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May 22, 2016 at 18:51 vote accept user91775
May 17, 2016 at 18:29 comment added Tom Goodwillie Let me restate this in possibly friendlier language. Let $G$ act freely on a space $\tilde X$ in such a way that this makes it a covering space of a space $X$. Also assume that $X$ has a CW structure (a triangulation, if you prefer). If $Y$ is a space with any action of the same $G$, and if $Y$ is contractible (or even weakly contractible: every map of a sphere $S^n$ to $Y$ extends to the ball $B^{n+1}$, for all $n\ge -1$). Then any two $G$-equivariant maps $\tilde X\to Y$ are equivariantly homotopic. The proof is basically by constructing the homotopy one cell at a time in $X$.
May 17, 2016 at 12:43 comment added user91775 This sounds amazing, but do you know a more elementary way to show this, without the language of equivariant homotopy theory?
May 17, 2016 at 10:20 history answered Neil Strickland CC BY-SA 3.0