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How can I show that a suspension of a CW complex is a CW complex? I tried to start with showing first that a product of CW complexes is a CW complex but I didn't know how to construct the attaching maps.

J. W. Tanner
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BERTI2020
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    For a product of CW complexes to be a CW complex, instead of looking at attaching maps it's a little bit easier to look at the characteristic maps. If $e^n$ is an $n$-cell of $X$ and $e^m$ is an $m$-cell of $Y$, then $e^n \times e^m$ is an $(n+m)$-cell of $X \times Y$, and its characteristic map will be the product of the characteristic maps for the $n$- and $m$-cells. Since a choice of characteristic maps is equivalent to a choice of attaching maps, this is really all you need. – kamills Dec 29 '20 at 23:47
  • what is a characteristic map? – BERTI2020 Dec 30 '20 at 00:04
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    When you attach an $n$-cell to a space $X$, you are attaching via a map $\varphi: S^{n-1} \to X$. The result is commonly denoted $X \cup_{\varphi} e^n$. This I imagine you know. One way to write this result, however, is as the pushout of the diagram $D^n \xleftarrow{} S^{n-1} \xrightarrow{\varphi} X$. If you write down the pushout square, the attaching map is on the top edge of the square, and the characteristic map is the one on the bottom. – kamills Dec 30 '20 at 00:09
  • Also, https://math.stackexchange.com/q/1125445/497007 gives a way to do this with actually looking at the attaching maps. (I found it easier for this exercise to think of characteristic maps, but the attaching map situation is not that complicated as indicated in that question) – kamills Dec 30 '20 at 16:19

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Hint: The product of two CW complexes can (often) be made into a CW complex. Same for the quotient of one.

Apparently the result does not always hold. See @Tyrone's comment below.

Correction to the correction: according to @SteveD it is true.

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    The product of two CW complexs can fail to be a CW complex. It is true that $X\times Y$ admits a natural cell structure if $X,Y$ are CW complexes, but the CW topology is in general strictly finer than the product topology. It is true that if one of $X,Y$ is locally-finite (i.e. locally compact), then the two topologies agree. – Tyrone Jan 04 '21 at 15:38
  • Similarly, any non-Hausdorff quotient of a CW complex is not going to be a CW compelx. – Tyrone Jan 04 '21 at 15:38
  • Okay. @Tyrone Is the result not always true, then? What I wrote was too general, in any case. –  Jan 04 '21 at 15:49
  • No. For products see Hatcher's Theorem A.6 for a statement. He gives Dowker's example of where it fails at the bottom of the page. The full story is quite interesting. Quotient spaces are always badly behaved. It is true that if $X$ is a CW complex and $A\subseteq X$ a subcomplex, then $X/A$ has a natural CW structure. – Tyrone Jan 04 '21 at 15:55
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    The result about suspension does always hold, since one of the factors in the product is the unit interval (which is a finite complex), and the quotient is by a sub complex. – Steve D Jan 04 '21 at 18:36