Let $\mathcal{D} \underset{G}{\overset{F}{\leftrightarrows}}\mathcal{C}$ with $F\dashv G$ and $G$ fully faithful. I want to show that this implies that the counit is an isomorphism. I was trying to do this in a similar way as in this answer to a similar question. I ran into some issues. This is what I tried:

$$\mathcal{D}(A,B)\cong \mathcal{C}(GA,GB)\cong\mathcal{D}(FGA,B)$$

(First natural iso from fully faithfulness of $G$ and second natural iso from adjunction). Now by a corollary of the Yoneda Lemma there is a natural isomorphism $i:A\overset{\cong}{\rightarrow}FGA$. I would like to show that this is precisely the component $\epsilon_A$ of the counit of the adjunction. A way to do it would be to find $\eta$ such that $\eta,i$ respect the triangle inequalities. It seems that a good candidate for $\eta_X$ would be the image of the identity under the natural isomorphism $\mathcal{D}(FX,FX)\rightarrow \mathcal{C}(X,GFX)$.

But I failed to check that the triangle identities hold, and I'm afraid I chose the unit or the counit wrongly.

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    $\begingroup$ The easiest way to show what you want would be to explicitly describe the natural isomorphism $\mathcal{D}(FA,Y)\cong\mathcal C(A,GY)$ in terms of units and counits. Similarly for the Yoneda lemma. Then the result is immediate. $\endgroup$ – Derek Elkins left SE Apr 7 '18 at 9:30

Expanding a little bit on Derek Elkins' comment above.$\newcommand{\D}{\mathcal D}$ $\newcommand{\C}{\mathcal C}$

Assume that $$\varphi \colon \D(F(X),A) \stackrel{\sim}{\longrightarrow}\C(X,G(A))$$ is the natural isomorphism of the adjunction.

The natural isomorphism you described, namely $$\D(A,B) \stackrel{G}{\longrightarrow}\C(G(A),G(B)) \stackrel{\varphi^{-1}}{\longrightarrow} D(F(G(A)),B)$$ associates to each $f \colon A \to B$ the morphism $\epsilon_B\circ F(G(f)) \colon F(G(A)) \to B$ (this comes from the fact that $\varphi^{-1}(g)= \epsilon \circ F(g)$).

Using naturality of $\epsilon$ it follows that $$\epsilon_B \circ F(G(f))=f\circ\epsilon_A$$ that means the natural isomorphism is given by precomposition $\epsilon_A$, that is it is represented by $\epsilon_A$ via the Yoneda embedding.

Since embeddings reflect isomorphism, and since the Yoneda embedding is an embedding, it follows that $\epsilon_A$ must be an isomorphism.

  • $\begingroup$ i haven't studied representables yet, but i did study the Yoneda embedding. It seems to me that the natural isomorphism is $y_{\epsilon_A}(-)(id_{FGA})$, not simply $y_{\epsilon_A}$ (because $f\circ\epsilon_A = \mathcal{D}(f,\epsilon_A)(id_{FGA})=y_{\epsilon_A}(f)(id_{FGA})$), so it is not immediate to use that $y$ reflects isomorphisms (?).. $\endgroup$ – Soap Apr 8 '18 at 10:39
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    $\begingroup$ @Soap by $y_{\epsilon_A}$ do you mean the natural transformation $\mathcal D(\epsilon_A,-) \colon \mathcal D(A,-) \to \mathcal D(F(G(A)),-)$? In that case I do not understand the extra argument. $\endgroup$ – Giorgio Mossa Apr 8 '18 at 13:47
  • $\begingroup$ Oh yes, I was confused. Now it is clear, thank you. $\endgroup$ – Soap Apr 8 '18 at 14:02
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    $\begingroup$ @Soap no problem, you're welcome ;-) $\endgroup$ – Giorgio Mossa Apr 8 '18 at 14:37
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    $\begingroup$ @Brofessor it follows by a simple Yoneda-like argument: it follows from the fact that $g=1_Y\circ g=\mathcal C[g,G(Y)](1_Y)$ and from the naturality of $\varphi^{-1}$ (here $Y \in \mathcal C$ and $g \in \mathcal C[X,Y]$). $\endgroup$ – Giorgio Mossa Oct 24 '20 at 19:25

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