Coarse moduli space and rational points

Question

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 $h_M(\bar{K})$. So, the natural morphism from $\mathrm{Spec}(\bar{K})$ 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 $h_M(\bar{K})$ 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?

Answer 1

Yes. Even without assuming $M$ is a coarse moduli space, the existence of a natural transformation $\mathcal{F}\to h_M$ gives rise to a set function $\mathcal{F}(\mathop{\text{Spec}}K)\to h_M(\mathop{\text{Spec}} K)$. In particular, if the domain of this function is non-empty, then so is the co-domain.