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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 $n\neq m$?

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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.

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  • $\begingroup$ Thank you for your answer. That's true. Indeed you use the fact that $\mathbb{Z}_{p^r}$ for some prime $p$ and $r\in \mathbb{N}$, is indecomposable. But what is the answer for other cases like $\mathbb{Z}_6$ or such ones? $\endgroup$ – M.Ramana Dec 2 '17 at 10:34
  • $\begingroup$ I don't know how much can be said for the general case. Are you aware of the papers by Olum and Rutter detailing the self-equivalences of the pseudo-projective planes? These self-equivalence groups have a very rich structure. Although in favourable cases there are obvious maps $P_m\rightarrow P_n\rightarrow P_m$ which could potentially compose to something homotopic to the identity, proving in the general case that it is not some non-trivial homotopy self-equivalence would be difficult. $\endgroup$ – Tyrone Dec 2 '17 at 16:01
  • $\begingroup$ Yes. I read this paper but I coudn't use it properly to find my answer. I understood. Thank you very much for the comment and your help. $\endgroup$ – M.Ramana Dec 3 '17 at 4:07

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