Let $\{ X_i\}$ ($i=1,2,\ldots $) be a family finite CW-complexes such that $X_{i+1}$ is homotopy domintaed by $X_i$, i.e. there exists contionuos maps $g_i:X_i \to X_{i+1}$ and $f_i :X_{i+1} \to X_i$ such that $g_i f_i \simeq id_{X_{i+1}}$ for $i=1,2,\ldots$. If $\pi_1 (X_1 )$ is a free group of finite rank, then is there an $i_0$ such that $X_{i}\simeq X_{i_0}$ for $i\geq i_0$? (I mean $X_i$ and $X_{i_0}$ have the same homotopy type).
What I've tried: We have $\pi_1 (g_i)\pi_1 (f_i)=id_{\pi_1 (X_{i+1})}$ and $H_i (\tilde{g})H_i (\tilde{f})=id_{H_i (\tilde{X}_{i+1})}$ for all $i$. Then we have a sequence of retracts $\pi_1 (X_1)\geq \pi_1 (X_2)\geq \cdots $ and $H_i (\tilde{X}_1)\geq H_i (\tilde{X}_2)\geq \cdots $. Since $\pi_1 (X_1 )$ is a free group of finite rank, the sequence $\pi_1 (X_1)\geq \pi_1 (X_2)\geq \cdots $ will stop. But I don't know whether $H_i (\tilde{X}_1)\geq H_i (\tilde{X}_2)\geq \cdots$ will stop or not. The only thing that I know is that $H_i (\tilde{X}_j)$'s are all $\mathbb{Z}\pi_1 (X_1)$-modules. If it stops, then by the Whitehead theorem we have $X_{i}\simeq X_{i_0}$ for some $i_0$.
