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Mar 7, 2013 at 9:21 comment added Russell @Chris, it looks like I'm wearing my dunce cap tonight, having forgotten that $2 id =0$ in $\mathbb{Z}_{2}$.
Mar 7, 2013 at 9:19 history edited Russell CC BY-SA 3.0
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Mar 7, 2013 at 8:33 vote accept Juan
Mar 7, 2013 at 8:48
Mar 7, 2013 at 8:33 comment added Chris Gerig @Russell, this still doesn't work because $|\mathbb{Z}_2|$ will kill the 2-torsion, and so the conclusion would be satisfied trivially.
Mar 7, 2013 at 8:31 comment added Chris Gerig @Juan that is not enough, because his original example had $S^{2k}$ orientable while $\mathbb{R}P^{2k}$ is nonorientable! You need a condition on the action (luckily for $S^{2k+1}$ the antipodal action has orientable quotient).
Mar 7, 2013 at 7:49 history edited Russell CC BY-SA 3.0
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Mar 7, 2013 at 7:46 comment added Juan Thank you for the answer, Russell. I just added orientability in my question. I would appreciate your counter-example.
Mar 7, 2013 at 7:42 comment added Russell Ahh, good point! I'll modify the example so that everything is orientable.
Mar 7, 2013 at 7:37 comment added Chris Gerig Be careful, we're invoking Poincare-duality and so your spaces have to be orientable. I assume the question wants $X$ oriented and $\pi$ to be orientation-preserving.
Mar 7, 2013 at 7:32 history answered Russell CC BY-SA 3.0