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).

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.

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:

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  • $\begingroup$ To clarify, when he says "natural isomorphism," he doesn't mean that there is a natural transformation lurking there. Rather, "natural" in this context implies that the correspondence is functorial in both variables, in some sense. $\endgroup$ – Matt Feller Sep 17 '19 at 23:40

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.

  • $\begingroup$ I'm looking at naturality condition (2.3) which I added to the question. As far as I can see, in our case it is $\overline{G(g)\circ f}=\overline{G(g)}\circ F(f)$. This doesn't seem to agree with what you have. Did you apply the same equation (2.3)? $\endgroup$ – user634426 Sep 18 '19 at 0:21
  • $\begingroup$ You should use (2.2) to get $\overline{g\circ\bar f}=G(g)\circ f$, and then apply the bar again to get $g\circ\bar f=\overline{G(g)\circ f}$. Sorry for the notation conflict, but I didn‘t have Leinster‘s book to hand. $\endgroup$ – Andrew Hubery Sep 18 '19 at 6:18
  • $\begingroup$ By the way, the same reasoning also gives that $G$ is faithful if and only if the counit is always epi, and there are analogous results for $F$ and the unit. $\endgroup$ – Andrew Hubery Sep 18 '19 at 6:23
  • $\begingroup$ In the last paragraph, do you mean isomorphism in $\textbf{Set}$? That is, is this the right statement: "$G$ is fully faithfull iff the map of sets $g\mapsto g\epsilon_B$ is a bijection"? And then I guess you're using that $\epsilon$ is known to be a natural transformation, so it's a natural isomorphism iff each $\epsilon_B$ is an isomorphism in $\mathscr B$; and the latter happens iff $g\mapsto g\epsilon_B$ is a bijection. $\endgroup$ – user634426 Sep 18 '19 at 18:16
  • 1
    $\begingroup$ You’re right, this deserved a little more explanation. Assume $G$ is fully faithful. Taking $B’$ to be $FGB$ shows that $\epsilon_B$ has a left inverse $g$. Then both $\mathrm{id}_B$ and $\epsilon_Bg$ are sent to $\epsilon_B$, and since it’s a bijection we deduce that $g$ is also a right inverse. $\endgroup$ – Andrew Hubery Oct 7 '19 at 21:18

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.


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