I am struggling with the following exercise from Emily Riehl's Category Theory in Context, regarding adjunction morphisms: paraphrasing, let $F : C \to D, G: D \to C$ and $F' : C' \to D', G' : D' \to C'$ be functors such that $F \dashv G$ and $F' \dashv G'$. We now consider functors $H : C \to C'$ and $K : D \to D'$ such that the squares of adjunctions commute, that is so that $HG = G'K$ and $KF = F'H$. The exercise then asks to prove that the following are equivalent:
- if $\eta, \eta'$ are the units of the adjunctions, then $H \eta = \eta' H$, i.e. $H\eta_c = \eta'_{Hc}$.
- if $\epsilon, \epsilon'$ are the counits of the adjunctions, then $K\epsilon = \epsilon'K$.
- the arrow compositions $$ D(Fc,d) \xrightarrow{\simeq} C(c,Gd) \xrightarrow{H} C(Hc,HGd) = C(Hc,G'Kd) $$ and $$ D(Fc,d) \xrightarrow{K} D(KFc,Kd) = D(F'Hc,Kd) \xrightarrow{\simeq} C(Hc, G'Kd) $$ are equal.
I've though about this for a fair amount of time, trying to use the adjunction relations between transposes/adjunts and the commutativity relations, with no luck: the only implication I have managed to prove is $(3) \Rightarrow (1)$, via tracking $1_{Fc}$ along both arrows, which give $H\eta_c$ and $\eta'_{Hc}$ respectively.
Any hints?
Following Derek Elkins' hint, I think I have managed to solve the problem: first off, by noting that the commutative diagram of $(3)$ is equivalent to the commutative diagram with the isomorphism legs reversed, from the technique used to solve $(1) \iff (3)$ one can similarly prove $(2) \iff (3)$.
Having said that, $(1) \iff (3)$ follows from the fact that the transpose of an arrow $f : Fc \to d$ can be defined (or characterized) as the composition $\eta_cGf$. Suppose that $(1)$ holds and let $f : Fc \to d$ be an arrow. Now, commutativity of the diagram amounts to showing $$ H(\eta_cGf) = \eta'_{Hc}G'Kf \tag{$\star$} $$ since the transpose of $Kf : F'Hc = KFc \to d$ is, via the same remark, $\eta'_{Hc}G'Kf$. In effect, $$ H(\eta_cGf) = H\eta_c HGf \stackrel{(1)}{=} \eta'_{Hc}HGf = \eta'_{Hc}G'Kf $$ as desired.