Recall that the pseudo projective plane $\mathbb{P}_n =\mathbb{S}^1 \cup_f e^2$ is obtained by attaching a 2-cell $e^2$ to $\mathbb{S}^1$ via the map $f:\mathbb{S}^1 \longrightarrow \mathbb{S}^1$ of degree $n$.
Also, recall that a space $X$ is homotopy dominated by a space $Y$ if there are maps $f:X\longrightarrow Y$ and $g:Y\longrightarrow X$ such that $g\circ f\simeq id_X$?
My question is that:
Is $\mathbb{P}_m$ homotopy dominated by $\mathbb{P}_n$ when $m|n$ and $n\neq m$?
No. Firstly We calculate that $\pi_1P_n=\mathbb{Z}_n$ is cyclic of order $n$. Now assume that $m|n$ and $m\neq n$ and $P_m$ is homotopy dominated by $P_n$ with maps $i:P_m\rightarrow P_n$, $r:P_n\rightarrow P_m$. Then the composite $\pi_1P_m\xrightarrow{i_*}\pi_1P_n\xrightarrow{r_*}\pi_1P_m$ is the identity. But this is a composite $\mathbb{Z}_m\rightarrow \mathbb{Z}_n\rightarrow\mathbb{Z}_m$ which must be trivial when, say $m=p$, $n=p^r$ for some prime $p$. Hence $P_n$ does not homotopy dominate $P_m$ in this case.