A space homotopy dominated by a wedge of spheres

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?

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.
• Excuse me. Could you explain more? Why $\pi_2 (\phi )$ is a direct summand of $\pi_2 (\phi ')$? Then if this is true, is $\pi_2 (\phi ')$ $\mathbb{Z}\pi_1 (A)$-module which be comparable with $\pi_2 (\phi )$ as $\mathbb{Z}\pi_1 (A)$-module? Sep 13 '17 at 14:34
• (1) Because $\phi$ is a retract of $\phi'$. (2) They are both ${\mathbb Z}[\pi_1(K)]$-modules, and $\pi_1(K)\cong \pi_1(A)$. Sep 13 '17 at 14:55
• I understood that $\phi$ is a retract of $\phi '$. Also, I agree with the fact that $\pi_2 (\phi )$ is a $\mathbb{Z}\pi_1 (K)$-module. Sep 13 '17 at 15:02
• But I don't understand how does $\pi_2 (\phi ')$ is a $\mathbb{Z}\pi_1 (K)$-module? Since maybe $\pi_1 (K)$ and $\pi_1 (X)$ are not isomorphic! Sep 13 '17 at 15:04
• In general, $\pi_1(K)$ acts on the relative homotopy groups $\pi_1(M_f, K)$ for any map $f\colon K \to X$. See, for example, Hatcher's book page 345 Sep 13 '17 at 15:09