A chain of adjoint functors between $\mathbf{Set}$ and the category of presheaves on a fixed space $X$

Question

Fix a (topological) space $X$, write $\mathscr{O}(X)$ for the poset of open sets of $X$ under inclusion, and consider $\mathbf{P} = [\mathscr{O}(X)^{op}, \mathbf{Set}]$, the category of presheaves on $X$.

An exercise in Tom Leinster's Basic Category Theory (2.1.17) gives a simple example of a functor $\Delta : \mathbf{Set} \rightarrow \mathbf{P}$ that assigns to a set $A$ the constant presheaf with value $A$. That is, $(\Delta A)(U) = A$ and $(\Delta A)(\rightarrow) = 1_A$ for all arrows.

He then asks to find a chain of adjoint functors as follows:

with $\Delta$ in the middle and $\Lambda \dashv \Pi \dashv \Delta \dashv \Gamma \dashv \nabla$.

I've managed to scrape together all of the intuition I've got to find suitable choices for $\Pi$ and $\Gamma$.

$\Gamma$ is the 'evaluation at the whole space' functor. That is, for an $R \in \mathbf{P}$, we let $\Gamma(R) = R(X)$, and for a natural transformation $\varepsilon : R \rightarrow S$, we let $\Gamma(\varepsilon) = \varepsilon_X$, the component of $\varepsilon$ at $X$. The bijection between $\mathbf{Set}(A, \Gamma R)$ and $\mathbf{P}(\Delta A, R)$ can be written down quite naturally. For the simplest direction, given a natural transformation $g : \Delta A \rightarrow R$, we assign a map $\overline{g} : A \rightarrow \Gamma R$ i.e. $\overline{g} : A \rightarrow R(X)$ as $\overline{g} = g_X$.

Similarly, $\Pi$ is the 'evaluation at the empty set' functor and the construction is exactly dual to that of $\Gamma$.

I'm struggling to come up with ideas for the last two $\Lambda$ and $\nabla$. My intuition suggests that since $\Pi$ is "forgetful" in that it evaluates at a particular open set ($\varnothing$), it should have a left adjoint that is "free", but I can't seem to write down what a "free presheaf from a set" should be.

Is this a useful way of thinking about this? I could use some guidance or intution on what these final functors could be.

Answer 1

Your intuition about $\Lambda$ is the right idea, but to understand what "free" means here you need to think about what it means to think of $\Pi$ as "forgetful". More precisely, you can think of a presheaf $F$ as a bunch of sets $F(U)$ together with a bunch of operations (restriction maps) relating the sets, as a sort of multi-sorted algebraic structure. Your "forgetful functor" $\Pi$ considers the "underlying set" of this structure to be just the set $F(\emptyset)$. So the left adjoint $\Lambda$ should take a set $A$ to the presheaf $F$ that is obtained by letting $A$ sit inside $F(\emptyset)$ and then "freely" applying all the operations (i.e., restriction maps).

Now note that the only restriction map with domain $F(\emptyset)$ is the restriction $F(\emptyset)\to F(\emptyset)$, which is required to be the identity. So there are actually no operations you can apply to your elements of $A\subseteq F(\emptyset)$ to get new elements of the presheaf. So the presheaf freely generated by $X$ is just given by $F(\emptyset)=A$ and $F(U)=\emptyset$ for $U\neq\emptyset$ (since there are no operations you can use to get elements of $F(U)$). I'll leave it to you to verify that this really does define a functor which is left adjoint to $\Pi$.

As for $\nabla$, I'm afraid I don't have as nice of an intuition for how to construct it. But here's one way to find it. Note that none of the constructions or proofs here really use anything special about $\mathbf{Set}$. So we can dualize the whole picture, replacing $\mathscr{O}(X)^{op}$ and $\mathbf{Set}$ by their opposite categories, and left and right adjoints will be swapped so $\Lambda$ will swap with $\nabla$. Now we constructed $\Lambda$ so that it sends a set $A$ to the presheaf $F$ which takes value $A$ on the terminal object of $\mathscr{O}(X)^{op}$ and takes value $\emptyset$ (the initial object of $\mathbf{Set}$) on all other objects. Dually, then, $\nabla(A)$ will take value $A$ on the initial object of $\mathscr{O}(X)^{op}$ and will take value the terminal object of $\mathbf{Set}$ on all other objects. That is, $\nabla(A)$ is the presheaf $F$ defined by $F(X)=A$ and $F(U)=\{*\}$ for all $U\neq X$.

Answer 2

You don't necessarily need a lot of intuition to construct adjoints.

A good way to find a left adjoint $\mathcal L: \mathbf C \to \mathbf D$ to a know functor $\mathcal R:\mathbf D \to \mathbf C$ is to look at each comma category $(d\downarrow \mathcal R)$ and try to find a initial element in it (formally you are looking for the unit of the adjunction).

So in your case, looking at the category $(A\downarrow \Pi)$ whose objects are maps $A \to F(\emptyset)$ for all presheaves $F$ and whose morphisms are commutative triangles $$ \require{AMScd} \begin{CD} A @>>> F(\emptyset) \\ @| @VV{\alpha_\emptyset}V \\ A @>>>G(\emptyset) \end{CD} \qquad \alpha:F \to G $$ An initial object in that category is then a presheave $\Lambda A$ together with a map $\eta_A : A \to \Lambda A(\emptyset)$ having the property that for any $f: A \to F(\emptyset)$ there exists a unique $\alpha:\Lambda A \to F$ such that $f = \alpha_\emptyset \circ \eta_A$. The most obvious thing to try it to take $\Lambda A$ such that $\Lambda A(\emptyset) = A$, define $\eta_A = \mathrm{id}_A$ and then for any $f : A \to F(\emptyset)$ try to find $\alpha : \Lambda A \to F$ with component $f$ at $\emptyset$. So in short we need a presheaf $\Lambda A$ with value $A$ at $\emptyset$ and the ability to "expand" any map defined only at $\emptyset$ to a full presheaf map. We don't really have a choice : $\Lambda A(U)$ for $U \neq \emptyset$ need to be initial in $\mathbf{Set}$, i.e. $\Lambda A(U) = \emptyset$. Then any $f : A \to F(\emptyset)$ is expandable as $\bar f : \Lambda A \to F$ by defining $\bar f_U$ as the unique $\emptyset \to F(U)$. Moreover $\bar f$ is indeed a natural transformation because all naturally squares are vacuously commutative.

You can do the same to find a right adjoint to $\Gamma$ by looking for a terminal object in each comma category $(\Gamma \downarrow A)$.