Let $X$ be a variety over a field $k$ (we always assume $k=\bar k$, but I think this doesn't matter here), and let $k_p$ denote the skyscraper sheaf on $X$ w.r.t. the point $p\in X$. I want to find all the morphisms from the structure sheaf $\mathcal{O}_X$ to $k_p$.
I know that the answer is that there are only the evaluation morphisms, i.e., morphisms of the form $\phi\mapsto\phi(p)$ on open subsets that contain $p$, but I don't really see why.
First of all, wouldn't I need to know which kind of morphisms I am looking for? That is, morphisms of sheaves of abelian groups, or $\mathcal{O}_X$-modules, or... But from the very short solution I have here, I think that the $\mathcal{O}_X$-module one is being searched for.
So let $U$ be an open subset which contains $p$ (otherwise $k_p(U)=0$ and there is only the zero map). Then we took $\frac{g}{h}\in\mathcal{O}_X(U)$, and if $f$ is the wanted morphism we must have $f_U(\frac{g}{h})=\frac{g}{h}(p)\cdot f_U(1_U)=\frac{g}{h}(p)\cdot f_X(1_X)$. I really don't quite understand this calculation.
My guess: the $\mathcal{O}_X$-module structure on $k_p$ should be given by $\mathcal{O}_X(U)\to k_p(U)=k$, $\phi\mapsto\phi(p)$, and then we'd have to pull out $\frac{g}{h}$ as above. But why is this the 'canonical' structure (I searched my notes for something regarding this, but there is nothing which mentions this structure)? And why do we have to use that $f_U(1)=f_X(1_X)$ at all?
Thanks in advance!