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Let $\mathcal{C}$ and $\mathcal{D}$ be two categories and $F$ and $G: \mathcal{C}\to \mathcal{D}$ be two functors. Suppose $F$ and $G$ have right adjoints $F^{\wedge}$ and $G^{\wedge}: \mathcal{D}\to \mathcal{C}$.

Now let $T:F\Rightarrow G$ be a natural transformation.

My question is: does $T$ induce a natural transformation $T^{\wedge}:G^{\wedge}\Rightarrow F^{\wedge}$?

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    $\begingroup$ Yes. Let $\eta: 1 \to F^\wedge F$ denote the unit and $\varepsilon: G G^\wedge \to 1$ the counit. Then form the evident composite $$G^\wedge \stackrel{\eta G^\wedge}{\to} F^\wedge F G^\wedge \stackrel{F^\wedge TG^\wedge}{\to} F^\wedge G G^\wedge \stackrel{F^\wedge \varepsilon}{\to} F^\wedge$$ $\endgroup$
    – Todd Trimble
    Commented Nov 19, 2020 at 5:43
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    $\begingroup$ The evident composite in Todd's comment obviously agrees with the transformation you would get by using the adjunction : $F\to G$ gives $id \to F^\wedge G$ which gives $G^\wedge \to F^\wedge$, as precomposition by $G$ becomes right adjoint to precomposition by $G^\wedge$ (although of course the easiest way to prove this is with the unit and co-unit, so in the end we really get down to Todd's comment) $\endgroup$
    – Maxime Ramzi
    Commented Nov 19, 2020 at 10:56
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    $\begingroup$ I seem to recall that the natural transformation between the adjoints is called its mate. $\endgroup$
    – Paul Taylor
    Commented Nov 19, 2020 at 20:29
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    $\begingroup$ What Paul says is correct, and it resonates most convincingly when you hear an Australian category theorist say it. (I tend to be somewhat lax [haha] in my usage, where any morphism you derive from an adjunction in similar fashion could also be called a "mate". For example, in informal chat, I might also refer to Maxime's $id \to F^\wedge G$ as a mate.) $\endgroup$
    – Todd Trimble
    Commented Nov 19, 2020 at 22:26
  • $\begingroup$ Some non-Australians refer to such a transformation as a "conjugate". The general definition of mate involves a square of four functors, two with adjoints, and taking a couple to be identities in different ways we obtain both your $G^\wedge \Rightarrow F^\wedge$, Todd's $\mathrm{Id} \to F^\wedge G$, and others. $\endgroup$
    – Mike Shulman
    Commented Jan 17, 2021 at 4:39

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I should have made my comment an answer earlier. The answer is yes:

Let $\eta: 1 \to F^\wedge F$ denote the unit of the adjunction and $\varepsilon: G G^\wedge \to 1$ the counit. Then form the composite

$$G^\wedge \stackrel{\eta G^\wedge}{\to} F^\wedge F G^\wedge \stackrel{F^\wedge TG^\wedge}{\to} F^\wedge G G^\wedge \stackrel{F^\wedge \varepsilon}{\to} F^\wedge.$$

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