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In this post, Joe Johnson 126 mentioned the above fact, which I'm skeptical of. It is well-known that $\pi_n(X^{n+1})=\pi_n(X)$, but being a $K(G,1)$ space doesn't seem to imply the identity in the title. Also, he alluded that using Hurewicz one obtains $\pi_n(X^n,X^{n-1})\simeq\tilde{H}_n(X^n/X^{n-1})$. But this doesn't necessarily follow according to Hatcher's relative version of Hurewicz theorem, as $X^{n-1}$ isn't simply connected.

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This isn't true at all. For instance, let $X=\Delta^3$. Then $X^2=\partial\Delta^3\cong S^2$, so $\pi_3(X^2)=\mathbb{Z}$. But $X$ is contractible, so it is a $K(G,1)$ (with $G$ trivial).

Eric Wofsey
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  • Is it true that $\pi_n(X^{n-1})\rightarrow\pi_n(X^n)$ is trivial? – Enigma Mar 09 '16 at 23:36
  • Thanks for your answer as well as the one posted in the link. – Math.StackExchange Mar 09 '16 at 23:38
  • @Enigma: It's definitely not true for arbitrary $X$ (consider $n=3$ and $X=S^2$, for instance). When $X$ is a $K(G,1)$, it is true. Indeed, passing to the universal cover, you may assume $X$ is contractible, so the inclusion $X^{n-1}\to X$ is nullhomotopic. By cellular approximation, this nullhomotopy can be taken to contained in $X^n$, so the inclusion $X^{n-1}\to X^n$ is nullhomotopic. – Eric Wofsey Mar 09 '16 at 23:48