Let $k$ be an Algebraically closed field. Let $\mathcal C_k$ be the category of integral $k$-schemes of finite type over $k$ (the morphisms between two objects being the morphism of schemes that also commutes with the structure morphism to $k$) . Let $\mathcal D_k$ be the full subcategory (so the morphisms remain the same) of $\mathcal C_k$ whose objects have Krull dimension $0$. So $Z \in \mathcal D_k$ if and only if $Z \cong Spec (k)$.
Now let $X, Y\in \mathcal C_k$ be such that the functors $Hom_k (-,X) $ and $Hom_k (-,Y)$ are isomorphic when considered from $\mathcal D_k \to Set$ . Then is it true that $X \cong Y$ ?
Now I know that $Hom_k (-,X), Hom_k (-,Y) : \mathcal D_k \to Set$ are isomorphic means that $ Hom _k (Spec (k), X)$ and $Hom_k (Spec(k), Y)$ are bijective as sets.
Now if I knew that $ Hom _k (Spec (k), X) \cong Hom_k (Spec(k), Y)$ as locally ringed spaces (where I identify $ Hom _k (Spec (k), X) $ with the set of closed points $X(k)$ of $X$ and the structure sheaf I give is the restriction of that of $X$) , then I would be able to say $X \cong Y$ . Unfortunately, I'm not sure whether I can say that ... Please help
Also note that Yoneda Lemma doesn't apply since $X,Y$ are not necessarily in $\mathcal D_k$
Your $\mathcal D_k$ is equivalent to the category with one object and one morphism, since the only endomorphism of $\mathrm{Spec} k$ over itself is the identity and all objects of $\mathcal D_k$ are isomorphic. Thus an isomorphism of functors on $\mathcal D_k$ is nothing more than a bijection of sets; any two schemes with the same cardinality of their sets of $k$-points are isomorphic in this sense, which is certainly much weaker than an isomorphism of schemes.