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Your notation is rather nonstandard; in my experience with these situations, $\eta$ is used for the unit of
Suppose $F \dashv G$ F^{op}$with components unit$\eta_M: M \eta: 1_C \to GFM$(which you have) F^{op} F$ and $\varepsilon$ for the counit with components $\varepsilon: F G N F^{op} \to N$ (which you don't have; yours is in the opposite direction)1_{C^{op}}$. I'm going to stick with the standard notation, so that the question becomes whether, in our situation, we have Since$\eta^{-1} = \varepsilon^{op}$. Given \varepsilon$ is the unique transformation $F: C^{op\theta: F F^{op} \to C$ such that $F \dashv F^{op}$ and 1$such that both unit and counit are invertible, we have$\eta^{-1} = \varepsilon^{op}$if and only if $$\eta F^{op} 1_{F^{op}} = (F^{op} \stackrel{\eta F^{op}}{\to} F^{op} F F^{op} \stackrel{F^{op} \varepsilon}{\to} F^{op} \stackrel{F^{op}\eta^{op}}{\to} F^{op} F F^{op}) = F^{op} \eta^{op}$$ where in the second equation we used a triangular identity. So stackrel{F^{op}\theta}{\to} F^{op})$$the question becomes is whether we have \eta F^{op} F^{op}\eta^{op} = F^{op} \eta^{op}. As I say, no(\eta F^{op})^{-1}. Consider a Take C to be an commutative group G, considered as a one-object category G; choose with one object G so that it contains a central element \bullet. Here we may simply identify g such that G^{op} with g \neq g^{-1} (e.g., G, i.e., the identity G = 1_G \mathbb{Z} would do). Take F: colon G \to G^{op} to G may be the functor that sends seen as a morphism h to h^{-1}. The contravariant functor F^{op} also takes because we have h to 1_G(g h) = 1_G(h)1_A(g) by commutativity. Now take h^{-1}, F and the composite therefore F^{op} F is F^{op} to be the identity . Denote the object of G by on \bullet. We can use the central G. Any element g: \bullet \to u \bullet in G as a unit morphism 1_G \bullet \to F^{op} F because g: \bullet \to F^{op}F = F^{op} F \bullet is natural (since g is central), and g can serve as a universal arrow the unit transformation (naturality also follows from commutativity). But then, as soon as 1_G \u is not equal to F^{op}F for trivial reasons. Then$$\eta F^{op} \bullet = \eta \bullet = g$$and u^{-1}, we have$$F^{op} \eta^{op} \bullet F^{op}u^{op} = F^{opu \neq u^{-1} g = g^{-1}$$(u F^{op})^{-1}$. Therefore taking $G$ to be the additive group $\mathbb{Z}$ and therefore $\eta F^{op} u = 1 \neq F^{op} in \eta^{op}$. mathbb{Z}\$, we reach a counterexample.