# One-point-functors and the Yoneda Lemma

Let $\mathfrak{Sets}$ be the category of sets, and $\mathfrak{Sch}$ the category of schemes. For any scheme $X$, consider the functor $h_{X}(-)=\mathsf{Hom}_{\mathfrak{Sch}}(-,X):\mathfrak{Sch}\longrightarrow \mathfrak{Sets}$. Consider also the following lemma:

Lemma (Yoneda): Let $C$ be a category, $A$ an object of $C$ and $F:C^{opp} \longrightarrow \mathfrak{Sets}$ a functor. Let $\mathsf{Nat}(h_{A},F)$ be the set of all natural transformations between $h_{A}$ and $F$. Then there exists a one-to-one correspondance $\mathsf{Nat}(h_{A},F)\cong F(A)$.

I would like to show that any scheme $X$ can be recovered from $h_{X}$ up to unique isomorphism. Can this fact be proven using the Yoneda lemma?

• What do you mean by "recovered"? The Yoneda embedding is fully faithful and so conservative in particular. Therefore $X$ is unique up to isomorphism, and even up to equality if you know how to recognise $\mathrm{id}_X$. Commented Oct 10, 2013 at 22:04
• Indeed, I have several times heard Yoneda's lemma casually summarized as, "You are who your friends say you are." The fact that a scheme is completely described by its functor of points is so significant that several important generalizations of schemes (such as algebraic spaces) are defined simply as functors on the category of schemes. Commented Oct 12, 2013 at 1:28
• Why do you restrict to the category to schemes?! Or do you want to construct the ringed space of $X$ from the functor $\hom(-,X)$? Commented Oct 12, 2013 at 10:24

This should come immediately from the Yoneda Lemma. Suppose you have objects $X$ and $Y$. First from Yoneda we know that given any functor $F: C^{opp} \to \textbf{Set}$ that
$$\mathsf{Nat}(h_X,F) = F(X).$$
If you now put $F= h_Y$ then Yoneda in particular says $$\mathsf{Nat}(h_X,h_Y) = h_Y(X) = \mathsf{Hom}(Y,X).$$
So if you had an isomorphism between the functors $h_X$ and $h_Y$ can you use this to construct an isomorphism between $X$ and $Y$?