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.