# Universal covering that induces zero on homologies

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.

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

A counter-example: $$\mathbb{RP}^n$$