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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 $\eta$.

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    $\begingroup$ What is $U$ here? $\endgroup$ – Arnaud D. Feb 25 '19 at 13:58
  • $\begingroup$ Thanks for your comment! Indeed it deserves a clarification: $U$ is an arbitrary member of the homset $X(x, G'a)$ that I substitute to $id_{G'a}$ to try to generalize. $\endgroup$ – Almeo Maus Feb 25 '19 at 14:10
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    $\begingroup$ Then your left-hand side should really by $G \epsilon'_a \circ G F' (U) \circ G \sigma_{x} \circ \eta_{x}$, no? $\endgroup$ – Arnaud D. Feb 25 '19 at 14:17
  • $\begingroup$ Ah you're right, thanks! :) Let me fix that. That said, that doesn't solve my problem, right? $\endgroup$ – Almeo Maus Feb 25 '19 at 14:25
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Use the naturality of $\sigma$ first, and then the naturality of $\eta$ : $$\begin{align}G \epsilon'_a \circ G F'U \circ G \sigma_{x} \circ \eta_{x} & = G (\epsilon'_a) \circ G \sigma_{G'a}\circ GFU \circ \eta_{x}\\ & = G\epsilon'_a \circ G \sigma_{G'a} \circ \eta_{G'a} \circ U.\end{align}$$

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  • $\begingroup$ Thank you so much!! This seems simple and yet I was stuck on that for so long!! $\endgroup$ – Almeo Maus Feb 25 '19 at 14:38
  • $\begingroup$ Maybe could you just fix the last $f$ of your answer into an $U$? This was the whole point of my attempt to edit your answer (and some tricks trying to workaround the rule that all edits must be at least 6 characters). $\endgroup$ – Almeo Maus Feb 28 '19 at 4:07
  • $\begingroup$ @AlmeoMaus Ah sorry, I rejected the edit because I thought the tricks were not necessary but I missed that $f$... I should have read the note more carefully ! It's done now. $\endgroup$ – Arnaud D. Feb 28 '19 at 8:42

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