How to interpret "opposite functor" for a dual statement Consider the following result:

For every pair of categories $(C,D)$ and every functor $F:C\rightarrow
 D$, $F$ has a right adjoint if and only if, for every
$Y\in\text{Ob}(D)$, the functor
$\text{Hom}_D(F(-),Y):C^{\text{op}}\rightarrow\textbf{Set}$ is
representable.

I want to use the above to deduce that

For every pair of categories $(D,C)$ and every functor $G:D\rightarrow
 C$, $G$ has a left adjoint if and only if, for every
$X\in\text{Ob}(C)$, the functor
$\text{Hom}_C(X,G(-)):D\rightarrow\textbf{Set}$ is
representable.

First of all, if we have a morphism $f:X\rightarrow Y$ in some category $B$, then $f:Y\rightarrow X$ is a morphism in $B^{\text{op}}$.
Now, suppose $G:D\rightarrow C$ is a morphism in $\textbf{Cat}$. By the same logic $G:C\rightarrow D$ is a morphism in $\textbf{Cat}^{\text{op}}$. What is the difference between this $G:C\rightarrow D$ and $G^{\text{op}}:D^{\text{op}}\rightarrow C^{\text{op}}$? Is the latter in $\textbf{Cat}$ or $\textbf{Cat}^{\text{op}}$?--When I want to dualize a statement do I take the former or the latter?
 A: There is no need to invert the directions of functors. Adjointness means that $C(FA, B) \simeq D(A, GB),$ so if you can speak of the left adjoint and want to speak about the right one, you are to replace $C$ and $D$ with $C^{op}$ and $D^{op}$: then the right adjoint becomes the left one and vice versa: $C^{op}(B, FA) \simeq D^{op}(GB, A)$ (note that a functor $C \to D$ is equivalently a functor $C^{op} \to D^{op}$). So if you denote those categories by $C'$ and $D'$ and write the first result for them, you get:

For every $F:C'\rightarrow
 D'$, $F$ has a right adjoint if and only if, for every
$Y\in\text{Ob}(D')$, the functor
$\text{Hom}_{D'}(F(-),Y):C'^{\text{op}}\rightarrow\textbf{Set}$ is
representable.

Rewriting this in terms $C$ and $D$ gives

For every $F:C\rightarrow
 D$, $F$ has a left adjoint if and only if, for every
$Y\in\text{Ob}(D^{op}) = \text{Ob}(D)$, the functor
$\text{Hom}_D(Y, F(-)):C\rightarrow\textbf{Set}$ is
representable.

Finally, replacing "$F$" with "$G$", "$C$" with "$D$" and "$D$" with "$C$" here gives what you want to deduce.
A: Thinking about $G$ as a morphism in $\mathbf{Cat}^{\mathrm{op}}$ is not helpful, because the result you want to use is about morphisms in $\mathbf{Cat}$! Instead:

*

*Start with a functor $G : D \to C$.

*$G$ can also be viewed as a functor $G^{\mathrm{op}} : D^{\mathrm{op}} \to C^{\mathrm{op}}$

*Apply the result, to get the following statement:


$G^{\mathrm{op}}$ has a right adjoint if and only if, for every $Y \in \mathrm{Ob}(C^{\mathrm{op}})$, the functor $\mathrm{Hom}_{C^{\mathrm{op}}}(G^{\mathrm{op}}({-}), Y) : (D^{\mathrm{op}})^{\mathrm{op}} \to \mathbf{Set}$ is representable.



*Unwind the definition of "$\mathrm{op}$":


$G^{\mathrm{op}}$ has a right adjoint if and only if, for every $Y \in \mathrm{Ob}(C)$, the functor $\mathrm{Hom}_C(Y, G({-})) : D \to \mathbf{Set}$ is representable.



*Show that a right adjoint for $G^{\mathrm{op}}$ is the same as a left adjoint for $G$.

