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]

  • $\begingroup$ Someone may hit this with the CW-hammer if you think a “one property/condition” per answer style would be best. $\endgroup$ – jmc Jun 24 '14 at 8:53
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    $\begingroup$ Do dominant non-surjective morphisms of smooth projective varieties really exist? I would have thought some valuative criterion would get violated. $\endgroup$ – S. Carnahan Jun 24 '14 at 9:21
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    $\begingroup$ 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). $\endgroup$ – Francesco Polizzi Jun 24 '14 at 10:02
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    $\begingroup$ @FrancescoPolizzi: over any field, the image of a projective variety is closed, and that's all you use, right? (Or did I misunderstand something?) $\endgroup$ – user5117 Jun 24 '14 at 11:34
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    $\begingroup$ @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! $\endgroup$ – jmc Jun 24 '14 at 19:36

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

For posterity, and as it was a bit tricky to find a reference, note also that the statement holds for possibly non-projective, quasi-projective smooth varieties, or, more generally for non-compact smooth Kähler manifolds:

Theorem (Wells): Let $f:X\to Y$ be a proper surjective morphism of smooth complex manifolds and assume that $X$ is Kähler. Then the following induced homomorphisms are injective

  1. $f^*:H^q(Y,\Omega^p_Y)\to H^q(X,\Omega^p_X)$
  2. $f^*:H^k(Y,\mathbb{C})\to H^k(X,\mathbb{C})$
  3. $f^*:H^k(Y,\Omega^p_Y(E))\to H^k(X,\Omega^p_X(f^*E))$ for a holomorphic vector bundle $E$.

The reference is Wells Comparison of de Rham and Dolbeault cohomology for proper surjective mappings.

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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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  • $\begingroup$ Thanks for your answer! This is indeed a criterion, but not exactly the flavour that I was looking for (no offense!). $\endgroup$ – jmc Jun 24 '14 at 8:50
  • $\begingroup$ Sorry, but I don't understand the last paragraph. Are there some missing hypotheses? $\endgroup$ – Donu Arapura Jun 24 '14 at 16:10
  • $\begingroup$ 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. $\endgroup$ – Donu Arapura Jun 24 '14 at 18:03

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