This question is related to https://mathoverflow.net/questions/149301/attaching-cells-of-different-dimensions-at-once-in-a-cw-complex There, I didn't manage to formalize the idea I had in mind, and ended up with a question whose answer was obviously *no*, as Jeff Strom and Tom Goodwillie showed. This question here is hopefully more meaningful. In particular, Tom's concrete counterexample to the former doesn't apply here.

Let $X$ be a *connected* 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

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

were $Y$ is an $(n+r-1)$-dimensional CW-complex with $Y^n=X^n\vee (\vee_{C_{n+1}}S^n)$, 
the quotient $Y/X^n$ is a desuspension of $X^{n+r}/X^{n}$, the arrows out of $Y$ restrict to the indentity on $X^n$, and the two maps $X^n\rightarrow X^{n+r}$ are the inclusion. 

For $r=1$, the previous square would be equivalent to 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 about uniqueness.

---------

In order to prove that this question is not completely trivial, let me show that the answer is *yes* for $r=2$. 

Let $C_*(X)$ be the cellular chain complex of the universal conver of $X$, which is a complex of $\mathbb Z[\pi_1X]$-modules. Recall that $C_m(X)=\pi_m(X^m, X^{m-1})$ for $n>2$ is a free $\mathbb Z[\pi_1X]$-module, whose basis we denote $C_m$. The differential out of $C_{m+1}(X)$ is the composite
$$d_{m+1}\colon\pi_{m+1}(X^{m+1}, X^{m})\longrightarrow \pi_mX^m\longrightarrow\pi_m(X^m, X^{m-1})$$
which uses homomotphisms of the long exact sequences of the pairs $(X^{m+1},X^m)$ and $(X^m, X^{m-1})$. One can easily check that $C_m(X)=\ker[\pi_k(0,1)\colon \pi_k(\vee_{C_m}S^k\vee X^l)\rightarrow \pi_kX^l]$, $k,l\geq 2$, hence we can represent $d_m$ by a map $\bar d_m\colon \vee_{C_{m+1}}S^n\rightarrow \vee_{C_m}S^n\vee X^n$.

The composite
$$\pi_{n+2}(X^{n+2}, X^{n+1})\stackrel{d_{n+2}}\longrightarrow\pi_n(X^{n+1}, X^{n})\longrightarrow\pi_nX^n$$
is trivial because it factors through to consecutive homomorphisms of the long exact sequence of $(X^{n+1},X^{n})$. Hence, the composite
$$\vee_{C_{n+2}}S^n\stackrel{\bar d_{n+2}}\longrightarrow \vee_{C_{n+1}}S^n\vee X^n\stackrel{(f, 1)}\longrightarrow X^n,$$
where $f$ is the attaching map of $(n+1)$-cells, is nullhomotopic. Let $Y$ be the mapping cone of $\bar d_{n+2}$. A null-homotopy yields a map $Y\rightarrow X^n$, which will be the upper horizontal map in the square.

The composite
$$\vee_{C_{n+2}}S^n\stackrel{\bar d_{n+2}}\longrightarrow \vee_{C_{n+1}}S^n\vee X^n\stackrel{(0,1)}\longrightarrow X^n,$$
is also nullhomotopic, since $\bar d_{n+2}$ represents a map in $\ker\pi_{n}(0,1)$, hence we get another map $Y\rightarrow X^n$, which will be the left vertical map.

We can form a square as above, commutative up to a certain homotopy comming from the fact that $f$ composed with the inclusion $X^n\subset X^{n+1}$ is nullhomotopic. Therefore we obtain a map $Z\rightarrow X^{n+2}$ from the homotopy push-out $Z$ of the upper left corner. By construction, this map induces an isomorphism on $\pi_1$ and on cellular chain complexes of universal covers, so it is a homotopy equivalence by the homological Whitehead theorem.