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 $Y\rightarrow X^n$.