7
$\begingroup$

I don't know how to show this.

Do I assume $G$ acts on $S^{2n}$ by homeomorphisms? Then, since $S^{2n}$ is Hausdorff I'd know $G$ acts freely and properly discontinuously, and since $\pi_1(S^{2n})={1}$ I'd have $\pi_1(X/G)\cong G$. But I'm not sure whether this is useful.

  • 0
    Why does $S^{2n}$ being Hausdorff imply that $G$ acts freely? Why does $\pi_1(X/G) \cong G$? I don't see how either of your conclusions follow. Anyway, this can be solved by considering the degree of the maps in $G$.2012-04-24
  • 8
    Euler characteristic is multiplicative for covers.2012-04-24
  • 0
    Use Lefschetz fixed point theorem, you'll get all fixed-point-free homomorphism from $S^{2n}$ to itself must be orientation reversing.2012-04-24
  • 3
    Any map $S^{2n}\to S^{2n}$ without fixed points is homotopic to the antipodal map $x\mapsto -x$, and this map has degree $-1$. Since degrees are multiplicative, the only way how all non-$1$ elements of $G$ have degree $-1$ is that there is at most one such element.2012-04-24

2 Answers 2