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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 retractsclosed 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.

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.

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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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 has a homotopicallythere is an incompressible deformation retractCW-complex (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desiredwe 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.

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 each CW-complex has a homotopically incompressible deformation retract (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desired)?

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.

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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CW-complexes that cannot be homotopically contractedcompressed

Definition. A CW-complex $A$ can be elementary contractedcompressed 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. ContractibilityCompressible is the transitive closure of "elementary contractility"compressible".

It is clear that contractilitycompressible 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 each CW-complex has a homotopically incompressible deformation retract (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desired)?

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.

CW-complexes that cannot be homotopically contracted

Definition. A CW-complex $A$ can be elementary contracted 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. Contractibility is the transitive closure of "elementary contractility".

It is clear that contractility 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 each CW-complex has a homotopically incompressible deformation retract (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desired)?

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.

CW-complexes that cannot be homotopically compressed

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 each CW-complex has a homotopically incompressible deformation retract (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desired)?

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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