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Let $p:\tilde{X}\rightarrow X$ be the universal covering space such that $p_*$ is zero on all homologies of dimension greater than zero. Does this imply that $X$ is $K(\pi_1(X),1)$? Working with the second homology groups implies that Hurewicz homomorphism $\pi_2(X)\rightarrow H_2(X)$ is zero while $\pi_2(\tilde{X})\rightarrow H_2(\tilde{X})$ is an isomorphism. I cannot derive any contradictions from here. So maybe it is not true.

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    $\begingroup$ excuse me, would you please explain what is your notation $K(\pi_1(X),1)$ means? $\endgroup$ – yoyo May 18 '19 at 3:42
  • $\begingroup$ It is the Eilenberg-Maclane space. The first homotopy group is $\pi_1(X)$ the higher ones are zero. $\endgroup$ – user127776 May 18 '19 at 3:44
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    $\begingroup$ Oh~~is it Eilenberg–MacLane space ? $\endgroup$ – yoyo May 18 '19 at 3:44
  • $\begingroup$ Ok~I got it, thanks $\endgroup$ – yoyo May 18 '19 at 3:45
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    $\begingroup$ I think I found the counterexample. $\mathbb{RP}^{2n}$. $\endgroup$ – user127776 May 18 '19 at 4:01
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A counter-example: $\mathbb{RP}^n$

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