Given an adjunction $F\colon \mathcal C \leftrightarrows \mathcal D\colon G$ I would like to call "sharp" a pair of objects $(C,D)$ if the bijection $$ \hom_{\cal D}(FC,D) \cong \hom_{\cal C}(C, GD) $$ restricts to a bijection $\text{Iso}_{\cal D}(FC,D)\cong \text{Iso}_{\cal C}(C, GD)$, i.e. $f : FC\to D$ is an isomorphism in $\cal D$ if and only if its mate $Gf\circ \eta_C\colon C\to GD$ is an isomorphism in $\cal C$; as a consequence, there are as many isomorphisms $FC\to D$ as there are $C\to GD$.

It's easy to see that this notion is linked to that of a fixed point for the adjunction $F\dashv G$, or more precisely to that of a fixed point for the monad $GF$ and the coomnad $FG$; more precisely, it's immediate to see that if $(C,D)$ is sharp then $GFC\cong C$ and $FGD\cong D$ via unit and counit of the adjunction.

I suspect sharp pairs are strictly contained in $Fix(GF)\times Fix(FG)$; is it true?

Is there a way to characterize sharp objects of a generic adjunction? Obviously, equivalences are all and the only adjunctions for which all objects are sharp.

Isomorphisms have nothing special: take any 3-for-2 class $\cal W$ of morphisms (e.g., weak equivalences somewhere) and consider the notion of $\cal W$-sharpness: if $(C,D)$ is $\cal W$-sharp, then $\eta_C,\epsilon_D\in \cal W$. Now take your favourite $\Omega$-spectrum $E_*$, endowed with homotopy equivalences $E_{n-1}\to \Omega E_n$; I wonder if "sharp" $\Omega$-spectra, for which also the mates $\Sigma E_{n-1}\to E_n$ are homotopy equivalences, have interesting topological properties among all spectra.

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    $\begingroup$ Regarding your last musing about spectra. You are asking to consider spectra which are both suspension spectra and Omega-spectra at the same time. I believe the zero spectrum is the unique example. $\endgroup$ – Chris Schommer-Pries Aug 15 '14 at 11:19
  • $\begingroup$ According to your definition of "sharp", isn't it possible for $(C,D)$ to be a sharp pair but for no isomorphisms $FC\to D$ or $C\to GD$ to exist at all? Then there isn't any reason for $C$ and $D$ to be fixed points. $\endgroup$ – Owen Biesel Aug 15 '14 at 12:03
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    $\begingroup$ It looks like Owen is right, and that you meant something slightly different. Take the free-forgetful adjunction between rings and sets; let $C$ be the empty set and let $D$ be the terminal ring. Then $Iso(FC, D)$ and $Iso(C, GD)$ are both empty. But neither $C$ nor $D$ is a fixed point. $\endgroup$ – Todd Trimble Aug 15 '14 at 12:20
  • $\begingroup$ Yeah, good point. I must rethink my original idea; sharpness seems a property of the adjunction, not of the objects $\endgroup$ – Fosco Loregian Aug 15 '14 at 12:23

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