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Jul 31, 2019 at 11:37 comment added Sunny Can you explain me why the functor $Coh(Y)/M_{Z}(Y) \rightarrow Coh(U)$ is full? For that you should start with morphism $ f $: $\mathcal{F}_{1}$ $\rightarrow$ $\mathcal{F}_{2}$ on $X$ such that $f_{\lvert U} = \phi $. But what are $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ in your proof?
Jul 31, 2019 at 5:14 comment added Sunny It seems very nice proof. Thanks
Jul 31, 2019 at 3:50 history answered dorebell CC BY-SA 4.0