# Left and right adjoint of the presheaf evaluation $\widehat{C} \to \mathbf{Set}$, $F \mapsto F(c)$

I would like to find the left and right adjoint of the presheaf evaluation functor $F \colon \widehat{C} \rightarrow \mathbf{Set}$, $X \mapsto X(c)$ for a fixed object $c \in \operatorname{ob}(C)$, where $\widehat{C} := \mathbf{Set}^{C^{\mathrm{op}}}$.

I could not understand this proof.

To show $F$ is right adjoint we define $L_c \colon \mathbf{Set} \rightarrow \widehat{C}$ such that $$\operatorname{Hom}(L_c(X), D) \cong \operatorname{Hom}(X, F(D) = \operatorname{Hom}(X,D(c)).$$

The cited proof states that we define $L_c(X)= X \cdot L_c(1) := \bigsqcup_{x \in X} L_c(1)$.

I don't know what this actually means—what is $L_c(1)$ anyways? Secondly, I don't see how this defines an element in $\widehat{C}$. I am equally confused with the construction of right adjoint. Elaboration is really appreciated.

So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $X\simeq \coprod_{x\in X}1$, where $1$ is the singleton. So we necessarily have that $L_c(X) \simeq \coprod_{x\in X}L_c(1)$ in the presheaf category $\widehat{C}$. So it suffices to determine $L_c(1)$ to determine the entire functor.
The adjunction $\operatorname{Hom}_{\widehat{C}}(L_c(X),D) \simeq \operatorname{Hom}_{\mathbf{Set}}(X,D(c))$ specialised to $X = 1$ gives then $\operatorname{Hom}_{\widehat{C}}(L_c(1),D)\simeq \operatorname{Hom}_{\mathbf{Set}}(1,D(c)) \simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :
If we denote $y: C \to \widehat{C}$ the Yoneda embedding defined by $y(c) = \operatorname{Hom}_C(-,c)$, then the Yoneda lemma states that $\operatorname{Hom}_{\widehat{C}}(y(c),D)\simeq D(c)$, which is exactly what we wanted.
We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)\simeq \coprod_{x\in X}y(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms: $$\operatorname{Hom}_{\widehat C} \left( \coprod_{x\in X} y(c), D \right) \simeq \prod_{x\in X} \operatorname{Hom}_{\widehat{C}}(y(c),D) \simeq \prod_{x\in X} D(c) \simeq D(c)^X \simeq \operatorname{Hom}(X,D(C))$$
Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=\bigsqcup_{x\in X} 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=\hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=\bigsqcup_{x\in X} \hom(-,c)$, where the coproduct is taken in the presheaf category.