4
$\begingroup$

This question is related to 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.

$\endgroup$
  • 1
    $\begingroup$ My motivation is to construct the non-fringed homotopy spectral sequence of Bousfield from a purely cellular point of view. Actually, this question can probably be answered in an easy way in other contexts where one has more control on maps, such as rational homotopy theory. I've actually made computations with nice examples DGAs and DG-operads were everything is easy to describe. That would maybe enough for my current purposes, but I think that general improvements of Bousfield's SS of a cosimplicial space (or rather of its applications) could be interesting. $\endgroup$ – Fernando Muro Nov 19 '13 at 9:27
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Great answer! Homology truncation sounds very interesting. $\endgroup$ – David White Nov 19 '13 at 14:16
  • $\begingroup$ Indeed, David. Tom, thanks a lot, I'll run right now onto your paper with John Klein! $\endgroup$ – Fernando Muro Nov 19 '13 at 14:21
2
$\begingroup$

In a recent paper I wrote with John Oprea (Oprea, John; Strom, Jeff Lusternik-Schnirelmann category, complements of skeleta and a theorem of Dranishnikov. Algebr. Geom. Topol. 10 (2010), no. 2, 1165–1186) we proved that if $X$ is an $n$-dimensional simplicial complex, then the complement of the $r$-skeleton $X_r$ is a complex $Y_{n-r-1}$ of dimension at most $n-r-1$ and there is another subcomplex $L$ of $X$ such that $$ \begin{array}{ccc} L&\rightarrow &X_{r}\\ \downarrow&&\downarrow\\ Y_{n-r-1}&\rightarrow &X_n \end{array} $$ is a homotopy pushout square.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.