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If you read the question entirely, it is not a duplicate. The first time I asked the question, I already gave the link to the similar question and explained why the answer is not satisfying. If you read well the answer, it starts by stating that it is a well known fact that the product of CW complexes is a CW complex, exactly what I don't want.

In a few references I have been looking at (even here), everytime the proof of:

Let $X,Y$ be CW-complexes and let $f:X\rightarrow Y$ be a cellular map, then the mapping cylinder $M_f$ has a natural CW-complex structure,

comes up, there is a hidden argument such as the pushout of two CW-complexes is a CW-complex or other technical arguments which are not clear. Does someone know/can find a neat proof of this, without using implicitly any other major results about CW-complexes?

Thanks a lot,

N.

What I have done:

We want to show that $M_f$ is a CW-complex:

$$\require{AMScd} \begin{CD} X @>{f}>> Y\\ @Vi_0VV @V\hat{\iota}VV \\ X\times I @>\widehat f>>M_f, \end{CD} $$ with $M_f=Y\sqcup_f(X\times I)$.

I started by constructing a CW-decomposition of $X\times I$ by using the CW-decomposition $\partial I\subset I$ of $I$. That is, given a CW-decomposition $X_0\subset X_1 \subset \cdots $ of $X$: $$\partial I \cup X_0\subset \partial I \times X_1 \cup I\times X_0\subset\ldots \subset \partial I \times X_n \cup I\times X_{n-1}\subset \ldots,$$ unfortunately, I get stuck very soon. Hope someone will be able to help me from there on.

Nre
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  • See http://math.stackexchange.com/questions/61370/the-mapping-cylinder-of-cw-complex – Igor Rivin Dec 29 '13 at 20:46
  • This is the reference which is mentionned, it does not answer the question. – Nre Dec 29 '13 at 21:08
  • @Nre: There is no need to state your explanation in the question. A comment and perhaps a minor edit to bump the question up to the front page is sufficient. – Eric Stucky Dec 30 '13 at 07:54
  • ok, thanks. I don't know exactly how it works once a question has been declared a duplicate by five people and it is in fact not... – Nre Dec 30 '13 at 08:04
  • Have you seen section 'Topology of Cell Complexes' in Hatcher? At least product of CW-complexes is discussed in detail there. – Grigory M Dec 30 '13 at 11:16
  • @Grigory M: Yes, but I don't understand exactly what is happening. As we are only working with $X\sqcup I$ here, I thought it would maybe give me some intuition understanding this particular case first. – Nre Dec 30 '13 at 18:11
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    So what you want is: $X$ and $Y$ CW-complexes, $A\subset X$ a sub complex, and $f\colon A\to Y$ cellular, then $Y\cup_f X$ has a natural structure of a CW-complex. Well, the every cell in $Y$ remains as it is, the cells in $A$ are gone, the cells in $X$ which are not in $A$ get new attaching maps: Everything that was formerly attached to something in $A$ is now attached to the corresponding part in $Y$. So the $k$-skeleton of $Y\cup_f X$ is $Y^{(k)}\cup_f X^{(k)}$. Does that help? – Carsten S Jan 09 '14 at 18:17
  • @Carsten Schultz: I see what is happening, but I can't prove formally that the k-skeleton really is the one you describe, could you help me with that please? – Nre Jan 09 '14 at 21:27
  • Try showing that there is a homeomorphism between $Y^{(k)} \cup_f X^{(k)}$ and the pushout given by attaching the cells described above, it should be clear that they are the same set and you just need to see that the topologies are the same. – Justin Young Jan 10 '14 at 10:32
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    Since they are both pushouts it suffices to find compatible maps from the components, this is relatively easy with the cell structure chosen above. – Justin Young Jan 10 '14 at 10:42
  • @JustinYoung: Thanks, I will try to do as you tell! – Nre Jan 10 '14 at 22:26

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