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I was reading the second solution here:

Suspension of a product - tricky homotopy equivalence but the author said that the situation of well-pointed spaces is considered a more general situation than of $X$ and $Y$ being CW complexes.

Could someone clarify to me why this is correct please?

Emptymind
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  • A CW-complex is well-pointed with respect to any of its points. Is that what you're interested in? – Thorgott Mar 22 '23 at 22:17
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    Any point $x \in X$ is in the interior of some $k$-cell, then show that $x \to X^{(k)}$ is a cofibration. – ronno Mar 22 '23 at 22:26
  • @Thorgott I am interested in more details (the proof that every well-pointed space is a CW-complex) – Emptymind Mar 23 '23 at 11:40
  • @ronno how can I show that? – Emptymind Mar 23 '23 at 11:40
  • Consider a small disk neighborhood $D$ of $x$ in the $k$-cell, now to extend homotopies, find an equivalence between $D$ and $D \cup_x x \times [0,1]$ rel $\partial D$ that takes $x$ to $(x, 1)$. One way to do this is to imitate the Alexander trick, at time $t$ map the disk of radius $t/2$ to the segment $x \times [0, t]$ and the annulus with inner radius $t/2$ to $D \setminus x$ by scaling the radius. – ronno Mar 23 '23 at 12:22
  • @Emptymind That's the opposite direction of what I said. – Thorgott Mar 23 '23 at 12:45
  • @Thorgott so I think I am interested in showing the equivalence between them then – Emptymind Mar 23 '23 at 13:47
  • There is no equivalence either. I don't understand what it is that you think I'm saying, because I think it was pretty clear. – Thorgott Mar 23 '23 at 14:58
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    @Thorgott I am not saying that you told me that there is an equivalence ...... I just understood wrongly that there is one ...... so you are saying now no, and the only correct direction is the direction you mentioned, am I right in my understanding? – Emptymind Mar 23 '23 at 20:09

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