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just a simple question:

if I have an abelian variety $A$ (choose the base field) and consider the inversion morphism $i: A \rightarrow A$ on it, then this map induces a map on cohomology

$H^{n}(A,\mathbb Z) \rightarrow H^{n}(A,\mathbb Z)$.

by pullback for each n.

As I dont know much topology, it would be nice if someone could explain why this is just the multiplication by $(-1)^{n}$ map.

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For complex cohomology for example, this is easily seen. Take $z_1, \ldots, z_n$ a system of coordinates, then $dz_1\wedge \ldots dz_n$ generates the one-dimensional space $H^n(A, \mathbb C)$, and $i^*dz_1\wedge \ldots dz_n=d(-z_1)\wedge \ldots d(-z_n) = (-1)^n dz_1\wedge \ldots dz_n$. –  Henri Sep 13 '11 at 8:33

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