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Let $f: X \rightarrow Y$ be continuous, X,Y pathwise connected and aspherical (i.e. trivial higher homotopy groups). Then $\pi_1(X)$ acts on the universal cover of $X$ via deck transformations, and on the universal cover of $Y$ via

$$([\sigma],y)\mapsto [\sigma \circ f]y$$

If g is another continuous map whose lift to the universal covers is equivariant under these actions, is g homotopic to f?

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  • $\begingroup$ What happens if only Y is aspherical? $\endgroup$ – user91775 May 17 '16 at 17:37
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In the language of equivariant homotopy theory, your question is as follows. You have a group homomorphism $\phi\colon G\to H$, which you use to make $EH$ into a $G$-space, and then you ask whether $[EG,EH]^G$ is a singleton. In fact, it is true more generally that $[EG,Y]^G$ is a singleton whenever $Y$ is a $G$-space that is nonequivariantly contractible. The reason is as follows. We can write $EG$ as the colimit of a sequence $X_k$, where $X_k$ is obtained from $X_{k-1}$ by attaching $G\times B^{n+1}$ along $G\times S^n$ for some $n$. Thus, to extend an equivariant map $X_{k-1}\to Y$ equivariantly over $X_k$ is the same as to extend a nonequivariant map $S^n\to Y$ nonequivariantly over $B^{n-1}$, and this can be done in an essentially unique way because $Y$ is assumed to be nonequivariantly contractible. (This kind of argument is called "obstruction theory".)

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  • $\begingroup$ This sounds amazing, but do you know a more elementary way to show this, without the language of equivariant homotopy theory? $\endgroup$ – user91775 May 17 '16 at 12:43
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    $\begingroup$ 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 $X$. $\endgroup$ – Tom Goodwillie May 17 '16 at 18:29

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