For local ring $R$, does funcotor $\operatorname{Hom( Spec}R, X)$ characterize scheme $X$? Let $\bf{Sch, Sets, Ring}$ be a category of schemes, sets, commutative rings.
By Yoneda's lemma, scheme $X$ is characterized by contravariant functor
$$\operatorname{Hom}(*, X): \bf{Sch}^{op}\to Sets$$
Now thinking  glueing of schemes by affine schemes, $X$ can be characterized by covariant functor
$$\operatorname{Hom}({\operatorname{ Spec} }(*), X): \bf{Ring}\to Sets$$
I want to know that whether $X$ is characterized by only local rings?
i.e. for schemes $X, Y$, if for all local ring $R$, $\operatorname{Hom}({\operatorname{ Spec} }(R), X)\cong \operatorname{Hom}({\operatorname{ Spec} }(R), Y)$, then $X\cong Y$?
 A: After (a lot of) thinking about it, I think the answer to the specific question you ask is no – I think there are non-isomorphic schemes that yield isomorphic functors $\textbf{LocCRing} \to \textbf{Set}$ – but there is a sense in which the answer to your general question is yes.
First, let me sketch an argument that suggests why the answer to your specific question is no.
Let $X$ be a scheme.
For every point $x$ of $X$, let $\langle x \rangle$ denote $\operatorname{Spec} O_{X, x}$ considered as a scheme over $X$ in the canonical way.
Let $P (X) = \coprod_{x \in X} \langle x \rangle$ and let $R (X) = P (X) \times_X P (X)$.
We have a cofork:
$$R (X) \rightrightarrows P (X) \rightarrow X$$
If this cofork were a coequaliser diagram for every scheme $X$, then it would be the case that $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ embeds the category of schemes fully faithfully into the category of functors $\textbf{LocCRing} \to \textbf{Set}$.
I think the converse is also true but I haven't checked carefully.
Anyway, this is all counterfactual, because the cofork is not always a coequaliser diagram.
Consider $\mathbb{A}^1_k$ and $\mathbb{P}^1_k$ for an algebraically closed field $k$.
It is well known that $P (\mathbb{A}^1_k) \cong P (\mathbb{P}^1_k)$.
Furthermore $R (\mathbb{A}^1_k) \cong R (\mathbb{P}^1_k)$, and we can arrange for these isomorphisms to be compatible with the two projections $R \rightrightarrows P$.
So if both coforks were coequaliser diagrams we would have $\mathbb{A}^1_k \cong \mathbb{P}^1_k$, which we know to be false.
So at least one of these coforks is not a coequaliser.
But your specific question is not quite the same as asking whether $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ is a fully faithful embedding, and your general question can be interpreted in other ways.
For instance, you might ask whether $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ is at least a faithful embedding, and the answer is yes:
Theorem.
The class of local schemes is a separating class in the category of schemes, i.e. given morphisms $f_0, f_1 : X \to Y$ of schemes, if for every local ring $T$ and every morphism $t : \operatorname{Spec} T \to X$ we have $f_0 \circ t = f_1 \circ t$, then $f_0 = f_1$.
Proof.
Let $\mathfrak{m}$ be the maximal ideal of $T$.
So $\mathfrak{m}$ is also the unique closed point of $\operatorname{Spec} T$.
Since the only open neighbourhood of $\mathfrak{m}$ is $\operatorname{Spec} T$ itself, the image of $t : \operatorname{Spec} T \to X$ is contained in every open neighbourhood of $t (\mathfrak{m})$.
Furthermore, every point of $X$ is $t (\mathfrak{m})$ for some choice of $T$ and $t : \operatorname{Spec} T \to X$.
Thus, it is enough to prove the theorem in the case where both $X$ and $Y$ are affine.
But this is an easy consequence of the following lemma.
Lemma.
Let $A$ be a commutative ring and let $a \in A$.
The following are equivalent:

*

*$a = 0$ in $A$.

*For every prime ideal $\mathfrak{p} \subset A$, $a = 0$ in $A_\mathfrak{p}$.

*For every maximal ideal $\mathfrak{m} \subset A$, $a = 0$ in $A_\mathfrak{m}$.

Proof.
The downward implications are trivial.
We verify the upward implication to complete the cycle.
Suppose $a = 0$ in $A_\mathfrak{m}$.
That means there is a $b_\mathfrak{m} \in A \setminus \mathfrak{m}$ such that $a b_\mathfrak{m} = 0$.
Choose such a $b_\mathfrak{m}$ for every maximal ideal $\mathfrak{m} \subset A$.
Consider the ideal generated by $\{ b_\mathfrak{m} : \mathfrak{m} \in \operatorname{MaxSpec} A \}$.
It is not contained in any maximal ideal of $A$, so it must be the unit ideal.
On the other hand, by construction, this ideal is annihilated by $a$.
So $a = 0$.　◼
