I am new to algebraic topology and I need some help. I am having some difficulties to prove the following statement, which I did not find in different standard textbooks:
Let X, Y be a CW-Complexes such that X has a filtration into subcomplexes:
and similarly for Y. Let f: X --> Y be a continuous map such that for each n the restriction f: Xn --> Yn is an homotopy equivalence. Then f is an homotopy equivalence.
Could you please help me or tell me where I could find this?
It is useful to have the following result, which is a version of 7.4.2 and its Addendum of Topology and Groupoids. This result was found in the 1960s by generalising the well known result that if $f:Y \to Z$ is a homotopy equivalence, then $f$ induces an isomorphism of homotopy groups: the issue is that homotopy groups are defined only for pointed spaces, but we do not assume $f$ is a homotopy equivalence of pointed spaces.
Let $(f,g):(Y,A) \to (Z,B)$ be a map of pairs of spaces such that the inclusions $i:A\to Y$, $j:B \to Z$ are cofibrations, and $f:Y \to Z, g:A \to B$ are homotopy equivalences with homotopy inverse $g': B \to A$ of $g$ and homotopies $H: g'g\simeq 1$, $K:gg' \simeq 1$. Then $f$ has a homotopy inverse $f':Z \to Y$ extending $g'$, and there are homotopies $H':ff' \simeq 1$, $K':f'f \simeq 1$ which extend respectively the homotopies $H$ and $$ K+ gHg' - gg'K. $$
It does not seem possible to simplify the curious "conjugate" homotopy given above. For example, one might expect that you can choose a homotopy $K'$ extending $K$, but I do not have a counterexample. (There is such a counterexample in the dual case, for fibrations, for which a relevant discussion is here.)
Going back to the question, you can use the above to work by induction to get maps and homotopies which fit together.