Let ‎$‎‎X\_1$ ‎‎be ‎the suspension of ‎$‎‎‎\mathbb{R}P^2‎$ and $X\_2=\bigvee\_{1\leq i\leq n} (\vee\_{r\_i} \mathbb{S}^i)$.


Is $\pi\_2 (X\_i)$ a projective (or a free) $\mathbb{Z}\pi\_1 (X\_i)$-module for $i=1,2$?


I was wondering if someone could help me about my question. I don't have much information about $\pi\_n (X)$ as a $\mathbb{Z}\pi\_1 (X)$.



Is $\pi_2 (X_i)$ a free $\mathbb{Z}\pi_1 (X_i)$-module for $i=1,2$?
No and yes, respectively.


For $X\_1$, the suspension is simply connected, so $\pi\_2(X\_1) = H\_2(X\_1) = H\_1(\mathbb{R}P^2) = \mathbb{Z}/2$ which is not a free $\mathbb{Z}$ module.


For the other one, $\pi\_2(X\_2) = H\_2(\tilde{X}\_2)$ where $\tilde{X}\_2$ is the universal cover. Since $\pi\_1(X\_2)$ is a free group on $r\_1$ letters (if I've understood your notation correctly), $\tilde{X}\_2$ is an infinite tree with a wedge of spheres (of dimension at least $2$) at each lift of the vertex of $X\_2$. Those spheres are permuted by the covering group action of $\pi\_1(X\_2)$, and so the homology of $\tilde{X}\_2$ is free as a $\mathbb{Z}[\pi\_1(X\_2)]$-module, with a fixed lift of each of the spheres (of dimension at least 2) as basis. In particular, $\pi\_2(X\_2)$ is a free $\mathbb{Z}[\pi\_1(X\_2)]$-module on $r\_2$ generators.


Because you can compute $\pi\_2$ as $H\_2$ of the universal cover, it's a lot easier than the higher homotopy groups.


