# Adjoints to the Evaluation functors

Let $C$ be a category. Consider $Psh(C)$ the category of presheaves on $C$. It is a well known fact that the evaluation functor at $c$ an object of $C$ of presheaves preserves both limits and colimits yet I can't find in the littérature nor compute right and left adjoints. If someone knows a formula for both adjoints it would be very helpful.

The left adjoint $L_c$ to evaluation-at-$c$ is very simple; left adjoints preserve colimits and every set is a coproduct of $1$ with itself. Consequently,

$$L_c(X) = X \cdot L_c(1)$$

where $\cdot$ means to take the $X$-fold coproduct of $L_c(1)$ with itself. Finally,

$$\mathbf{PSh}(\mathcal{C})(L_c(1), F) \cong \mathbf{Set}(1, F(c)) \cong F(c)$$

therefore, $L_c(1)$ is the functor $\mathcal{C}(-, c)$ represented by $c$. That is,

$$L_c(X) = X \cdot \mathcal{C}(-, c)$$

The right adjoint $R_c$ is even simpler:

$$R_c(X)(d) \cong \mathbf{PSh}(\mathcal{C})(\mathcal{C}(-, d), R_c(X)) \cong \mathbf{Set}(\mathcal{C}(c, d), X) \cong X^{\mathcal{C}(c,d)}$$

That is,

$$R_c(X) \cong X^{\mathcal{C}(c, -)}$$

(thanks to Andreas Blass for reminding me of the argument)

The evaluation-at-c functor $F \to F(c)$ is, incidentally, given by the functor

$$\mathbf{Set}^{\mathcal{C}^\circ} \to \mathbf{Set}^1$$

induced by the inclusion $1 \to \mathcal{C}^{\circ}$ that identifies $c$, so both adjoints are special cases of the fact this functor has adjoints. I don't remember how this works out off the top of my head, but it should probably be easier to find a reference for.

• Is 2 the coproduct of tow copies of 1 ? Mar 15, 2017 at 15:14
• Oh ! I think I understand the last paragraph. We have the inclusion functor $i_c$ of 1 into $C^o$. This functor induces naturally a functor between the two categories of functors. Then $i_c^*$ is the evaluation functor we can define $i_{c,*}$ and $i_{c,!}$ which are the adjoints, algebraic-geometry style. Mar 15, 2017 at 15:30
• Seems my last comment was not at all how this works... do you have any reference ? Mar 15, 2017 at 16:12
• The adjoints mentioned in the last paragraph of the answer are known as the left and right Kan extensions along the inclusion $1\to\mathcal C$. Mar 15, 2017 at 16:27

Hurkyl has computed the left adjoint. Here's a way to get the right adjoint $R_c$. For any set $X$ and any object $u$ of $\mathcal C$, the value of the presheaf $R(X)$ at the object $u$ is (up to isomorphism natural in both $X$ and $u$) given by $\mathbf{PSh}(\mathcal C(-,u),R(X))$ by Yoneda's Lemma. This in turn is naturally isomorphic to $\mathbf{Set}(\text{eval}_c(\mathcal C(-,u)),X)$, by adjointness, and this is just $\mathbf{Set}(\mathcal C(c,u),X)$, i.e., $X^{\mathcal C(c,u)}$.

• How embarrassing; that trick had completely fallen out of my head by the time I got to the right adjoint. I've consolidated this into my answer, and marked it CW.
– user14972
Mar 15, 2017 at 16:38
• [Andreas Blass][1] could you please shed some light on the kan extension argument you commented on hurkyl's answer ? Mar 15, 2017 at 17:26
• See my answer. I hope it is clear--I am by no means an expert. Mar 16, 2017 at 3:36

Mac Lane has a good treatment of the Kan extensions in $CWM$. The following is a synopsis for the right Kan extension:

given categories $A,M$ and $C$, the functor $K:M\to C$ induces a functor $A^{K}:A^{C}\to A^{M}$ defined by $A^KS=SK$ and $A^K(\sigma:S\overset{\cdot }{\rightarrow}S')=\sigma K:SK\overset{\cdot }{\rightarrow}S'K.$

In this situation, a right Kan extension is just a right adjoint to $A^K:$

given $T:M\to A,\$ Ran$_TK$ will satisfy $\mathcal {Nat}[S,$Ran$_KT] \cong \mathcal {Nat}[SK,T].$

In your problem, you have$M=1, C=C, A=\bf{Set}$ and $K:1 \to C$ which induces the functor eval: $\bf{Set}^{C} \to \bf{Set}^1$.

The payoff of all this is that with a little work, you can exhibit Ran$_KT$ as a pointwise limit so these problems become computational. The details are in CWM (2d edition) on p. 234.

Similar results hold for the left Kan extension.