Let $F: \mathcal C \to \mathcal D$ a functor. This induces $$\mathbb F_{C, C'} :\text{Hom}_\mathcal C(C, C') \to \text{Hom}_\mathcal D(FC, FC'): f \mapsto Ff,$$ which is natural in $C, C'$. Suppose that $F$ has a right adjoint $R: \mathcal D \to C$ and consider the associate unit and counit $\eta :1 \to RF, \varepsilon:FR \to 1$.
Now I want to show that there is a natural transformation $\sigma: RF \to 1$ such that $\sigma \circ \eta = 1$ if and only if there is $\Gamma_{C, C'}:\text{Hom}_\mathcal D(FC, FC')\to \text{Hom}_\mathcal C(C, C')$ such that $\Gamma_{C, C'} \circ \mathbb F_{C, C'} = 1$. The "only if" part is not too hard by defining $$\Gamma_{C, C'} = (- \circ \eta_C) \circ (\sigma_{C'} \circ -) \circ R,$$ we do indeed get $(\Gamma_{C, C'} \circ \mathbb F_{C, C'})(f) = f$ using the naturality of $\eta$ and $\sigma$. But I am really stuck at the "if" part: If we have $\Gamma_{C, C'}$ satisfying the above condition, the more natural way to define $\sigma$ is (I think) $$\sigma_C := \Gamma_{RFC, C} (\varepsilon_{FC}).$$ I've tried to show that $\sigma_C \circ \eta_C = 1_C$ by using some property of the unit and the counit but I haven't been able to conclude. Does anyone know how to do that ? Am I on the right path ?
Define $$ \sigma_C = \Gamma_{RFC,C}(\varepsilon_{FC}).$$ Then we have $$\sigma_C\circ\eta_C = \Gamma_{RFC,C}(\varepsilon_{FC})\circ \eta_C=\Gamma_{C,C}(\varepsilon_{FC}\circ F(\eta_C))=\Gamma_{C,C}(\operatorname{id}_{FC}) = \operatorname{id}_C.$$
where we used the naturality of $\Gamma$ and then the triangle identity.