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Suppose we have two functors $F:C\leftrightarrow D:G$ and a morphism $\varepsilon:FG\rightarrow\operatorname{Id}_D$. I am looking for a way to check whether $\varepsilon$ is the counit of an adjunction. Mostly I am interested in a way to show that $\varepsilon$ is not a counit, i.e. necessary conditions.

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    $\begingroup$ A tautological condition is that the morphism $Hom(A,GB) \stackrel{F}\to Hom(FA,FGB) \stackrel{\varepsilon}\to Hom(FA,B)$ is an isomorphism for all $A \in C$, $B \in D$. $\endgroup$
    – Sasha
    Commented Jan 27, 2014 at 23:28
  • $\begingroup$ There are conditions on $F, G$, of course, e.g. that $F$ preserves colimits and $G$ preserves limits. Are you in a situation where you suspect $F, G$ may be part of an adjunction but you aren't sure what the counit should be? $\endgroup$ Commented Jan 28, 2014 at 1:17
  • $\begingroup$ You meant unit? Yes, I have a candidate for a co-unit but no explicit unit, and I need to show that sometimes my candidate cannot be a co-unit $\endgroup$
    – Adam Gal
    Commented Jan 28, 2014 at 9:15

1 Answer 1

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A morphism $\varepsilon\colon FG\to Id_D$ is the counit of adjunction $F\dashv G$ iff for every $d\in D$ the morphism $\varepsilon_d\colon FGd\to d$ is universal from $F$ to $d$(i.e. is the terminal object of the comma category $(F\downarrow d)$). So we have a family of necessary conditions, parametrized by objects of $D$.

Also, there are interesting connections between $G$ and $\varepsilon$(if $\varepsilon$ is the counit):

  • $G$ is faithful iff for any $d\in D$ the morphism $\varepsilon_d$ is an epimorphism;
  • $G$ is full iff for any $d\in D$ the morphism $\varepsilon_d$ is a split monomorphism.

So, for example, if $G$ is full, and for some $d_0\in D$ the morphism $\varepsilon_{d_0}\colon FGd_0\to d_0$ does not have a left inverse, then $\varepsilon$ is not a counit.

Finally, another interesting property you may find useful:

  • If one of the functors $F$, $G$ is full, then the natural transformation $G\varepsilon$ is invertible.

References:

Maclane, CFWM, chapter IV.

Edit: As you said, your $\varepsilon$ is invertible. Assuming $\varepsilon$ is the counit, you can also define the unit of the corresponding adjunction on the image of $G$ on objects(it follows from the triangular identities): $$ \eta_{G(d)}=(G(\varepsilon_d))^{-1} $$ Maybe it will help you to find the whole unit morphism. For example, if $G$ is surjective on objects, you don't need to check anything(in this case $(F,G)$ is an equivalence).

Also you can use some additional data about your categories to find the adjunction. For instance, if $D$ is locally small, complete, well-powered, has a small cogenerating family, and $G$ is continuous, then you can find the left adjoint by the Special Adjoint Functor Theorem. Then you can try to prove that $F$ is naturally isomorphic to this left adjoint etc.

References:

Borceux, "Handbook of categorical algebra", vol.1.

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  • $\begingroup$ I looked at my morphism, and sadly this doesn't help me since it is given as an inverse, so is invertible... $\endgroup$
    – Adam Gal
    Commented Jan 28, 2014 at 11:49
  • $\begingroup$ @AdamGal Is your $G$ fully faithful? Also, have you checked, that every $\varepsilon_d$ is a terminal object in $(F\downarrow d)$(it was the main part of my answer)? Anyway, this is a vague question as long as you don't mention your concrete $F$, $G$ and $\varepsilon$. $\endgroup$
    – Oskar
    Commented Jan 28, 2014 at 12:56
  • $\begingroup$ I am trying to check if it is terminal, but this is not so straitforward - amounts to showing that some map doesn't exist or is not unique, so there is a lot of guesswork. I am waiting to see if maybe someone knows something more. $\endgroup$
    – Adam Gal
    Commented Jan 28, 2014 at 14:00
  • $\begingroup$ @AdamGal See also edit. $\endgroup$
    – Oskar
    Commented Jan 28, 2014 at 14:55

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