# Right adjoint is fully faithful iff the counit is an isomorphism (without Yoneda) [duplicate]

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: • 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. – 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.

• 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)? – user634426 Sep 18 '19 at 0:21
• 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. – Andrew Hubery Sep 18 '19 at 6:18
• 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. – Andrew Hubery Sep 18 '19 at 6:23
• 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. – user634426 Sep 18 '19 at 18:16
• 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. – 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.