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Definition. A CW-complex $A$ can be elementary compressed to a CW-complex $B$ if there is a deformation retraction $A \to B$ or a quotient map by a contractible subcopmlex $A \to B$ (meaning according to a subcomplex in any cellular structure)

Definition. Compressible is the transitive closure of "elementary compressible".

It is clear that compressible is a preorder (any homeomorphism is an elementary contraction). The minimal elements in it will be called homotopy incompressible.

  1. It is clear that among the 1-dimensional CW-complexes these are exactly bouquets of circles, and all the other one-dimensional ones are deformation retracted on them.

Questions:

  1. Is it true that $S^n$ are homotopically incompressible? It is known that no subspace $\mathbb{R}^n$ is equivalent to $S^n$, therefore they have no proper deformation retracts. Intuitively, it seems that factorization over any contractible subcomplex will leave a sphere a sphere, but I still don't understand how to prove this.

  2. In general, closed manifolds have no proper deformation retracts. Are they incompressible?

  3. Is it possible to classify homotopically incompressible 2-dimensional CW-complexes?

  4. Is it true that under each CW-complex there is an incompressible CW-complex (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as we like)?

Due to (3), it is interesting - is it true that every connected n-dimensional CW-complex is homotopy equivalent to a bunch of n-dimensional closed manifolds? What is the criterion for incompressibility for non-closed manifolds? Are homotopy incompressible spaces closed with respect to a bouquet, product ..

Update 1: of course, the homotopy types of CW-complexes are not limited to bouquets of closed n-manifolds since the latter have a restriction on n-dimensional homology (first comment).

Update 2: After the comment about the error in the first statement, I included contraction of subcomplex in the definition of homotopy incompressibility.

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    $\begingroup$ Concerning the question at the end of your post: a wedge of $n$-manifolds will have top cohomology group a direct sum of $\mathbb{Z}$'s and $\mathbb{Z}/2$'s, so a mod-$p$ Moore space for $p$ an odd prime won't be of the form that you suggest. $\endgroup$
    – IJL
    Commented Dec 1, 2021 at 9:11
  • $\begingroup$ @IJL Oh thanks, that was easy! Yes, I practically did not hope that CW-complexes are arranged so simply, but I did not think of how to refute it. $\endgroup$
    – Arshak Aivazian
    Commented Dec 1, 2021 at 9:14
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    $\begingroup$ I don't understand your claim 1. Consider, say, the theta graph or the barbell graph. (These are the two graphs of Euler characteristic -1 with every vertex of valence 3.) Neither contains the wedge of two circles as a subgraph (since neither contains a point of valence 4). Actually, my guess is that a graph is minimal in your sense if and only if it has no vertices of valence 1. $\endgroup$
    – HJRW
    Commented Dec 1, 2021 at 9:34
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    $\begingroup$ Yes, my first statement is incorrect (when I wrote, I was thinking about contracting a spanning tree - factorization by contractible subspaces is also some special homotopy equivalence, but of a different kind) $\endgroup$
    – Arshak Aivazian
    Commented Dec 1, 2021 at 10:04
  • $\begingroup$ I will add this type of contraction and edit the question accordingly - this seems to be a more natural formulation of the problem. $\endgroup$
    – Arshak Aivazian
    Commented Dec 1, 2021 at 10:14

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