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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)

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1 Answer 1

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.

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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

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