$F$ is an equivalence of categories implies that $F$ is fully faithful and essentially surjective

Solution 1:

Yes, your map is not an identity in general, but you can always get around that problem by replacing one of the isomorphisms of your equivalence in such a way that it becomes an adjoint equivalence (see here).

You don't need to do that, however. Take a morphism $g : FX → FY$. There is only one choice for a morphism $f : X → Y$ such that $g = Ff$: by the naturality of $η$, $f$ must equal $η^{-1} ∘ Gg ∘ η$. So let $f$ be given by the previous formula, from which it follows that $GFf = Gg$. But $G$ is an equivalence, and you've already proven that equivalences are faithful, so $g = Ff$.