Fundamental dimension of a finite polyhedron $P$ is defined as : $$Fd(P)=\min \{ \dim (X):X\; \text{and} \; P \; \text{have the same homotopy type}\}.$$ My question is that: if $A$ is homotopically dominated by $P$ (i.e. there exist $g:P\to A$ and $f:A\to X$ such that $g\circ f\simeq id_A$) which is denoted by $A\leqslant P$ but $P\not \leqslant A$, then $Fd(A)<Fd(P)$?
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$\begingroup$ en.m.wikipedia.org/wiki/Wall%27s_finiteness_obstruction $\endgroup$– Mark GrantCommented Apr 24, 2022 at 17:07
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$\begingroup$ If there exists a finitely dominated space A whose finiteness obstruction is nontrivial, that would give a counter example. $\endgroup$– Mark GrantCommented Apr 24, 2022 at 17:09
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1$\begingroup$ No. $A = S^1$ is a retract of $P = S^1 \vee S^1$, this $P$ is not a retract up to homotopy of this $A$, and the `fundamental dimension' is $1$ for both $A$ and $P$. $\endgroup$– John RognesCommented Apr 24, 2022 at 19:38
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$\begingroup$ @JohnRognes Thank you so much for the example. Is any condition on $P$ under which we have $Fd(A)<Fd(P)$? $\endgroup$– M.RamanaCommented Apr 25, 2022 at 3:49
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$\begingroup$ @MarkGrant Thanks a lot for your comment and the link. $\endgroup$– M.RamanaCommented Apr 25, 2022 at 3:49
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