Let $A$ be a topological space homotopy dominated by a space $X$; i.e. there exist continuous maps $f:A\longrightarrow X$ and $g:X\longrightarrow A$ so that $g\circ f\simeq id_A$.
Assume that $\tilde{X}$ denotes the universal covering space of $X$. By a well-known fact, $H_i (\tilde{X})$ and $H_i (\tilde{A})$ are $\mathbb{Z}\pi_1 (X)$-module and $\mathbb{Z}\pi_1 (A)$-module, respectively, for all $i$.
My question is that:
Is $Im\Big(H_i (\tilde{f})\Big)$ a $\mathbb{Z}\pi_1 (X)$-submodule of $H_i (\tilde{X})$?
Note that $\tilde{f}:\tilde{A}\longrightarrow \tilde{X}$ is an induced map so that $p\circ \tilde{f}=f\circ q$ in which $p:\tilde{X}\longrightarrow X$ and $q:\tilde{A}\longrightarrow A$ are covering spaces.
No. Take the example of $A=\Bbb R\Bbb P ^2$ and $\tilde A=S^2,$ while $X=\Bbb R\Bbb P ^2\!\vee\Bbb R\Bbb P ^2$ (wedge product) and $\tilde X$ is an infinite wedge chain of $S^2$ (indexed by $\Bbb Z$) where even spheres are mapped to one projective plane and odd spheres are mapped to the other.
Now Im$(H_2(\tilde f))$ is the $H_2$ of one sphere in $\tilde X$. However, the image of that one sphere by the action of $ab\in \pi_1(X)$ will be another sphere two spots further (calling $a$ and $b$ the generators of the fundamental groups of each projective plane in $X$). Hence Im$(H_2(\tilde f))$ is not stable by the action of $\pi_1(X)$.