In Wikipedia I saw the result that, when category $\mathcal{C}$ has all small copowers, a functor $\mathcal{C}\overset{K}{\rightarrow}\text{Set}$ has a left adjoint if and only if it is representable. I have the following questions:
(1) Can this result be dualized ?
(2) Is there an easy way to characterize the functors into $\text{Set}$ that have right adjoints ?
(3) When do representable functors have right adjoints ?
Best Answer
To dualise a theorem in category theory, simply replace some or all of the universally quantified variables ranging over categories with their opposites, then rewrite all statements so that $^\textrm{op}$ disappears (where possible). In this case, we obtain:
Or, if you prefer:
But it is more conventional to have $^\textrm{op}$ on the domain of a functor, and anyway representable functors should have codomain $\textbf{Set}$.
Notice the dual statement is not a statement about functors in $\textbf{Set}$ that have a right adjoint. I am not aware of any easy theorems to that effect. The fundamental reason for this asymmetry is that $\textbf{Set}$ is freely generated under colimits by $\{ 1 \}$, so to describe a colimit-preserving functor (as a left adjoint would be) it suffices to say $1$ goes, but $\textbf{Set}$ is not freely generated under limits in any meaningful sense.
An object such that the corresponding representable functor into $\textbf{Set}$ has a right adjoint is said to be tiny. These are quite rare in nature, but it is easy to create artificial examples: indeed, for any small category $\mathcal{B}$ and any object $B$ in $\mathcal{B}$, the representable functor $h_B = \mathcal{B} (-, B)$ is a tiny object in the category $[\mathcal{B}^\textrm{op}, \textbf{Set}]$, i.e. the representable functor $[\mathcal{B}^\textrm{op}, \textbf{Set}] (h_B, -) : [\mathcal{B}^\textrm{op}, \textbf{Set}] \to \textbf{Set}$ has a right adjoint, namely the functor that sends each set $X$ to the functor $X^{\mathcal{B} (B, -)}$.