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 $Y\rightarrow X^n$.



Attaching cells of different dimensions at once in a CW-complex
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. 


