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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.

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  • $\begingroup$ See math.stackexchange.com/questions/61370/… $\endgroup$
    – Igor Rivin
    Commented Dec 29, 2013 at 20:46
  • $\begingroup$ This is the reference which is mentionned, it does not answer the question. $\endgroup$
    – Nre
    Commented Dec 29, 2013 at 21:08
  • $\begingroup$ @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. $\endgroup$
    – Eric Stucky
    Commented Dec 30, 2013 at 7:54
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    $\begingroup$ 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? $\endgroup$
    – Carsten S
    Commented Jan 9, 2014 at 18:17
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    $\begingroup$ Since they are both pushouts it suffices to find compatible maps from the components, this is relatively easy with the cell structure chosen above. $\endgroup$
    – user17892
    Commented Jan 10, 2014 at 10:42

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