Is this 2-complex a $K(\pi,1)$? 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?
 A: 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 $\widetilde X$.
A: 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 $H_2(\widetilde X)=\pi_2(\widetilde X)=\pi_2(X)=\mathbb Z^{119}$.
A: $\newcommand{\Z}{\mathbb{Z}}$
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 $\pi_2(\text{quillen plus construction on } A_5)=\Z/2$, $\pi_2(X)=0$.
