Timeline for Is pullback map on sheaf cohomology injective for surjective morphisms?
Current License: CC BY-SA 4.0
9 events
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Apr 9 at 11:10 | comment | added | cll | sounds right @Anonymous, thank you! | |
Apr 9 at 11:03 | history | edited | cll | CC BY-SA 4.0 |
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Apr 9 at 10:47 | comment | added | Anonymous | By the projection formula, it is enough to show that $\mathbf{Q}_Y \to Rf_* \mathbf{Q}_X$ is split (i.e., has a left inverse). This follows, e.g., by the decomposition theorem (or by slicing to the case of generically finite $f$ and using Poincare duality as well as the fact that multiplication by the generic degree $f$ is an isomorphism on $\mathbf{Q}_Y$). | |
Apr 9 at 8:41 | comment | added | Ben | You're right, @LazzaroCampeotti, thank you. Then let's just say that the answer is yes for vector bundles. | |
Apr 9 at 8:36 | comment | added | Lazzaro Campeotti | @Ben: the theorem of Wells in the linked answer applies to the case where $L$ is a vector bundle. But the OP is asking about other kinds of sheaves. | |
Apr 9 at 8:25 | comment | added | Ben | Although the question itself is not a clear duplicate, there is an answer which applies to this question as well: mathoverflow.net/a/317691/15782 The answer is yes, by a theorem due to Wells. | |
Apr 9 at 7:44 | history | edited | gmvh | CC BY-SA 4.0 |
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Apr 9 at 7:08 | comment | added | David Roberts♦ | Say $i=1$. What if $H^1$ vanishes for $X$? But I'm not an algebraic geometer, so I don't have good intuition | |
Apr 9 at 2:26 | history | asked | cll | CC BY-SA 4.0 |