I have trouble finding the proof of MacLane's statement that "(6) implies (5)" at page 101 of his Categories for the Working Mathematician. This is part of the proof of his Theorem 2 in the "Transformations of Adjoints" chapter (7).


The definition of *conjugate* natural transformations goes:


Let be two adjunctions
$$ <F, G, \varphi, \eta, \epsilon> \text{ and } <F', G', \varphi', \eta', \epsilon'>$$
then the two natural transformations
$$\sigma : F \xrightarrow{.} F' \text{ and } \tau : G' \xrightarrow{.} G$$
are **conjugate** given the previous two adjunctions iff the following diagram commutes:


$\require{AMScd}$
\begin{CD}
A(F'x, a) @>\varphi'>\approx> X(x, G' a) \\
@V(\sigma\_x)^\*VV @VV(\tau\_a)\_\*V \\
A(Fx,a) @>\varphi>\approx> X(x,Ga)
\end{CD}


The previous diagram is labelled (5) in the book.


The statement I'm struggling to proove is that (5) is implied (allegedly without the need of further explanation) by the following two diagrams (labelled (6) as a pair):


\begin{CD}
G' @>\tau>> G \\
@VV\eta G'V @AA G\epsilon'A \\
G F G' @>G\sigma G'>> GF'G'
\end{CD}


and


\begin{CD}
F @>\tau>> F' \\
@VVF\eta'V @AA \epsilon F'A \\
FG'F' @>F\tau F' >> F G F'
\end{CD}


I see clearly the implication in the other direction, well explained by MacLane, as one only needs to add an arrow from the one-element set $\*$ to $X(x,G'a)$ that points to its identity element $id\_{G'a}$ and chase the diagram. However I don't see how we can, the other way around, generalize from the identity element to the full Homset.


Some compositions that arise from adjunctions would lead me to the point where I still need to show things like
$$ G \epsilon'\_a \circ G F' U \circ G \sigma\_{x} \circ \eta\_{x} = G\epsilon'\_a \circ G \sigma\_{G'a} \circ \eta\_{G'a} \circ U$$
which seems to me intractable, as the way to pass $U$ through from the right would be to use $\epsilon$ not $\eta$.



Characterization of conjugate natural transformations (MacLane)
Use the naturality of $\sigma$ first, and then the naturality of $\eta$ :
$$\begin{align}G \epsilon'\_a \circ G F'U \circ G \sigma\_{x} \circ \eta\_{x} & = G (\epsilon'\_a) \circ G \sigma\_{G'a}\circ GFU \circ \eta\_{x}\\ & = G\epsilon'\_a \circ G \sigma\_{G'a} \circ \eta\_{G'a} \circ U.\end{align}$$


