3
$\begingroup$

Consider a surjective map $f\colon X\to Y$ of smooth projective varieties. It is well known (see e.g. Voisin's Hodge theory I, Lemma 7.28) that the map $H^i(Y,\mathbb Q)\to H^i(X,\mathbb Q)$ is injective.

Question: Can we replace $\mathbb Q_Y$ with some other sheaf $L$ in this statement? Namely, is it true that the map $$ H^i(Y,L)\to H^i(X,f^*L) $$ is injective for, say, $L$ local system or of geometric origin, or constructible?

Remark: $X$ and $Y$ are over $\mathbb C$ and cohomology is Betti cohomology.

Improve this question
$\endgroup$
6
  • $\begingroup$ Say $i=1$. What if $H^1$ vanishes for $X$? But I'm not an algebraic geometer, so I don't have good intuition $\endgroup$
    – David Roberts ♦
    Commented Apr 9 at 7:08
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ben
    Commented Apr 9 at 8:25
  • 1
    $\begingroup$ @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. $\endgroup$
    – Lazzaro Campeotti
    Commented Apr 9 at 8:36
  • $\begingroup$ You're right, @LazzaroCampeotti, thank you. Then let's just say that the answer is yes for vector bundles. $\endgroup$
    – Ben
    Commented Apr 9 at 8:41
  • 5
    $\begingroup$ 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 $\mathbf{Q}_Y$). $\endgroup$
    – Anonymous
    Commented Apr 9 at 10:47

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .