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Nov 19, 2020 at 16:33 vote accept user127776
Nov 19, 2020 at 19:31
Nov 19, 2020 at 13:48 comment added user127776 Let us continue this discussion in chat.
Nov 19, 2020 at 4:37 comment added Sasha It seems I forgot to save edits. Now it is done.
Nov 19, 2020 at 4:35 history edited Sasha CC BY-SA 4.0
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Nov 18, 2020 at 22:28 comment added user127776 Oh Ok thanks. You mentioned that you added an explanation for the first sequence. (I cannot find it though, not even in the history of the post!) I understand that the morphism to each direct summand in the first ses is surjective. I don't understand why that map to the direct sum of both is also surjective,
Nov 18, 2020 at 20:35 comment added Sasha This is the morphism from the first to the pushforward of the second.
Nov 18, 2020 at 20:05 comment added user127776 Alright. But then, the two vector bundles $V_2^{\vee}|_{x=0}$ and $V_3^{\vee}|_{x=y=0}$ are not defined on the same space. How is there a map between them?
Nov 18, 2020 at 19:57 comment added Sasha I added an explanation about the first sequence. By $V_3^\vee\vert_{x = y = 0}$ I mean the pullback with respect to the composition $X \times \{x = y = 0\} \to X \times \mathbb{A}^2 \to X$. Note that this map is an isomorphism, so this is essentially you original bundle $V_3$.
Nov 18, 2020 at 19:09 comment added user127776 Thanks for you answer. How did you get the first exact sequence? I know dualizing is left exact and turns the pushout into a pullback. I'm not sure how the surjectivity of your short exact sequence works. By $V_3^{\vee}|_{x=y=0}$ do you mean pukkback from $x=y=0$ to $x=0$? because one is defined on $x=0$ and the other one on $x=y=0$.
Nov 18, 2020 at 17:43 history answered Sasha CC BY-SA 4.0