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Suppose $X$ and $Y$ are spaces that are homotopy equvialent to CW complexes, and let $f:X\to Y$ be a continuous map. I am trying to show that the pair $(M_f,X)$ is homotopy equivalent to a CW pair.

I've only shown that there is a CW pair $(C,D)$ and a homotopy equivalence $C\to M_f$ such that its restriction its restriction to $D$ is a homotopy equivalence $D\to X$ (proof is given below). Can we get the desired conclusion from this result? Or is there another way? (Maybe the homotopy extension property of $(M_f,X)$ and $(C,D)$ may help but I'm not sure.)

Let $g: A\to X$ be a CW approximation. Using the map $fg:A\to Y$, attach cells to $A$ and obtain a CW approximation $h:B\to Y$ extending $g$. Then $g,h$ are homotopy equivalences (Whitehead's theorem), and they define a map $\varphi:M_j\to M_f$ where $M_j$ is the mapping cylinder of the inclusion $j:A\to B$. $\varphi$ is a homotopy equivalence because in the commutative diagram below the three maps except for $\varphi$ are homotopy equivalences. Also $\varphi|_A=g:A\to X$ is a homotopy equivalence. $$\require{AMScd} \begin{CD} B @>>> M_j\\ @VVV @VV{\varphi}V \\ Y @>>> M_f \end{CD}$$

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  • $\begingroup$ "Can we get the desired conclusion from this result?" Yes. See May's A Concise Course in Algebraic Topology, pp.47-48. $\endgroup$
    – Ken
    Jun 11, 2022 at 0:45
  • $\begingroup$ @Ken Thanks, that was exactly what I needed $\endgroup$
    – blancket
    Jun 11, 2022 at 1:00

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