Given a $m-$sheeted covering map from $p:M^n\to N^n$, where $M,N$ are manifolds of dimension $n$. Suppose $M$ and $N$ are homotopy equivalent to finite CW complexes $X$ and $Y$ of same dimension $k$. Does there exists a $m-$sheeted covering map $q:X\to Y$ so that $pf$ and $gq$ are homotopic, where $f:X\to M$ and $g:Y\to M$ are homotopy equivalences?
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5$\begingroup$ If we get to choose X to be whatever we want, then yes- pull back along the equivalence from Y to N. If not, then no: take M=N=Y=* and take X to be a closed disk. $\endgroup$– Dylan WilsonCommented Sep 23, 2021 at 16:26
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1$\begingroup$ Dylan's X and Y are not of the same dimension, but this is easy to fix; take Y to be the wedge sum of X with a disk of lower dimension. $\endgroup$– Reid BartonCommented Sep 23, 2021 at 16:43
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$\begingroup$ @DylanWilson thanks!! $\endgroup$– piper1967Commented Sep 23, 2021 at 16:59
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