma says that $\epsilon$ is epic.

Continuing, if $G$ is full, then then there is an arrow $x : b\to FGb$ such that $Gx = \eta_{Gb}.$ An application of a triangular identity (which one?) and the naturality of $\epsilon$ show that $1_{FGb} =x\circ \epsilon_b$ so $\epsilon$ is a split monic.

We have now that if $G$ is fully faithful then $\epsilon$ is an isomorphism.

Finally, let $f:Gb'\to Gb$ and suppose that $\epsilon_b'$ is a split monic. Then, there is a morphism $x:b'\to FGb'$ such that $x\circ \epsilon_{b'}=1_{FGb'}$. Then, $\overline {Gx}=1_{FGb'}=\overline {\eta_{Gb'}}$ so $Gx= \eta_{Gb'}$ and another application of the triangular identities gives $G(\epsilon_b \circ F f\circ x)=f$; i.e. $G$ is full.