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Let $X$ be a CW-complex and $X^m$ it's $m$-skeleton. I think that for any $n\geq 2$ and $1\leq r\leq n-1$ it should be possible to obtain $X^{n+r}$ directly from $X^n$ via a homotopy push-out

$$\begin{array}{ccc} Y\vee X^{n}&\rightarrow &X^{n}\\ \downarrow&&\downarrow\\ X^{n}&\rightarrow &X^{n+r} \end{array}$$

Were $Y$ is a desuspension of $X^{n+r}/X^{n}$, the left vertical arrow is $(0,1)$, the two maps $X^n\rightarrow X^{n+r}$ are the inclusion, and the upper horizontal arrow is the identity on the second factor, so the only non-obvious part would be the restriction $Y\rightarrow X^n$ of the upper horizontal map.

For $r=1$, $Y\rightarrow X^n$ would just be the attaching map of $(n+1)$-cells.

I'd like to know if this is true, known (references?) or if there is a short argument to prove it. If true, I'd also like to know how unique is the map $Y\rightarrow X^n$.

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Map the function $Y\to *$ into the left edge to get two homotopy pushouts in a row; this would produce a cofiber sequence $Y\to X_n\to X_{n+r}$. But this can't be done in general.

One interesting way to see that this is not the case in general is that it would lead to cone decompositions of a generic space with length roughly $\log_2$ of the dimension. This is impossible for spaces with high ratio of cone length to dimension, such as projective spaces.

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  • $\begingroup$ Thanks, Jeff, I should have thought of this. All in all, I think I didn't manage, in this question, to state what I meant. Would you mind to take a look at mathoverflow.net/questions/149318/…? $\endgroup$ – Fernando Muro Nov 19 '13 at 9:16
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Adding some detail to Jeff Strom's answer:

If $X=\mathbb CP^{2k+2}$ and $n=4k$ and $r=4$ it is impossible. The cofiber $\mathbb CP^{2k+2}/\mathbb CP^{2k}=S^{4k+2}\cup_{\eta}e^{4k+4}$ is a suspension, but there is no suitable map $S^{4k+1}\cup_{\eta}e^{4k+3}\to \mathbb CP^{2k}$; such a map would lift to a map $S^{4k+1}\cup_{\eta}e^{4k+3}\to S^{4k+1}$ whose restriction to $S^{4k+1}$ has degree one, which is impossible.

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  • $\begingroup$ Thanks, Tom. I've accepted Jeff's answer just because it came before. As I say above, I think I didn't manage to state here what I really meant. Would you mind to take a look at mathoverflow.net/questions/149318/…? $\endgroup$ – Fernando Muro Nov 19 '13 at 9:17

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