Suppose that $Z$ is a finite wedge of spheres containing circles and there exist maps $f:Y\to Z$ and $g:Z\to Y$ so that $g\circ f\simeq 1\_Y$. Assume that there exists a map $h:X\to Y$ which induces an isomorphism on fundamental groups, where $X$ is a finite wedge of circles. 


Is $\pi\_2 (M\_{h},X)$ a projective (or a free) $\mathbb{Z}\pi\_1 (X)$-module? 


Here $M\_h =\frac{X\times I\sqcup Y}{(x,1)\sim h(x)}$ denotes the mapping cylinder of $h$. 



A projective (or free) $\mathbb{Z}\pi_1$-module
Based on your comments to this [answer](https://mathoverflow.net/a/303345/94123), you seem to be in the situation where $\pi\_2(Z)$ is a free $\mathbb{Z}\pi\_1(X)$-module, and where the map $\pi\_1(X)\to \pi\_1(Y)$ is an iso. Under those assumptions, the answer should be a consequence of the following diagram, coming from the map between the mapping cylinders :


$$\require{AMScd}\begin{CD}
0@>>>\pi\_2(Y)@>\varphi\_3>>\pi\_2(M\_h,X)@>\varphi\_2=0>>\pi\_1(X)@>\varphi\_1>>\pi\_1(Y)\\
@VVV@VVV@VVV@VVV@VVV\\
0@>>>\pi\_2(Z)@>>>\pi\_2(M\_{f\circ h},X)@>\varphi'\_2>>\pi\_1(X)@>>>\pi\_1(Z)\\
@VVV@VVV@VVV@VVV@VVV\\
0@>>>\pi\_2(Y)@>\varphi\_3>>\pi\_2(M\_h,X)@>\varphi\_2=0>>\pi\_1(X)@>\varphi\_1>>\pi\_1(Y)\\
\end{CD}$$


Now, by assumption, $\varphi\_1$ is an iso. This implies that its kernel is $0$, which means that $\varphi\_2=0$. This means in turn that $\varphi\_3$ is an iso. But then, by assumption $\pi\_2(Z)$ is a free $\mathbb{Z}\pi\_1(X)$-module, and so $\pi\_2(M\_h,X)\simeq\pi\_2(Y)$ is a projective $\mathbb{Z}\pi\_1(X)$-module since it is a direct summand of $\pi\_2(Z)$.


In addition, using the commutativity of the diagram, we get that $\varphi'\_2=0$, which means that $\pi\_2(M\_{f\circ h},X)$ is isomorphic to $\pi\_2(Z)$, and in particular it is a free $\mathbb{Z}\pi\_1(X)$-module.


