Recall that the space $A$ is homotopy dominated by $X$ if there are maps $f:A\longrightarrow X$ and $g:X\longrightarrow A$ such that $gf\simeq id\_A$. 


Suppose that $X$ is a wedge of some spheres and $A$ homotopy dominated by $X$. Let $K$ be a bouquet of circles so that map $\phi :K\longrightarrow A$ induces an isomorphism on fundamental groups. Put $\pi\_2 (\phi )=\pi\_2 (M,K\times 1)$ where $M$ is the mapping cylinder of $\phi$. I know that $\pi\_2 (\phi )$ is a finitely generated $\mathbb{Z}\pi\_1 (A)$-module. My question is that: how do I prove $\pi\_2 (\phi )$ is free $\mathbb{Z}\pi\_1 (A)$-module?


Thanks in advance.



A space homotopy dominated by a wedge of spheres
The case when $A$ is itself equivalent to a wedge of spheres is easy. In the general case, $\pi\_2(\phi)$ is a direct summand of $\pi\_2(\phi')$, where $\phi'\colon K \to X$ is the map $f\phi$. The result follows because projective modules over the group ring of a free group are free modules, by a [theorem of Hyman Bass](http://ac.els-cdn.com/002186936490016X/1-s2.0-002186936490016X-main.pdf?_tid=8518c33a-988d-11e7-9fb3-00000aab0f01&acdnat=1505312104_4cb454828a91f3fbb299616a3338cb2c).


