Condition in proof of unit of adjoint is isom iff F fully faithful.

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Let $C,D$ be categories and $F:C\to D$ and $G:D\to C$ be adjoint functors. Then $F$ is fully faithful iff the unit is an isomorphism?

Regarding this proof in the first comment, in the part where he proves that $F$ is full if $\eta$ is split epic, how does he go from $GF(f) = G(g)$ to $F(f) = g?$

This must require that $G$ is invertible?

Additionally in the final line of the proof, how do you do a similar thing to remove the $F$?

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In the first case, what we use is not $GF(f)=G(g)$ but really $GF(f)\circ \eta_X=G(g)\circ \eta_X$; and similarly, in the second case, what we use is $\epsilon_{FX} \circ F(\eta_X\circ s_X)=\epsilon_{FX}\circ F(id_{GFX})$. The reason these equations imply $F(f)=g$ and $\eta_X\circ s_X=id_{GFX}$ is that the maps $$\theta_{X,FY}^{-1}:h\in\operatorname{Hom}_D(FX,FY)\mapsto G(h)\circ \eta_X\in \operatorname{Hom}_C (X, GFY)$$ and $$\theta_{GFX,FX}:k\in\operatorname{Hom}_C(GFX,GFX)\mapsto \epsilon_{FX} \circ F(k)\in \operatorname{Hom}_C(FGFX,FX)$$ are two components of the natural isomorphism $\theta:\operatorname{Hom}_{C}(- ,G-)\to \operatorname{Hom}_{D}(F- ,-)$ defining the adjunction $F\dashv G$, and thus in particular they must be injective.