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Let $K$ be a field (not necessarily algraically closed). Let $\mathcal{F}$ be a contravariant functor from the category of schemes over $K$ to sets and $M$ be a coase moduli space for the functor. So, there is a morphism of contravariant functors from $\mathcal{F}$ to $h_M$ where $h_M:=Hom(−,M)$ such that $\mathcal{F}(\bar{K})$ is bijective to $h_M(\bar{K})$, where $\bar{K}$ is an algebraic closure of $K$. So, the natural morphism from $\mbox{Spec}(\bar{K})$ to $\mbox{Spec}(K)$ gives rise to a natural transformation. As far as I understand this means that if $\mathcal{F}(K)$ is non-empty which maps to something non-empty in $\mathcal{F}(\bar{K})$ the image of $\mathcal{F}(K)$ in $h _M(\bar{K})$ under the composition of the morphism $\mathcal{F}(K) \to \mathcal{F}(\bar{K}) \xrightarrow{\sim} h_M(\bar{K})$ is non-empty (due to the bijectivity over $\bar{K}$). By the commutativity of the diagram (coming from the natural transformation) this means that $h_M(K)$ is non-empty. Is this correct?

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closed as off-topic by Neil Strickland, Daniel Loughran, Steven Sam, Stefan Kohl, abx Jun 8 '14 at 21:23

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Can't you just say that you have a map $\mathcal{F}(K)\rightarrow h_M(K)$, so $h_M(K)$ is non-empty if $\mathcal{F}(K)$ is non-empty? – abx Jun 8 '14 at 8:06
Indeed, that was the answer given at the duplicate on MSE. – Zhen Lin Jun 8 '14 at 8:31
@Lin, abx: However, I do not entirely understand the answer. My confusion is that $h_M(K)$ contains the empty set. So, why cant $\mathcal{F}(K)$ be that? – user45397 Jun 8 '14 at 10:36