Suppose that a space $A$ homotopy dominated by a space $X$. i.e., there exist continuous maps $f:A\longrightarrow X$ and $g:X\longrightarrow A$ so that $g\circ f\simeq 1_A$. Also, let $\phi :K\longrightarrow A$ be a continuous map. Put $\phi'=f\circ \phi:K\longrightarrow X$. We know that $M_{\phi}\simeq A$ and $M_{\phi'}\simeq X$, where $M_{\phi}=\frac{K\times I \cup A}{(k,1)\sim \phi (k)}$ is the mapping cylinder of $\phi$. Also, it is a well-known fact that $\pi_1 (K)$ acts on the whole long exact sequence of homotopy groups for $(M_{\phi},K\times \{ 1\})$ and $(M_{\phi'},K\times \{ 1\})$, the action commuting with the various maps in the sequence.
My question is :
Is $\pi_2 (M_{\phi},K\times \{ 1\})$ a direct summand of $\pi_2 (M_{\phi'},K\times \{ 1\})$ as $\mathbb{Z}\pi_1 (K)$-module?
Consider $h: M_\phi \to M_{\phi’}$ which agrees with $f$ on $A$ and is the identity on $K \times I$; $h$ is well-defined by definition of $\phi’$. Then the homotopy equivalences of the mapping cylinders to their bases shows that $h$ has a left homotopy inverse, since $f$ does. Hence, $h$ induces an inclusion of a direct summand $h_\ast: \pi_2(M_\phi, K) \to \pi_2(M_{\phi’}, K)$.
This $h_\ast$ should be equivariant with respect to the action of $\pi_1(K)$, although I’m not sure how to show this at the moment.