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Are two maps from a path connected space to itself inducing the same automorphism on the fundamental group freely homotopic?

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    $\begingroup$ Any simply connected but not contractible space will give a counterexample: identity vs constant map. $\endgroup$ – freakish Jul 17 '18 at 14:16
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Not necessarily. For example, every map $f : S^2 \to S^2$ induces the same map on $\pi_1(S^2) = 0$ but $[S^2, S^2] \cong \mathbb{Z}$.

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  • $\begingroup$ And is it true if the fundamental group is free? $\endgroup$ – B.nia Jul 17 '18 at 14:17
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    $\begingroup$ @B.nia No, you take $X=S^2\vee S^1$ and $f, g:X\to X$ where $f$ is the identity and $g$ is identity on $S^1$ while constant on $S^2$. These two maps induce identity on $\pi_1(X)\simeq\mathbb{Z}$ but are not homotopic (they induce different maps on $\pi_2$). In a similar manner you construct a counterexample for any (fundamental) group $G$, by taking its Eilenberg-MacLane space and joining it with $S^2$. $\endgroup$ – freakish Jul 17 '18 at 14:25
  • $\begingroup$ More interesting (and a very hard) question would be: what if $f,g:X\to X$ induce the same morphism on every homotopy group. I do not have an answer to that question but I guess there's a counterexample as well. $\endgroup$ – freakish Jul 17 '18 at 14:30
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    $\begingroup$ @MichaelAlbanese Oh, the trivial counterexample is a topological space that has all homotopy groups trivial but is not contractible, e.g. the quasi-circle from Hatcher. $\endgroup$ – freakish Jul 17 '18 at 14:38
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    $\begingroup$ For an example within CW complexes, just modify Michael’s example slightly: Consider the composition $K(Z,2)\times K(Z,4) \to K(Z,2) \to K(Z,4) \to K(Z,2)\times K(Z,4)$ where the maps are projection, squaring, then inclusion. This map induces the zero map on all homotopy groups, but is not nullhomotopic. $\endgroup$ – Aleksandar Milivojevic Jul 17 '18 at 15:14

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