# limit functor isomorphic to global sections functor?

Let $$C$$ be a small category, and $$D$$ any category, and $$D^C$$ the functor category of functors from $$C$$ to $$D$$.

The diagonal functor $$\Delta\colon D\to D^C$$ takes an object $$a\in D$$ to the constant functor at that object: $$\Delta(a)(x) = a$$ for all $$x\in C.$$

The limit functor $$\lim\colon D^C \to D$$ takes a functor $$F\colon C\to D$$ to its limit, $$\lim F$$, the initial cone over $$F$$.

There is an adjunction $$\operatorname{Hom}_{D^C}(\Delta(a),F) = \operatorname{Hom}_{D}(a,\lim F).$$ This is more or less just a restatement of the universal property of the limit.

Now if we set $$D=\text{Sets},$$ then $$D^C=\text{Sets}^C=\hat{C}$$ is the category of presheaves on $$C$$. $$\Gamma(F)=\operatorname{Hom}_{\hat{C}}(1,F)$$ is the global sections functor. In this setting we again have an adjunction with the diagonal functor: $$\operatorname{Hom}_{\hat{C}}(\Delta(a),F) = \operatorname{Hom}_{\text{Sets}}(a,\Gamma(F)).$$

By the uniqueness of adjoint functors, we can conclude that $$\Gamma\cong\lim$$, yes? Assuming that's true, that seems rather odd, can I have some context to that isomorphism to make it seem more natural or less surprising? How does it look for enriched presheaves, when $$D$$ is not $$\text{Sets}$$?

Is there some significance to the fact that the limit of a functor is just the set of natural transformations from the constant functor at the terminal object?

## 1 Answer

Your situation is actually a special case of a general fact :

Proposition : If $$G:\mathcal{X}\to \mathbf{Sets}$$ is a functor that has a left adjoint $$F$$, then $$G$$ is represented by $$F(1)$$ (where $$1$$ is just a one-element set).

The proof is easy : just notice that $$\operatorname{Hom}_\mathbf{Sets}(1,\_)$$ is naturally isomorphic to the identity functor on $$\mathbf{Sets}$$, and then the adjunction gives you directly an isomorphism $$G(x)\cong \operatorname{Hom}_\mathbf{Sets}(1,G(x))\cong \operatorname{Hom}_{\mathcal{X}}(F(1),x)$$ natural in $$x$$.

In your case $$G=\lim:\mathcal{X}\to \mathbf{Sets}$$ is right adjoint to $$\Delta$$, thus it must be represented by $$\Delta(1)$$, i.e. is must be isomorphic to $$\operatorname{Hom}_{[C,\mathbf{Sets}]}(\Delta(1),\_)$$, which is how you've defined the global sections functor.

• I guess it was the choice to elide the presence of a diagonal functor in the sheaf notation. All we're really saying is, via the first adjunction, that $\operatorname{Hom}(\Delta 1,F)=\operatorname{Hom}(1,\lim F)$, and then for any set $S$, $\operatorname{Hom}(1,S)\cong S.$ This is just an obvious fact about sets that they are determined by their elements, which we might call the axiom of extensionality or the fact that Set is a cocomplete well-pointed category. I guess my question turned out kind of dumb. – ziggurism Dec 4 '19 at 16:45