I know that it holds even if $X$ has infinite dimension, but I am looking for a specific argument in the finite-dimensional case.
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$\begingroup$ Many such questions boil down to the following trick. Problem: Given an open subset of the $n$-skeleton. Can we construct an open subset of the $n+1$-skeleton such that the intersection with the $n$-skeleton is the given set? So we choose attaching pushouts and we leave out the midpoints of the $n+1$- cells. Then we have a retraction of the $n+1$-skeleton without the midpoints to the $n$-skeleton. Now we can take the preimage of the given set. $\endgroup$– HenrikRüpingNov 10, 2020 at 19:34
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The classical (and short) argument is the one in
A. Hatcher: Algebraic topology, Cambridge University Press (ISBN 0-521-79540-0/pbk). xii, 544 p. (2002). ZBL1044.55001,
see Proposition A.4 p. 523. I do not think there is any significant simplification taking $X$ finite-dimensional.
answered Nov 10, 2020 at 9:44