I encounter the following situation. That is, if I understand it correctly.
Let $\mathcal{C}$ be a small category with set of objects $\mathcal{C}_0$. Consider the category Fam$(\mathcal{C}_0)$ having as objects families $(X_C)_{C \in \mathcal{C}_0}$ of sets and as morphisms between objects $(X_C)_{C \in \mathcal{C}_0}$ and $(Y_C)_{C \in \mathcal{C}_0}$ families of functions $(f_C : X_C \to Y_C)_{C \in \mathcal{C}_0}$, with identities and composition component-wise.
There is a forgetful functor $U: \text{PSh}(\mathcal{C}) = \textbf{Set}^{\mathcal{C}^\text{op}} \to \text{Fam}(\mathcal{C}_0).$ Namely, it maps the objects, i.e., functors $F : \mathcal{C}^\text{op} \to \textbf{Set}$
in PSh$(\mathcal{C})$ to the family of sets $(F(C))_{C\in \mathcal{C}_0}$, which indeed forms a set by the smallness of $\mathcal{C}$, and the morphisms, i.e., natural transformations $\mu : F \Rightarrow G$
to the family of functions $(\mu_C : F(C) \to G(C))_{C \in \mathcal{C}_0}.$
I am concerned with whether $U$ has either right or left adjoints. Would these exist, and if so, would they be similar? For a right adjoint, we should give a family of sets $(X_C)_{C ∈ \mathcal{C}_0}$ the structure of a presheaf, which seems doable, but I am not entirely sure if I have the right and/or complete view of the situation yet.
Attempt: would the following assigment form a presheaf as required?
$$(X_C)_{C \in \mathcal{C}_0} \mapsto (F : \mathcal{C}^\text{op} \to \textbf{Set})
$$
for $F(C) = X_C$, and the assignment
$$(f_C : X_C \to Y_C)_{C \in \mathcal{C}_0} \mapsto (\mu : F \Rightarrow G).
$$
Which $µ$ to choose, though?
Best Answer
The category $\DeclareMathOperator{Fam}{Fam}$ $\Fam(C_0)$ is just $Set^{C_0}$.
For each $A \in C_0$, we have a functor $U_A : Set^{C_0} \to Set$ defined by $U_A(X) = X_A$. This functor has both a left and a right adjoint. The left adjoint is given by $L_A(S)_B = \{x \in 1 \mid A = B\} \times S$. The right adjoint is given by $R_A(S)_B = S^{\{x \in 1 \mid A = B\}}$.
Now note that for all $X$, we have $X \cong \coprod\limits_{A \in C_0} L_A(X_A) \cong \prod\limits_{A \in C_0} R_A(X_A)$. We wish to find a left adjoint $L$ and a right adjoint $R$ to $U$. If we had such adjoints, we would clearly have $R(X) \cong \prod\limits_{A \in C_0} R(R_A(X_A))$ and $L(X) \cong \coprod\limits_{A \in C_0} L(L_A(X_A))$. Thus, we really just need to figure out $R \circ R_A$ and $L \circ L_A$, which, assuming $L$ and $R$ exist, are the right and left adjoints of $U_A \circ U$.
So we now turn to finding the adjoints $L’_A$ and $R’_A$ of $U’_A = U_A \circ U$, which we abusively also write as $U_A$. This should be a generalisation of our work in finding $R_A$ and $L_A$. Note that in $C_0$, viewed as a discrete category, we have $Hom(A, B) = Hom(B, A) = \{x \in 1 \mid A = B\}$. This set appears in our definitions for both $R_A$ and $L_A$, so it seems only natural to try it in our more general case. This natural guess gives us $L’_A(S)_B = Hom(B, A) \times S$ and $R’_A(S)_B = S^{Hom(A, B)}$, and the turn out to be the adjoints of $U’_A$.
Thus, we can then deduce and verify that the adjoints to $U$ are $L(X)_B = \coprod\limits_{A \in C_0} Hom(B, A) \times X_A$ and $R(X)_B = \prod\limits_{A \in C_0} X_A^{Hom(A, B)}$.