Does the application functor have adjoints?

Question

For $K$ and $C$ categories, with $C$ complete and cocomplete, we define the application functor on a fixed $k:K$ as $\ @\ k : [K,C]\rightarrow C$

$$D\ @\ k = D\ k$$ $$\alpha\ @\ k = \alpha_k$$

for $D:[K,C]$ and $\alpha$ a natural transformation between two functors in $[K,C]$.

Does $\ @\ k$ have left or right adjoints? How do you go about finding one?

Thanks in advance!

Answer 1

By the Yoneda lemma, we have $$\mathcal{C} (c, c' (k)) \cong [\mathcal{K}, \textbf{Set}](\mathcal{K} (k, -), \mathcal{C} (c, c' (-)))$$ naturally in $c$ and $c'$. Assuming $\mathcal{C}$ has coproducts, or at least copowers, the functor $$\textbf{Set} (X, \mathcal{C} (c, -)) : \mathcal{C} \to \textbf{Set}$$ is representable for every set $X$; let $X \odot c$ be the representing object. (It is, equivalently, the coproduct of $X$-many copies of $c$.) Consider the diagram: $$\mathcal{K} (k, -) \odot c : \mathcal{K} \to \mathcal{C}$$ By definition, we have $$\mathcal{C} (\mathcal{K} (k, k') \odot c, c' (k')) \cong \textbf{Set} (\mathcal{K} (k, k'), \mathcal{C} (c, c' (k'))$$ naturally in $c'$ and dinaturally in $k'$. Hence, $$[\mathcal{K}, \mathcal{C}] (\mathcal{K} (k, -) \odot c, c') \cong [\mathcal{K}, \textbf{Set}] (\mathcal{K} (k, -), \mathcal{C} (c, c' (-))) \cong \mathcal{C} (c, c' (k))$$ naturally in $c'$. But that means the left adjoint of ${-} @ k$ evaluated at $c$ exists and is $\mathcal{K} (k, -) \odot c$. In particular, ${-} @ k$ has a left adjoint.

Dually, assuming $\mathcal{C}$ has products, or at least powers, ${-} @ k$ has a right adjoint.