I have a sheaf $\mathcal F$ over a topological space $X$ (valued in a category $\mathcal C$ over which I don't think I should assume anything except existence of kernels... feel free to assume anything except something like "in this particular category, it works") and $p \in X$. Let $A \in \mathrm{Ob}(\mathcal C)$ and $\varphi : A \to \mathcal F_p$ be a morphism. Is there a sheaf $\mathcal G$ such that
- The stalk of $\mathcal G$ at $p$ is isomorphic to $A$
- We have a morphism of sheaves $\Phi : \mathcal G \to \mathcal F$ such that $\Phi_p = \varphi$.
The point is I was trying to show that in the category of sheaves over $X$, if $\Psi : \mathcal F \to \mathcal H$, then $\ker(\Psi_p) = (\ker \Psi)_p$. So I have a morphism of sheaves $\Psi$ which gives me a map on the stalks $\Psi_p = \psi : \mathcal F_p \to \mathcal H_p$. If $\Psi$ has kernel $\ker \Psi$, then I can take the limit of the maps $\ker \Psi(U)$ for $U \ni p$ to get a map $\varphi : (\ker \Psi)_p \to \mathcal F_p$. I want to show that this arrow has the universal property of a kernel, so I have another object $A \to \mathcal F_p$ ; to make this arrow factor through my object, I wanted a sheaf with stalk $A$ which maps to $\mathcal F$ in a nice way. I got stuck there.