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.