Let $F: \mathscr{C}\to \mathscr{C}^{op}$, with an adjoint $G$, and $\eta: 1_\mathscr{C} \Rightarrow G\circ F $ and $\varepsilon: F\circ G\Rightarrow 1_{\mathscr{C}^{op}}$ with components (in $\mathscr{C}$):

$\eta_M: M\to G(F(M))$ and $\varepsilon^{op}_N: N\to F(G(N))$ such that the compositions (in $\mathscr{C}$):

$F(M) \xrightarrow{\varepsilon^{op}_{F(M)}} FGF(M) \xrightarrow{F^{op}(\eta_M)}F(M)$ and

$G(N) \xrightarrow{\eta_{G(N)}}GFG(N) \xrightarrow{G(\varepsilon_N)}G(N)\ $ are units.

Moreover suppose $G=F^{op}$ and $\eta$ a isomorphis transformation. Follow that $F$ is full, faithfull and reflect isomorphisms (consider $F(f)\circ \eta_M=\eta_{M'}\circ f$ for $f: M\to M'$), then also $G=F^{op}$ reflect isomorphisms, from above: $\varepsilon$ is Iso. Observe that the components of $\eta: 1_\mathscr{C} \Rightarrow F^{op}\circ F $ and $\varepsilon^{op}: 1_{\mathscr{C}}\Rightarrow F^{op}\circ F$ are of type: $\eta_M: M\to F\circ F(M)$ and $\varepsilon^{op}_M: M\to F\circ F(M)$.

**Question:** Is true that $\eta =\varepsilon^{op}$ ?