Let $X$ be a finite wedge of $m$-spheres containing some circles.
Is $\pi_n (X)$ a free $\mathbb{Z}\pi_1 (X)$-module, for all $n\geq 2$?
Concerning homotopy groups of a finite wedge of spheres
algebraic-topologyhigher-homotopy-groupshomotopy-theory
algebraic-topologyhigher-homotopy-groupshomotopy-theory
Let $X$ be a finite wedge of $m$-spheres containing some circles.
Is $\pi_n (X)$ a free $\mathbb{Z}\pi_1 (X)$-module, for all $n\geq 2$?
Best Answer
No, $\pi_n(X)$ is not even a free $\mathbb{Z}$-module for most values of $n$. Indeed, if $m>1$, then $\pi_n(S^m)$ has torsion for infinitely many values of $n$. Since each of the spheres in $X$ is a retract of $X$, this means that if any of those spheres have dimension greater than $1$, then $\pi_n(X)$ has torsion for infinitely many $n$.