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$\DeclareMathOperator\Hom{Hom}\newcommand\id{\mathrm{id}}\newcommand\C{\mathcal C}$Right adjoint $R \colon \C' \to \C$ and left adjoint $L$.

$R$ is fully faithful iff the counit $\eta \colon LR \to \id_{\C'}$ is an isomorphism. The proof is given below from the textbook (Categories and Sheaves) using the commutative diagram.

Adjunction diagram

Proof: The map $\Hom_{\C'}(Y,Y') \to \Hom_{\C}(RY,RY')$ is bijective iff the map $\Hom_{\C'}(Y,Y') \to \Hom_{\C}(LRY,Y')$. Therefore, $R$ is fully faithful iff the map $LRY \to Y$ is an isomorphism for all $Y$.

I know that this question has been answered in Right adjoint is fully faithful iff the counit is an isomorphism (without Yoneda).

I understand that, $R$ fully faithful, taking $Y'$ to be $LRY$ we get that $\eta_Y$ has a left inverse $g \colon Y \to Y'$. It was not clear how is $g$ also a right inverse. I somewhat know the explanation has to be simple but I am being foolish about it.

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    $\begingroup$ When you say $R$ is fully faithful, that means the arrow you've denoted by $R$ is injective (that's the faithful part) and surjective (that's the full part). So this arrow $R$ is bijective: is invertible. So $\eta_Y$ is a composition of two isomorphisms. $\endgroup$
    – Todd Trimble
    Commented Jul 23 at 20:23
  • $\begingroup$ Alright. I guess the approach I used with showing $g$ is also a right inverse is not needed? $\endgroup$
    – Siya
    Commented Jul 23 at 20:57
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    $\begingroup$ Oh, I see what you're trying to ask (I was thrown off by your notation). Since you've shown $\eta_Y$ has a left inverse, it will suffice to show it is an epimorphism. So suppose $f, g$ are distinct morphisms of the form $Y \to Y'$. Since the diagonal arrow in your diagram is injective, the two morphisms $f \eta_Y, g\eta_Y$ of the form $LRY \to Y'$ are also distinct. And this completes the proof. $\endgroup$
    – Todd Trimble
    Commented Jul 23 at 21:14
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    $\begingroup$ Make sense? We have $\eta_Y g \eta_Y = \eta_Y$ using the fact that $g$ is a left inverse of $\eta_Y$. So $\eta_Y g \eta_Y = 1_Y \circ \eta_Y$. Now conclude $\eta_Y g = 1_Y$ since $\eta_Y$ is an epimorphism. $\endgroup$
    – Todd Trimble
    Commented Jul 23 at 21:22
  • $\begingroup$ Yes! that makes sense. That's what I was looking for. Many Thanks! $\endgroup$
    – Siya
    Commented Jul 23 at 21:24

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