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.

3$\begingroup$ 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 $A \in C$, $B \in D$. $\endgroup$– SashaCommented 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? $\endgroup$– Qiaochu YuanCommented Jan 28, 2014 at 1:17

$\begingroup$ You meant unit? Yes, I have a candidate for a counit but no explicit unit, and I need to show that sometimes my candidate cannot be a counit $\endgroup$– Adam GalCommented Jan 28, 2014 at 9:15
1 Answer
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, wellpowered, 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.

$\begingroup$ I looked at my morphism, and sadly this doesn't help me since it is given as an inverse, so is invertible... $\endgroup$– Adam GalCommented 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 $\varepsilon$. $\endgroup$– OskarCommented 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 GalCommented Jan 28, 2014 at 14:00
