I'm not the first one asking for clarifications on the subject.
But my question is slightly different, and possibly easier to answer.
Suppose $\varphi\colon \mathcal{F}\to \mathcal{G}$ be a morphism of sheaves on a topological space $X$, so that $V\subseteq U$ (inclusion of open subsets) implies $\varphi(V)\circ \rho^\mathcal{F}_{UV}=\rho^\mathcal{G}_{UV}\circ \varphi(U)$, where $\rho^\mathcal{F}_{UV}\colon \mathcal{F}(U)\to\mathcal{F}(V)$, $\rho^\mathcal{G}_{UV}\colon \mathcal{G}(U)\to\mathcal{G}(V)$ restriction maps.
The kernel presheaf of $\varphi$ works as follows: $U\mapsto \mathrm{Ker}(\varphi(U))\subseteq \mathcal{F}(U)$.
This means that $V\subseteq U$ implies $\rho^\mathcal{F}_{UV}\colon \mathrm{Ker}(\varphi(U))\to \mathrm{Ker}(\varphi(V))$, restriction to the kernel of the above restriction map.
So I would like to prove that $\rho^\mathcal{F}_{UV}(\mathrm{Ker}(\varphi(U)))=\mathrm{Ker}(\varphi(V))$.
I easily proved $(\subseteq)$, but I have no idea for the converse $(\supseteq)$.
Can someone help me, please?
Best Answer
it's not true unless $\mathcal{F}$ is special since it's not you are just free not to verify