This question is related to [Attaching cells of different dimensions at once in a CW-complex](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.



Attaching cells of different dimensions at once in a CW-complex II
Yes. More generally, suppose you have a square


$$\begin{array}{ccc}
A&\rightarrow &B\\
\downarrow&&\downarrow\\
C&\rightarrow &D
\end{array}$$


and you want a space $Y$ between $A$ and the pullback of $B\to D\leftarrow C$ such that 


$$\begin{array}{ccc}
Y&\rightarrow &B\\
\downarrow&&\downarrow\\
C&\rightarrow &D
\end{array}$$


is a homotopy pushout. You can always get it if the maps are $2$-connected and the cohomological dimension of the map $B\cup\_A C\to D$ is less than approximately the sum of the connectivities of the maps $B\to D$ and $C\to D$. If the inequality is stronger then you get a uniqueness statement.


The proof uses something called homology truncation, which I learned from John Klein. We used it in our paper in Journal of Topology 2008, where there is a self-contained section on this.


