Let $C,D$ be categories and $F:C\to D$ and $G:D\to C$ be adjoint functors. Then $F$ is fully faithful iff the unit is an isomorphism?

Question

Let $C,D$ be categories and $F:C\to D,G:D\to C$ be such that $F$ is a left adjoint of $G$. Prove that $F$ is fully faithful iff the unit is an isomorphism.

(This is an exercise from the book by T. Leinster)

I think I can do one direction:

$\impliedby$: If we let $\eta$,$e$ be unit/counit then as $\eta:\text{id}_C \implies GF$ is an isomorphism, it follows that the composite

$$ \text{Hom}(x,y)\to\text{Hom}(F(x),F(y))\to\text{Hom}(GF(x),GF(y)) $$

for all $x,y\in C$ is an isomorphism, which implies that $F$ is faithful. But $F$ must be full since the composite

$$ G=G\circ \text{id}_D \implies GFG \implies G $$

is an identity transformation (I'm rather skeptical about this).

But I don't have any idea regarding reverse direction, in which I have to show that the arrow $x\to GF(x)$ has an inverse for all $x\in C$.

Answer 1

$F$ is fully faithful iff the map $\text{Hom}(x, y) \to \text{Hom}(Fx, Fy)$ is an isomorphism. By adjunction, the map $\text{Hom}(Fx, Fy) \to \text{Hom}(x, GFy)$ is always an isomorphism. Now use the Yoneda lemma; this proves both directions simultaneously.

Answer 2

It's actually possible to prove something a bit more general :

1. $F$ is faithful if and only if every component of $\eta $ is a monomorphism.
2. $F$ is full if and only if every component of $\eta$ is a split epimorphism.

The first statement holds because for every objects $X,Y$ and every arrows $u,v:X\to Y$ of $C$, $$\eta_Y\circ u=\eta_Y\circ v\Longleftrightarrow F(u)=F(v)$$ (because the two equalities correspond to one another through the natural bijection $\operatorname{Hom}_C(X,GFY)\simeq \operatorname{Hom}_D(FX,FY)$).

For the second statement, first assume that every $\eta_X$ is a split epimorphism, with some section $s_X$. Take an arrow $g:FX\to FY$, then $f=s_Y\circ G(g)\circ \eta_X$ is an arrow $X\to Y$, and $$GF(f)\circ \eta_X=\eta_Y\circ f = G(g)\circ \eta_X,$$which shows that $F(f)=g$. Assume now that $F$ is full; then for every $X$ there must be some arrow $s_X:GFX\to X$ such that $F(s_X)=\epsilon_{FX}:FGFX\to FX$. Now $$\epsilon_{FX} \circ F(\eta_X\circ s_X)=\epsilon_{FX} \circ F(\eta_X)\circ \epsilon_{FX}=\epsilon_{FX}=\epsilon_{FX}\circ F(id_{GFX}),$$hence $\eta_X\circ s_X=id_{GFX}$.