# Exhibit a chain of adjoints passing through the diagonal

A problem in Leinster's Basic Category Theory:

Fix a topological space $X$, and write $\mathscr{O}(X)$ for the poset of open subsets of $X$, ordered by inclusion. Let $$\Delta: \mathsf{Set} \to [\mathscr{O}(X)^{\text{op}}, \mathsf{Set}]$$ be the functor assigning to a set $A$ the presheaf $\Delta A$ with constant value $A$. Exhibit a chain of adjoint functors $$\Lambda \dashv \Pi \dashv \Delta \dashv \Gamma \dashv \nabla.$$

$\Gamma$ should be the global section functor (the limit of a diagram $F:\mathscr{O}(X)^{\text{op}} \to \mathsf{Set}$ is just $F(X)$, since $\mathscr{O}(X)$ has a final element). $\Pi$ should be the colimit of the diagram $F$ in $\mathsf{Set}$. (Why would they denote it by $\Pi$? Because it is the fundamental groupoid when $F$ is chosen properly?)

I'm a little stuck on the others. Perhaps someone can loosen the ketchup bottle for me.

For $∇$ you have $\mathrm{Hom}(FX, Y) ≅ \mathrm{Hom}(F, ∇Y)$, and are trying to find out what $∇Y$ is. A good way to do that is to probe it with other convenient objects. For example, by choosing $F = Δ1$ you get $(∇Y)X = Y$, which isn't that useful but illustrates the point. Fortunately, that there is a great collection of probes among presheaves is exactly the content of the Yoneda lemma. By taking $F = \mathrm{Hom}(-, U)$, you can immediately calculate $∇$: $$\mathrm{Hom}(Γ(\mathrm{Hom}(-, U)), Y) = \mathrm{Hom}(\mathrm{Hom}(X, U), Y) ≅ \mathrm{Hom}(\mathrm{Hom}(-, U), ∇Y) ≅ (∇Y)U,$$ where the last isomorphism is by the Yoneda lemma. In other words, $∇Y$ is $Y$ on $X$, $1$ otherwise, because $\mathrm{Hom}(X, U)$ is singleton for $U = X$, and empty otherwise.
For $Λ$ we have $\mathrm{Hom}(ΛY, F) ≅ \mathrm{Hom}(Y, F∅)$. Since $Λ$ is cocontinuous, and $\mathrm{Set}$ has a convenient property of being made up of coproducts of $1$, it suffices to calculate $Λ1$, but this again we easily get by Yoneda: $$\mathrm{Hom}(Λ1, F) ≅ \mathrm{Hom}(1, F∅) ≅ F∅ ≅ \mathrm{Hom}(\mathrm{Hom}(-, ∅), F),$$ for all $F$, so $Λ1 ≅ \mathrm{Hom}(-, ∅)$. In other words, $ΛY$ is $Y$ on $∅$, and $∅$ otherwise.
Of course, there is another way to find these functors: guessing. $\mathrm{Hom}(FX, Y) ≅ \mathrm{Nat}(F, ∇Y)$ tells us literally that we're looking for a presheaf $∇Y$ such that giving a natural transformation $F ⇒ ∇Y$ is the same thing as giving a function $FX → Y$. Since every $α : F ⇒ ∇Y$ already needs an $α_X : FX → (∇Y)X$, setting $(∇Y)X = Y$ seems inevitable, and setting $(∇Y)U$ to the terminal object will certainly force uniqueness. You just need to check that everything works out. The reasoning for $Λ$ is analogous.