Dual vs. opposite: Why isn't the dual of the Yoneda functor $Y': \text{Set}^D \rightarrow D^{\text{op}}$? From Mac Lane's Category theory:  

If $Y : D^{\text{op}} \rightarrow \text{Set}^D$ is the Yoneda functor, then the dual is $Y' : D \rightarrow (\text{Set}^D)^{op}$.

Why isn't the dual of the Yoneda functor $Y': \text{Set}^D \rightarrow D^{\text{op}}$, as I can take the statement "$Y$ is a functor with domain $D^{\text{op}}$ and codomain $\text{Set}^D$" and accordingly switch the domain and codomain?
 A: When you "switch the domain and codomain" in your statement, you are really taking dual of the functor $Y$ seen as an arrow in $\mathbf{Cat}$, so you end up with a "cofunctor", i.e. an arrow in $\mathbf{Cat}^{op}$. But this is not what we want here : what we want is a functor between the opposite categories, i.e. a functor $(D^{op})^{op}=D\rightarrow (\text{Set}^D)^{op}$. You can do this for any functor $F:A\to B$, and this makes "taking opposites" a covariant functor from $\mathbf{Cat}$ to itself (ignoring the size issues).
A: In a 2-category $C$ (a category with objects, 1-morphisms between objects, and 2-morphisms between 1-morphisms), there are multiple ways to dualize.
You can keep the same objects, reverse all the 1-morphisms, and keep the same 2-morphisms. This is $C^\text{op}.$
Or you can keep the same objects, keep the same 1-morphisms, and reverse all 2-morphisms. This is $C^\text{co}.$
Or you can keep the same objects, reverse both the 1-morphisms and the 2-morphisms. This is $C^\text{coop}=(C^\text{co})^\text{op}.$
If we wanted we could view the Yoneda map as an arrow in the category of categories, and then in $\text{Cat}^\text{op}$ or $\text{Cat}^\text{coop},$ it would be reversed in the way you describe. However what Mac Lane has in mind here is just the Yoneda embedding applied to opposite category. If you like you may think of it as $Y\circ\text{op}$, where $\text{op}\colon\text{Cat}\to\text{Cat}$ is the opposite category functor.
