What is the right adjoint to the functor $\sf{Psh}\to\sf{Set}$ which evaluates the presheaf on the whole space?

Question

$\newcommand{\O}{\mathcal{O}}\newcommand{\T}{\mathcal{T}}\newcommand{\op}{^{\sf{op}}}\newcommand{\set}{\sf{Set}}\newcommand{\ps}{\sf{Psh}_{\T}}$Let $\T$ be a topological space and $\O(\T)$ the poset category (ordered by $\subseteq$) of the open sets of $\T$. We define a (geometric) presheaf to be any functor $\O(\T)\op\to\set$, so let $\ps$ be the functor category of presheaves, $\ps=[\O(\T)\op,\set]$. Note that beyond this, I have no knowledge of algebraic geometry. To a student of algebraic geometry I'm sure this exercise is quite easy (but it is curious it was left in Leinster's book, which does not cover the subject).

I have been baffled by the following exercise for a while - Source of exercise, Leinster's Basic Category Theory, the first section of the "Adjoints" chapter:

Define $\Delta:\set\to\ps$ to be the "diagonal functor", where for any set $A$, $\Delta A$ is the constant presheaf which maps all inputs to $A$ and arrows to the identity arrow on $A$. $\Delta$'s action on functions $f:A\to B$ is to map them to the obvious constant natural transformations given by the components $\Delta f_U:=f$ for each open set $U$ of $\T$.

Exhibit a chain of adjoint functors $\Lambda\dashv\Pi\dashv\Delta\dashv\Gamma\dashv\nabla$.

So far in the course, he has not taught us about the other ways of thinking about adjunctions (units/counits and initial/terminal morphisms), only the definition via "equivalence of arrows" has been covered. I tried this exercise after having read ahead and learning about adjoint functors from other sources, so I was able to figure out $\Pi$ and $\Gamma$ by examining terminal/initial morphisms. The answer is that $\Gamma$ maps a presheaf $F$ to the set $F(\T)$, and maps natural transformations $\alpha:F\to G$ to the component function $\alpha_\T$. Dually $\Pi$ maps a presheaf $F$ to the set $F(\emptyset)$, and maps natural transformations $\alpha$ to $\alpha_\emptyset$.

I was able to find these functors since $\Delta$ has the nice property that, in the search for a terminal morphism $\varepsilon_F:\Delta\Gamma F\to F$, we just need to find arrows that complete the wedge $F(U)\leftarrow\Gamma F\rightarrow F(U')$ whenever $U'\subseteq U$ since $\Delta$ collapses a presheaf into one object. This made the problem significantly easier!

However, even thinking of what kind of presheaf $\nabla X$ might be is proving too much for me. I've toyed with $\nabla X(U):=\{\text{functions $U\to X$}\}$ (or the other way round) but this leaves it impossible to choose a terminal morphism $\varepsilon:\Gamma\nabla X\to X$ in a natural way. Beyond sets of functions, I have no idea how to construct a presheaf from a set (I have seen no examples). What's worse is that, when $U'\subseteq U$, I need to find arrows $\nabla_{U,U'}$ that entail the uniqueness criterion of the terminal morphism. What do I mean by that? Well, for any presheaf $F$ and any function $f:F(\T)\to X$, there needs to be a unique natural transformation $\alpha:F\to\nabla X$ such that $\varepsilon_X\circ\alpha_\T=f$. So, $\alpha$ needs to be uniquely specified by $\alpha_\T$. Of course every $U\in\O(\T)$ is a subset of $\T$ itself, so we need $\alpha_U$ to be uniquely specified by the naturality criterion $\alpha_UF_{\T,U}=\nabla_{\T,U}\alpha_\T$ (uniqueness up to $F$). So, this $\nabla_{\T,U}$ arrow needs to be somewhat special.

Does anyone have any ideas? The nature of this exercise is that $\nabla X$ is going to be quite an "obvious" construction, but I can't see it.

It is worth noting that I heard from a fairly reliable source that the left and right adjoints to $\Pi,\Gamma$ respectively do not exist for every topological space $\T$. They did not say anymore sadly, but that's left me even more confused since my construction cannot be too general. I'd appreciate any useful hints or full answers - I am totally stuck.

Answer 1

Let $\nabla$ be the functor which associates to a set $A$ the following presheaf: $\nabla(A)(X) = A$, and $\nabla(A)(U)$ is a singleton for all proper open subsets $U$, with the restriction maps being the only ones possible. Then we have the desired adjunction: $$\mathsf{Hom}_{\sf{Set}}(F(X),A)\simeq\mathsf{Hom}_{\sf{Psh}}(F,\nabla(A)). $$ Indeed, the map from right to left is given by taking the induced map on global sections. But since a singleton is a final object in the category of sets, the map on global sections will uniquely determine a map of presheaves.

Maybe your reliable source was talking about the category of sheaves? The global sections functor on the category of sheaves is almost always left but not right exact, so it can't be a left adjoint.

Answer 2

I will denote your topological space by $X$, because otherwise I get confused. A morphism of presheaves $F\to \nabla A$ should be the same as a morphism of sets $FX \to A$. If $\phi$ is a morphism of presheaves, then its $X$th component is a set map $FX \to (\nabla A)X$. So it would be nice if $(\nabla A)X = A$ and all the other components of $\phi$ contain no information. To make this happen, we can define $\nabla A$ as follows. We set $(\nabla A)X= A$ and $(\nabla A)U =\textrm{pt}$ for all other open subspaces $U$ of $X$. Here $\textrm{pt}$ denotes a one-point set (the terminal object of Set).

In the same way we can find out how $\Lambda A$ for a set $A$ should look like. We define $(\Lambda A)\varnothing = A$ and $(\Lambda A)U = \varnothing$ for all other subspaces $U$ of the topological space $X$. Now a map of presheaves $\phi: \Lambda A\to F$ contains exactly as much information as its component $\phi_\varnothing :( \Lambda A) \varnothing \to F\varnothing$, and hence is the same as a set map $A \to \Pi F$.

Answer 3

As you have already figured out, the functor $Γ$ is the “global sections” functor that evaluates a presheaf $F$ on the entire space $\mathcal{T}$. The desired right adjoint $∇$ of $Γ$ needs to assign to every set $X$ a presheaf $∇ X$ on $\mathcal{T}$, so that natural transformations $$ α \colon F \Longrightarrow ∇ X $$ are “the same” as functions $$ Γ(F) \longrightarrow X , $$ i.e., functions $$ F(\mathcal{T}) \longrightarrow X \,. $$ The natural transformation $α = (α_U)_U$ consists of a map $$ α_U \colon F(U) \longrightarrow (∇X)(U) $$ for every open subset $U$ of $\mathcal{T}$.

• By choosing $$ (∇ X)(\mathcal{T}) := X \,, $$ the component $α_{\mathcal{T}}$ is precisely a map $F(\mathcal{T}) \to X$.

• For every other open subset $U$ of $\mathcal{T}$ with $U ≠ \mathcal{T}$ we would like there to be only one choice for the map $α_U$. We therefore choose $$ (∇ X)(U) := \{ \ast \} \qquad \text{for $U ≠ \mathcal{T}$} \,. $$

We have now figured out how to define the functor $∇ X$ on objects. The action on a morphism $i \colon U \to V$ in $\mathcal{O}(\mathcal{T})$ is given by $∇(X)(i) = \mathrm{id}_X$ if $U, V = \mathcal{T}$, and otherwise by the unique map from $(∇ X)(V)$ to $(∇ X)(U) = \{ \ast \}$.

At this point, we are only left with a bunch of calculations to check that

• $∇ X$ is indeed a contravariant functor from $\mathcal{O}(\mathcal{T})$ to $\mathsf{Set}$,
• $∇ X$ is functorial in $X$ (which requires us to define the action of $∇$ on morphisms),
• the mapping $α \mapsto α_{\mathcal{T}}$ is a natural bijection between natural transformations $F \Rightarrow ∇ X$ and functions $Γ(F) \to X$.

Answer 4

Here is a method to find the answer without "having to guess". The essential point is that the functor of sections over an open subset $U \subseteq X$ (where $X$ is a topological space) is representable, and the representing object is given by the presheaf $h_U$ where $$h_U(V) := \begin{cases} \{ * \}, & V \subseteq U; \\ \emptyset, & V \not\subseteq U. \end{cases}$$ (And the restriction maps are the unique maps possible.)

Therefore, if you have a right adjoint $\nabla$ to $\Gamma(X, {-})$, then for each set $S$ and open subset $U \subseteq X$, we must have $$(\nabla S)(U) \simeq \operatorname{Hom}_{\mathsf{Psh}}(h_U, \nabla S) \simeq \operatorname{Hom}_{\mathsf{Set}}(\Gamma(X, h_U), S).$$ Now, we see that if $U = X$, then $\Gamma(X, h_U) = \{ * \}$, so the right hand side is canonically isomorphic to $S$. On the other hand, if $U \ne X$, then $\Gamma(X, h_U) = \emptyset$, so the right hand side is a one-point set.

From here, you can see that there is only one possible choice for each restriction map in $\nabla S$, and verifying the details that $\nabla$ does indeed form a right adjoint to $\Gamma(X, {-})$ should be straightforward (or you could consult one of the other answers).