# 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?

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:

1. Start with a functor $$G : D \to C$$.
2. $$G$$ can also be viewed as a functor $$G^{\mathrm{op}} : D^{\mathrm{op}} \to C^{\mathrm{op}}$$
3. 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.

1. 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.

1. Show that a right adjoint for $$G^{\mathrm{op}}$$ is the same as a left adjoint for $$G$$.
• Thanks for your help! Nice youtube channel btw; hope you upload more videos soon! Commented Feb 10, 2022 at 17:01

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.

• Thank you for your help! Commented Feb 10, 2022 at 17:01