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Does anyone have any references on the following type of thing, which one might call "adjunction up to autoequivalences"?

We have functors $F \colon C \to D$ and $F' \colon D \to C$, but they are not necessarily adjoint (and they don't necessarily possess any adjoints, either). Instead, there is an autoequivalence $G_C$ of $C$ such that we have an adjunction $G_C F' \dashv F G_C$. Or $G_C F' \dashv F$. Or maybe there is another autoequivalence $G_D$ of $D$ such that $G_C F' \dashv G_D F$, or maybe ... [insert variations on the theme]

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  • $\begingroup$ Oh, from the related questions I learned about "lax 2-adjunctions". Is this maybe a specific case of such a thing? $\endgroup$
    – Jo Mo
    Commented Feb 1, 2022 at 7:53
  • $\begingroup$ No, lax 2-adjunctions is not the right place. Instead, for the simple case with $G_C F' \dashv F$, it is true that F is a $G_C^{-1}$ -relative right adjoint of $F'$. Similarly the adjunction $F' \dashv G_D F$ means that $F'$ is a $G_D^{-1}$-relative left adjoint of $F$. But the situation $F' \dashv F G_C$ does not seem to fit into that scheme. Also, I don't know why I only considered equivalences and not arbitrary endofunctors. I think because of some example. $\endgroup$
    – Jo Mo
    Commented Feb 1, 2022 at 8:54
  • $\begingroup$ This question is pretty similar to the recent one on non-adjoint equivalences. It's also rather confused. Sorting out the confusion and answering the question are grad-student exercises. $\endgroup$
    – Paul Taylor
    Commented Feb 1, 2022 at 10:19
  • $\begingroup$ @PaulTaylor Uh. This question is not related to (non-)adjoint equivalences at all, and surely not similar to the linked question. Would you care to elaborate why you think it's confused? And calling a "reference request" a grad-student exercise is imo pretty unfair... $\endgroup$
    – Jo Mo
    Commented Feb 1, 2022 at 10:29
  • $\begingroup$ @varkor: I think you meant $F' \dashv F$ in my notation. In any case, it does not seem possible that from the hom-set adjunction $C(GF' X, Y) = D(X, FY)$ one can construct an iso $C(F'X, Y) = D(X, FY)$ or $D(FX, Y) = C(X, F'Y)$. The proof would actually work in any (rigid) monoidal category, if given via unit-counit, and then it would mean that $X \otimes i = Y^*$ implies $X = Y^*$ for any invertible object $i$, and that seems quite wrong. Note that I am not asking the easy question: "Are duals in a monoidal category, if they exist, unique?" $\endgroup$
    – Jo Mo
    Commented Feb 1, 2022 at 13:17

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