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After this answer, the following question comes :

What's the classification (up to homeo.) of CW complexes formed by gluing a 2-cell to a circle ?

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Sebastien Palcoux
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    "Up to homeo" is something that we usually do not care much about, really :-) – Mariano Suárez-Álvarez Aug 29 '13 at 16:03
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    It might help you to know (if it's enough for you to characterize them up to homotopy type) that homotopic attaching maps yield homotopy equivalent adjunction spaces since $(D^2,S^1)$ is a cofibration – Stefan Hamcke Aug 29 '13 at 16:24
  • @StefanH. : you say that they are all "homotopy equivalent" to the closed disk, isn't it? – Sebastien Palcoux Aug 29 '13 at 16:48
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    No, not all gluing maps are homotopic to the identity $S^1\to S^1$. If for example the attaching map isn't surjective, then you get a wedge of a circle and a sphere since a non-surjective map $S^1\to S^1$ is homotopic to a constant map. – Stefan Hamcke Aug 29 '13 at 16:52
  • @StefanH. : so there are one classification "up to homeomorphism" and one classification "up to homotopy equivalence", the last being strictly weaker than the first thank to this answer. So what you say is the classification "up to homotopy-equivalence" is already far away to be obvious, isn't it ? – Sebastien Palcoux Aug 29 '13 at 16:59
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    No, the classification up to homotopy type is very easy. If you know about the fundamental group of the circle, then you know that the homotopy class of a map $S^1\to S^1$ is uniquely determined by an integer number, called the degree. – Stefan Hamcke Aug 29 '13 at 17:04
  • @StefanH. : ok thank you, it's much more clear now. So the first classification is partitioned by subclasses with same degree, and perhaps every object of degree $n$ is generically the deformation of an object of degree $0$, so that the classification reduce to degree $0$, isn't it ? – Sebastien Palcoux Aug 29 '13 at 17:15
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    I don't quite get what you mean by perhaps every object of degree n is generically the deformation of an object of degree 0. The homotopy type is determined by the degree of the gluing map. It basically tells you how often the circle wraps around itself. If you have two CW-complexes with the same degree, they will be homotopy equivalent. On the other hand, an attaching of the disk by a gluing map of different degree will give you a complex of a different homotopy type (which you can proof by considering the cellular homology, for example.) – Stefan Hamcke Aug 29 '13 at 19:25
  • @StefanH. : yes I agree. I'm sorry my word "deformation" is not clear. Let me explain my point : in order to do this classification "up to homeomophism", we can start by classifying the objects (with attaching map) at degree 0. Then perhaps we can deduce the classification at degree n, from the classification at degree 0. – Sebastien Palcoux Aug 29 '13 at 19:30
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    Do you really want to classify them up to homeomorphism? As @MarianoSuárez-Alvarez said: We don't usually care about homeomorphism classes. Usually one is interested in the homotopy type, a way coarser classification, but spaces of different homotopy type can be distinguished by algebraic invariants like homotopy/homology groups, cohomology rings, but they won't tell you when hom. equivalent spaces aren't homeomorphic. And since you tagged your question algebraic-topology, I assume you want to use those invariants. – Stefan Hamcke Aug 29 '13 at 19:54
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    I think a classification up to homeomorphism can be very daunting, since it will matter how often and when the map from the circle to itself changes direction, how often a point is met by this map, and so on... All these subtleties do not matter for the homotopy type. There would be uncountably many possibilities and it would not be easy to prove that they some are really homeomorphic or that they are not. – Stefan Hamcke Aug 29 '13 at 19:59
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    If the attaching map $f: S^1 \to S^1$ is of degree $n$ then the fundamental group of the space with the $2$-disk attached is the cyclic group of order $n$. This shows you get different homotopy types with attaching maps of degree $m,n$ if $m \ne \pm n$. – Ronnie Brown Aug 29 '13 at 20:01
  • @StefanH. : I have posted this question (borderline to algebraic topology) because I was impressed by this answer. Before this answer I thought every CW complex as homeomorphic to a regular CW complex, but it's not true, it's only homotopy equivalent to a regular CW complex. – Sebastien Palcoux Aug 29 '13 at 21:22
  • @StefanH. : Now if (as you say) algebraic topology is only interested in homotopy equivalence, then it only needs regular CW complex. So why algebraic topologists defined CW complex in a way that it can be non-regular, this seems useless ? Perhaps there is something interesting in the "up to homeo." classification, that's why I ask this question. – Sebastien Palcoux Aug 29 '13 at 21:29
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    It may be true that every CW complex is homotopy equivalent to a regular Cw complex, but sometimes we'd like to represent a given space as a CW complex and not need to replace it by another equivalent space. And even if we could just change the number of cells for the particular space to make it regular, this would require several additional cells, which would make, for example, the computation of cellular homology more work. – Stefan Hamcke Aug 29 '13 at 21:40
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    What I'm saying is that a more general definition gives you more flexibility, and it allowes us to define the restricted one as a special case of the general one. Think about Delta-complexes compared to regular Delta-complexes or even simplicial complexes. We can make the torus of just two triangles, three edges, and one vertex. To put a simplicial complex structure on the torus, one needs at least $14$ triangles, $21$ edges, and $7$ vertices. – Stefan Hamcke Aug 29 '13 at 21:47
  • @StefanH. : We agree (thank you for these examples). But if we don't have general methods for knowing if two gluing maps $f$, $g : \mathbb{S}^{1} \to \mathbb{S}^{1} $ give homeomorphic CW complexes or not, then we're faced to an open problem, and it's interesting for itself as an open problem (for finding new invariants, making new mathematics etc...). – Sebastien Palcoux Aug 29 '13 at 21:58
  • @StefanHamcke I have a serious doubt that the degree determines the resulting space. Consider this "triple" winding: it goes two times clockwise and once counterclockwise. The degree is $1$ but you claim that the quotient space is contractible (homotopy equivalent to $D^2$ = the result of attaching via the identity)? Where exactly does this come from? How homotopy class of the attaching map corresponds to the homotopy type of the resulting space? – freakish Oct 03 '19 at 11:37

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