1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.

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: $$ \operatorname{Hom}_{\widehat C} \left( \coprod_{x\in X} y(c), D \right) \simeq \prod_{x\in X} \operatorname{Hom}_{\widehat{C}}(y(c),D) \simeq \prod_{x\in X} D(c) \simeq D(c)^X \simeq \operatorname{Hom}(X,D(C)) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.