Let $\mathcal{C}$ and $\mathcal{D}$ be two categories. Suppose that

  • $F\colon\mathcal{C}\to\mathcal{D}$ is a functor from $\mathcal{C}$ to $\mathcal{D}$;
  • $U\colon\mathcal{D}\to\mathcal{C}$ is a functor from $\mathcal{D}$ to $\mathcal{C}$;
  • $F$ is left adjoint to $U$; and
  • $\eta\colon\mathrm{id}_\mathcal{D}\to UF$ is the natural isomorphism forming the unit of the adjunction $(F, U, \eta)$.

Let $\epsilon$ be the inverse of $\eta$. My question is:Is $\epsilon$ the counit of the adjunction $(F, U, \eta)$ (which should now be a natural equivalence)?

  • 4
    $\begingroup$ A unit of an adjoint pair $(F,U)$ is a natural transformation $\eta\colon\text{id}_{\mathcal{C}}\to UF$; a counit of an adjoint pair $(F,U)$ is a natural transformation $\epsilon\colon FU\to\text{id}_{\mathcal{D}}$. So, in general, the inverse of unit cannot be a counit of the same adjoint pair. Btw, a unit of an adjoint pair $(F,U)$ is a natural isomorphism if and only if $F$ is fully faithful (not an equivalence, in general). $\endgroup$ – Oskar Dec 28 '18 at 17:55
  • 2
    $\begingroup$ If the unit of an adjunction is a natural isomorphism, then the same need not be true for the counit. For example, let $U$ be the forgetful functor $\operatorname{\underline{Top}} \to \operatorname{\underline{Set}}$, and let $F$ be the indiscrete topology functor. Then $\eta \colon \operatorname{id}_{\operatorname{\underline{Set}}} \to UF$ is an isomorphism but $\varepsilon \colon FU \to \operatorname{id}_{\operatorname{\underline{Top}}}$ is not. $\endgroup$ – R. van Dobben de Bruyn Dec 28 '18 at 17:56
  • 2
    $\begingroup$ In fact there are examples when $F$ is left adjoint to $U$, at the same time $U$ is left adjoint to $F$, and then counit of the second adjunction may be inverse for the unit of the first adjunction. Google for bireflectivity. On the other hand all this might happen but the second counit might be not inverse for the first unit, even if both are isomorphisms. $\endgroup$ – მამუკა ჯიბლაძე Dec 28 '18 at 18:03

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