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I would like to apologize for this rather stupid abstract nonsense question.

Let $h=f\circ g$ for composable functors $f,g$; assume that there exist left or right adjoints to $f$ and $g$. Then it seems natural to say that the composition of adjunction transformations for $f$ and $g$ equals the one for $h$. Yet I am not quite sure that one can speak about a "strict" equality here since adjunction transformations (as well as adjoint functors) are only defined up to a (canonical) equivalence.

Now, suppose that $h$ is only isomorphic to $f\circ g$ (this is my situation; I would not like to enforce an equality here since I actually treat a lax commutative square of functors). Then the adjunction transformation for $h$ remains to be equivalent to the composition mentioned. Yet I want to "make" this equivalence a strict equality. What ("words") should I write to indicate that I make compatible choices in the corresponding equivalence classes of transformations?

Does there exist a (classical?) reference where questions of this sort are treated carefully?

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    $\begingroup$ What do you mean by “adjunction transformations”? I’m not familiar with that terminology (and neither is Google, apparently). $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Mar 31, 2015 at 7:30
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    $\begingroup$ Do you prefer "unit/counit of the adjunction"? $\endgroup$
    – Mikhail Bondarko
    Commented Mar 31, 2015 at 11:52
  • $\begingroup$ I don’t know, because I still don’t know what you mean by “adjunction transformation”! I’m guessing that you mean or the natural isomorphism $\mathbb{C}(fx,y) \cong \mathbb{D}(x,gy)$ (often called transposition). If you mean this, or the unit/counit, then they are determined up to strict equality; i.e. if $f \vdash g$ is an adjunction, with unit $\eta$ (or counit $\epsilon$, or transposition $\varphi$), and $g'$ is another functor with $\alpha:g'\cong g$, then there are unique $\eta'$ ($\epsilon'$, $\varphi'$) making $g'$ an adjoint to $f$, and commuting with $\alpha$. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Mar 31, 2015 at 16:19
  • $\begingroup$ @Peter is $f \vdash g$ the same as $g \dashv f$, meaning that $g$ is left adjoint to $f$? $\endgroup$
    – Tim Campion
    Commented Mar 31, 2015 at 16:29
  • $\begingroup$ Yes; if we fix $f$ and its adjoint then the unit and the counit will be fixed also. Yet I am interested in the case when the adjoint to $f$ is not fixed. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 31, 2015 at 20:01

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