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Consider $CW$-complex $X$ obtained from $S^1\vee S^1$ by glueing two $2$-cells by the loops $a^5b^{-3}$ and $b^3(ab)^{-2}$. As we can see in Hatcher (p. 142), abelianisation of $\pi_1(X)$ is trivial, so we have $\widetilde H_i(X)=0$ for all $i$. And if $\pi_2(X)=0$, we have that $X$ is $K(\pi,1)$.

My question is: how can one compute $\pi_2(X)$? Computing homotopy groups is hard, what methods may i use?

Andrey Ryabichev
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    WHY $\pi_2=0$ IMPLIES $X$ is $K(\pi,1)$ ? – Anubhav Mukherjee Jan 17 '16 at 16:19
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    @Anubhav.K, 'cause universal covering of $X$ is 2-dimensional without $\pi_2$, so it's aspherical – Andrey Ryabichev Jan 17 '16 at 16:23
  • not true. see my answer below(its an answer because I couldnt fit it into a comment0 – Hari Rau-Murthy Aug 06 '16 at 02:23
  • @HariRau-Murthy mayby, in your example $X$ is not $2$-dimensional $CW$-complex? in such case we really have $X=K(\pi,1)$ because of $\widetilde X$ has the same homology as a point – Andrey Ryabichev Aug 06 '16 at 09:03
  • Sorry for being thick :(. Yeah. The $X$ in your example, is a 2-dimensional CW complex so $\pi_i(X)$ is automatically 0 for $i\geq 3$. So $\pi_2(X)=0 \Rightarrow X$ is aspherical. – Hari Rau-Murthy Aug 06 '16 at 19:03
  • I somehow thought, for reasons that are no longer clear, that you had thought that this space being acyclic+(another condition) would imply that $X$ is an eilenberg maclane space. But obviously you would never do that. I'll leave my answer up though, because I still like my own example :). – Hari Rau-Murthy Aug 06 '16 at 19:04
  • relevant: http://math.stackexchange.com/questions/84422/finite-dimensional-eilenberg-maclane-spaces – Hari Rau-Murthy Aug 07 '16 at 16:44

3 Answers3

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Letting $\widetilde X$ be the universal cover of $X$ we have $$\pi_2(X) \approx \pi_2(\widetilde X) \approx H_2(\widetilde X,\mathbb{Z}) $$ The first isomorphism comes from the long exact sequence of homotopy groups of a fibration, using discreteness of the fiber of the universal covering map. The second isomorphism comes from the Hurewicz theorem, using simple connectivity of $\widetilde X$.

Lee Mosher
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  • but why $H_2(\hat{X} , \mathbb Z)=0$ ?? here OP asked whether $X$ is $KG,1)$ or not?? How can we conclude anything about $H_2(\hat{X} , \mathbb Z)$ – Anubhav Mukherjee Jan 17 '16 at 17:06
  • dear ee Mosher, here you don't answer my question, but i already can do it (after weeks of trying). maybe, equality that i think out is actually what you mean? – Andrey Ryabichev Jan 17 '16 at 17:21
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    This an answer to the question of how to compute $\pi_2(X)$ in this situation. The general answer is, calculate $H_2(\widetilde X, \mathbb{Z})$. Whether this is helpful, will depend on your ability to work constructively in $\widetilde X$, as in the other answer @Anubhav.K – Lee Mosher Jan 17 '16 at 17:47
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No, $\pi_2(X)=\mathbb Z^{119}$, is is obvious by considering Euler characteristic.

Denote universal cover of $X$ by $\widetilde X$. In Hatcher we see (and i don't know how to prove it) that $\pi_1(X)$ has order $120$. So $\widetilde X$ has 120 $0$-cells, 240 $1$-cells, and 240 $2$-cells; therefore $\chi(\widetilde X)=120$.

We know that $H_0(\widetilde X)=\mathbb Z$, $H_1(\widetilde X)=0$ and $H_2(\widetilde X)$ has no torsion. So $H_2(\widetilde X)=\pi_2(\widetilde X)=\pi_2(X)=\mathbb Z^{119}$.

Andrey Ryabichev
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$\newcommand{\Z}{\mathbb{Z}}$ The statement $\pi_2(X)=0$ would imply that $X$ is a $K(\pi,1)$, is not true. I will construct for you now an acyclic space,(the space $X$ below) with $\pi_2=0$, that is not an eilenberg maclane space.

Let $2I$ be the $\Z/2$ extension of the icosahedral group, $I$, given by the preimage of $I$ under $spin(3) \to SO(3)$(FYI spin(3) is the connected double cover of $SO(3)$). Since $I$ is perfect,i.e. $H_1(I)=0$, the hoschild-serre spectral sequence of this extension implies that $H_1(2I)$ is perfect. Therefore by the Quillen plus construction there is a space with $X$ with $\tilde H_*(X)=0$, and $\pi_1(X)=2I$. Since $2I$ is not an acyclic group, $K(2I,1) \neq X$ in the homotopy category. But if $X$ were an eilenberg maclane space it would have to be $K(2I,1)$. Therefore $X$ is not an eilenberg maclane space. Furthermore, since $\pi_2(\text{quillen plus construction on } A_5)=\Z/2$, $\pi_2(X)=0$.

Hari Rau-Murthy
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