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Let $k$ be a field, let $f \colon X \to Y$ be a morphism of $k$-varieties, and assume $X$ and $Y$ are smooth and projective. Let $H(\_)$ be a classical Weil cohomology theory (i.e. one of $\ell$-adic étale, Betti, algebraic de Rham, crystalline). Of course it depends on $k$ which of the theories are applicable, but I do not want to spell all the cases out here.

What conditions can one impose on $f$ to assert that $f^{*} \colon H^{i}(Y) \to H^{i}(X)$ is injective/surjective?

Currently I can only think of the rather trivial: if $f$ admits a section (resp. retraction), then $f^{*}$ is injective (resp. surjective).

[Edit] For example, is it true that $f^{*}$ is injective if $f$ is dominant? [/Edit]

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Someone may hit this with the CW-hammer if you think a “one property/condition” per answer style would be best. – jmc Jun 24 '14 at 8:53
Do dominant non-surjective morphisms of smooth projective varieties really exist? I would have thought some valuative criterion would get violated. – S. Carnahan Jun 24 '14 at 9:21
Well, over $\mathbb{C}$ I would say that they do not exist. Using the analytic topology, the image of the morphism must be compact (since $X$ is compact and $f \colon X \to Y$ is continuous), hence closed (since $Y$ is Hausdorff). On the other hand, the image contains a dense open set (since the map is dominant), so the image is actually the whole of $Y$ (by a dimension argument). – Francesco Polizzi Jun 24 '14 at 10:02
@FrancescoPolizzi: over any field, the image of a projective variety is closed, and that's all you use, right? (Or did I misunderstand something?) – user5117 Jun 24 '14 at 11:34
@ArtiePrendergast-Smith – Thanks for your comment. In the end I should have realised this myself. Anyway, I didn't know Kleiman's result, so Timo's answer + your comment really answers my question! – jmc Jun 24 '14 at 19:36
up vote 9 down vote accepted

Kleiman, Algebraic Cycles and the Weil Conjectures, Proposition 1.2.4: Let $f: X \to Y$ be surjective. Then $f^*: H^*(Y) \to H^*(X)$ is injective.

Let me recall the proof.

Let $x$ be a closed point of the generic fibre of $f$ and set $Z := \overline{\{z\}} \subseteq X$ and let $z = cl(Z)$ be the cycle class of $Z$. Since the cycle class map commutes with $f_*$, one has $f_*(z) \neq 0$.

Now assume $a \in H^*(Y)$ is in the kernel of $f^*$. Then one has $f^*(ab)z = f^*(a)f^*(b)z = 0$ for every $b \in H^*(Y)$, so by the projection formula $0 = f_*(f^*(ab)z) = abf_*(z)$, so by Poincaré duality $a = 0$.

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Ok, cool. Do you know if dominant is enough? – jmc Jun 24 '14 at 8:55
If $X$ and $Y$ are projective, then dominant implies surjective. – Dan Petersen Jun 24 '14 at 11:45
@DanPetersen – Of course, I should have realised this myself. Thanks! – jmc Jun 24 '14 at 19:32

I do not know any effective, general criterion. But the Lefschetz hyperplane theorem for complex algebraic varieties tells you that if you have a hyperplane section $X\subset Y$ where $Y$ is a complex projective algebraic smooth variety of dimension $n$ such that $Y-X$ is smooth then the induced map $$H^*(Y,\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$$ is an isomorphism for $*<n-1$ and an injection for $*=n-1$.


Edit: Let $Y$ be a smooth complex projective variety and let $D$ be a normal crossing divisor ($dim_{\mathbb{C}}(Y)=n$). If the contraction $Y/D$ is algebraic projective and if the map $Y\rightarrow Y/D$ is also algebraic then $$H^i(Y,\mathbb{C})\rightarrow H^i(D,\mathbb{C})$$ is surjective for $i\geq n$.

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Thanks for your answer! This is indeed a criterion, but not exactly the flavour that I was looking for (no offense!). – jmc Jun 24 '14 at 8:50
You are welcome :-) – David C Jun 24 '14 at 8:56
Sorry, but I don't understand the last paragraph. Are there some missing hypotheses? – Donu Arapura Jun 24 '14 at 16:10
OK, I see. I misunderstood your notation. In algeometer-speak, you're saying: suppose $D$ can be contracted to a point, then... And you're getting this from the decomposition theorem. – Donu Arapura Jun 24 '14 at 18:03

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