MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:X \to Y$ be a finite map of (smooth, compact) complex algebraic varieties.

Then a map $f^*$ is defined at the level of Chow and cohomology rings. Say that for simplicity we work with rational coefficients.

Question: is $f^*$ injective? I know this is true when $f$ is the natural map onto the quotient via the action of a finite group. Indeed, it is simply the inclusion of the $G$-invariants.

In the case when $X$ and $Y$ are smooth and projective, is $f^*$ injective at the level of each individual $H^{p,q}$?

(in the case of cohomology, Proposition 2.2 of answers affirmatively my question, but I would anyway be curious of seeing a simpler proof in my generality)

share|cite|improve this question
up vote 11 down vote accepted

Let $f:X\to Y$ be a finite surjective map of degree $d$ between smooth projective varieties of dimension $n$. Then $f$ is flat (apply EGA IV 2 Prop 6.1.5), so Example 1.7.4 of Fulton's Intersection Theory shows that $f_{\ast}\circ f^{\ast}$ is multiplication by $d$ on the Chow groups $CH^*(Y)$. Hence the kernel of $f^{\ast}$ on $CH^{\ast}(Y)$ is of $d$-torsion.

It remains true that $f_{\ast}\circ f^{\ast}$ is multiplication by $d$ on the integral cohomology of $Y$, hence that the kernel of $f^{\ast}$ on $H^{\ast}(Y,\mathbb{Z})$ is of $d$-torsion. It may be seen as follows. Recall that if $\alpha\in H^{k}(Y)$ and $\beta\in H^{k}(X)$, their Poincaré-dual classes are $\alpha\cap [Y]\in H_{n-k}(Y)$ and $\beta\cap [X]\in H_{n-k}(X)$ and that $f_*:H^{k}(X)\to H^{k}(Y)$ is defined to be the push-forward at the level of Poincaré-dual classes. Combining the projection formula $f_{\ast}(f^{\ast}\alpha\cap[X])=\alpha\cap f_{\ast}[X]$ and the fact that $f_{\ast}[X]=d[Y]$ because $f$ is a degree $d$ map, we get the formula $f_{\ast}\circ f^{\ast}(\alpha)= d\alpha$ we wanted to prove.

This shows that $f^{\ast}$ is injective on Chow groups modulo torsion, on integral cohomology modulo torsion, on rational cohomology, on cohomology with complex coefficients... This last point moreover implies that $f^{\ast}$ is injective on the pieces $H^{p,q}(Y)$ of the Hodge decomposition.

However, $f^{\ast}$ need not be injective on the integral cohomology. If $Y$ is an Enriques surface and $X$ is its universal cover (a K3 surface), $H^2(Y,\mathbb{Z})_{tors}=\mathbb{Z}/2\mathbb{Z}$ and $H^2(X,\mathbb{Z})$ has no torsion so that $f^{\ast}$ is not injective on $H^2$.

Torsion in cohomology is not the only reason why $f^{\ast}$ may fail to be injective on $CH^{\ast}(Y)$. For example, let $X$ be an elliptic curve. Suppose that its $2$-torsion is generated by $\sigma$ and $\tau$, and let $Y=X/\sigma$. We still denote by $\tau$ the image of $\tau$ in $Y$. Then, working in $CH^1=Pic$, we get $f^*(\tau-0)=\tau-0+\tau+\sigma-0-\sigma=2\tau-2.0=0$, so that $f^{\ast}$ is not injective on $CH^1$.

When dealing with $0$-cycles, it is possible to apply the theorem of Roitman. For example, if $H^1(Y,\mathbb{C})=0$, $f^*$ is injective on $CH_0$. Indeed, the Albanese variety of $Y$ is then trivial, and, by Roitman, $CH_0$ has no torsion.

share|cite|improve this answer
Thanks for the very nice answer. Is it also easy to see that $f^*$ is injective at the level of rational cohomology and at the level of rational algebraic cohomology? – calc May 13 '12 at 19:18
Yes it is : the kernel of $f^{\ast}$ is torsion, so that if you look at a "cohomology theory" without torsion, such as rational cohomology, $f^{\ast}$ will be injective. – Olivier Benoist May 14 '12 at 8:17
As for rational cohomology, the injectivity of $f$ is automatic as soon as $f$ is surjective (not necessarily finite), cf the second paragraph of… – Henri May 14 '12 at 9:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.