# 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?

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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 $\mathrm{id}_X$. – Zhen Lin Oct 10 '13 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. – Slade Oct 12 '13 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 $\hom(-,X)$? – Martin Brandenburg Oct 12 '13 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 $Y$?