Talk:Homotopy

message: Is the definition set up correctly for homotopy equivalence? It seems if you have a disk, X, and a point, Y = {y}, there is only one function f:X->Y, f(x) = y. Take a, b in X such that a /= b. Thus, g(f(a)) = g(y) = g(f(b)), so if g*f is the identity function, we have a contradiction that a = b.
63.73.199.69 (talk) 14:03, 28 July 2008 (UTC)
replies: text: And actually a topological space is call contractable if (and only if) it is homotopic to a point. Dharma6662000 (talk) 17:37, 25 August 2008 (UTC)
replies: text: Continuous functions have homotopies, but not topological spaces. Orthografer (talk) 14:20, 26 August 2008 (UTC)
thread_title:  Correct Definition? 

message: What topology do we usually assume on functions X->Y? The definition in the article says H must be continuous, but it is not explained when H is continuous. 
63.73.199.69 (talk) 14:03, 28 July 2008 (UTC)
replies: text: You are asking two nice questions here. I think that one of the questions you don't realise your asking. So first of all I'll answer the question I think you're asking, i.e. "What hoes it mean for  H : X \times [0,1] \to Y to be continuous</math>. Well Y is a topolgical space so we have a collection of open sets which cover Y and satisfy some axioms. For H to be continuous we require that the preimage of every open set of Y is open in X \times [0,1]. For the open sets in X \times [0,1] we consider the product topology given by the topology on X and the topology on [0,1]. The second question which I don't think you meant to ask was "What is a topology on the space of maps X \to Y?" This is a very nice question since this space is an infinite dimensional space. If we restrict our attention to the space of smooth functions, i.e. C^{\infty}(X,Y) then we can use the Whitney C^{\infty}-topology. This compares k-jets and successive partial derivatives to give a notion of proximity it this infinite dimensional space. This is a very interesting question, and I urge you to study it in depth. Dharma6662000 (talk) 17:57, 25 August 2008 (UTC)
replies: text: A simple answer is to give a topology in Xx[0,1] and then H:Xx[0,1]->Y is continuous if for each open set U in Y, we have H^(-1)U is open in Xx[0,1], that easy--kmath (talk) 19:59, 25 August 2008 (UTC)
replies: text: That's what my rely said (and I quote: "For H to be continuous we require that the preimage of every open set of Y is open in X \times [0,1]. "), why bother to say the same thing twice? The key problem is what are the open sets on X \times [0,1] ? And my reply answers that question too by saying that the product topology should be used. Dharma6662000 (talk) 20:28, 25 August 2008 (UTC)

text: Now our  friend 63.73.199.69 is going to be focused on that matter--kmath (talk) 02:10, 26 August 2008 (UTC)
thread_title:  When is ''H'' continuous? 

message: I think it is not true as stated in the article, that when X and Y are homotopy equivalent, and X is locally path connected, then Y is also locally path connected. For an example, take the subspace of a unit square consisting of the interval [(0,0),(0,1)] together with all line segments from (0,1) to the points (1/n,0) where n runs through all natural numbers. The space is contractible, but it is not locally path connected at (0,0).
I am not sure that the statement that two homotopies can be
rotated makes much sense (but I know what is meant..:) )
:Yes, locally path-connected isn't a homotopy invariant.  
It seems to me the stuff on homotopy groups belongs in its own article.
--gorlim 16:52, 24 Apr 2004 (UTC)
replies: text: The stuff on homotopy groups has its own page.  It seems to me that the stuff on isotopy belongs in its own article. —Blotwell 08:40, 23 July 2005 (UTC)
thread_title:  Incorrect statment? 

message: The illustration with caption "An illustration of a homotopy between the two bold paths" is somewhat confusing for two reasons:
1. first, which subsets of the bold loop correspond to the "two bold paths" could be misinterpreted, making the isolines even more confusing
2. second, at first glance, the illustration makes you think that all homotopies are necessarily relative to some none-empty subset, since the endpoints of the curves are fixed.
I'm changing the caption for reason #2, feel free to change it back...
I think the etimology here is wrong: homo-topy comes from the greek verb "topao"(= to melt, or to deform), and not from the name "topos" (= place).
(in reference to above) was the etymology clarified? 68.239.20.246 00:29, 22 June 2007 (UTC)
thread_title:  Misleading picture? 

message: Maybe a note should be added that says that all homeomorphisms are homotopies. (Or have I missed it?)
Two continuous maps  \phi, \psi : X \to Y are called homotopic if (and only if) there exists a continuous map \chi : X \times [0,1] \to Y such that \chi(x,0) = \phi(x) and \chi(x,1) = \psi(x) for all x \in X. Now, two topological spaces X and Y are homotopic if (and only if) there exist continuous maps f : X \to Y and g: Y \to X such that g \circ f : X \to X and f \circ g : Y \to Y are both homotopic to the identity maps. Now, assume that X and Y are homeomorphic, then we can choose a homeomorphism f : X \to Y and let  g := f^{-1} : Y \to X. Then g \circ f = f^{-1} \circ f = \text{id}_X and f \circ g = f \circ f^{-1} = \text{id}_Y. It follows that X and Y are homotopic. Dharma6662000 (talk) 17:46, 25 August 2008 (UTC)
replies: text: I think you're imagining these topological spaces as embedded in other spaces. A homeomorphism is not necessarily a homotopy.Orthografer (talk) 14:22, 26 August 2008 (UTC)
replies: text: What's wrong with the above proof? It makes no mention of an ambient space and deals only with abstract topological spaces. The proof seems correct to me. The proof shows that two spaces that are homeomorphic are necessarily homotpic. See pages 5 and 6 of 
* H. Osborn's Vector Bundles, Volume 1: Foundations and Stiefel-Whitney Classes. Academic Press 1982. Declan Davis (talk) 14:53, 26 August 2008 (UTC)

text:  In fact the article says that all homeomorphisms are homotopies. In homotopy equivalence and null-homotopy the article says: "Clearly, every homeomorphism is a homotopy equivalence, but the converse is not true: for example, a solid disk is not homeomorphic to a single point, although the disk and the point are homotopy equivalent." Declan Davis (talk) 16:01, 26 August 2008 (UTC)
replies: text: I agree that a homeomorphism is a homotopy equivalence. My point is the following: saying that a homeomorphism is a homotopy is, strictly speaking, not true. Further, it distracts the reader from the main idea, which is that a homotopy is simply a 1-parameter family of maps. Certainly a trivial homotopy can be constructed from any continuous map. What is special about such a map being a homeomorphism? I also think some of the people here are confusing the word "homotopic" with "deformation equivalent." Orthografer (talk) 18:43, 24 October 2008 (UTC)
replies: text: I completely agree with Orthografer. There is a clear confusion supra: an homotopy equivalence is not an homotopy.  I think this should be clarified explicitly in the article.Moreover, I'm a bit uncomfortable with the coffey cup <---> doughnut illustration; it does illustrate an homotopy (and indeed even an isotopy), provided that the cup and the torus are considered not as topological spaces, but as two different embeddings of the same space into (e.g.) R3. We should not be too formal; but also not incorrect. JoergenB (talk) 20:38, 31 October 2008 (UTC)
thread_title:  Homeomorphism <math> \implies </math> Homotopy 

message: I object to the following as a definition of a homotopy:
:An alternative notation is to say that a homotopy between two continuous functions f , g : X → Y is a family of continuous functions ht: X → Y for t ∈ [0,1] such that h0= f and h1= g and the map t → ht is continuous from [0,1] to the space of all continuous functions X → Y. The two versions coincide by setting ht(x) = H(x,t).
This is meaningless so long as no topology has been put on the set of continuous functions from X to Y.
The definition would be correct if this set were given the compact-open topology and X was locally compact. This is too complicated, though. I think the best thing to do is to say that the family ht is "continuous" only in the sense that the original definition is satisfied, where H(x,t) = ht(x). 67.150.252.66 (talk) 12:45, 14 November 2008 (UTC)
replies: text: I do not recall any mention of a topology on the function space Ω(X,Y) when I was learning about homotopies. The (usual) definition of homotopy says that two maps f : X → Y and g : X → Y are homotopic if there exists a continuous map H : X × [0,1] → Y such that H(x,0) = f(x) and H(x,1) = g(x) for all x in X. This definition only requires that we have a topology on the spaces X × [0,1] and Y. I've never seen any mention of the map [0,1] → Ω(X,Y) needing to be continuous. Maybe if we seek a special kind of homotopy then this extra condition may be imposed.  Δεκλαν Δαφισ   (talk)  22:42, 14 November 2008 (UTC)
thread_title: Alternative definition appears incorrect or incomplete

message: I would find it really useful to have a dummies' explanation of the differences between all these things, including such subtleties as distinguishing homotopies vs. homotopy equivalents vs homeomorphisms. Examples would be good. At present the article reads little better a student's notes. — Cheers, Steelpillow (Talk) 22:29, 25 January 2011 (UTC)
replies: text: Completely agree. Will try to work on this. -  Subh83 (talk | contribs) 04:43, 22 November 2011 (UTC)
thread_title:  Homology, homotopy and homeomorphism 

message: I would like to suggest a separate article for "Homotopy equivalence" since it is significantly more advanced than the notion of simple homotopy. Moreover it has applications in may different context that need special mention. For example, homology groups and homotopy groups are invariant up to homotopy equivalence. Moreover its relation with and distinction from deformation retract is worth mentioning.
Finally, and most importantly, a clear clarification of the difference between two spaces being "homotopic" vs. them being "homotopy equivalent" is important to avoid confusion.
If these make sense, I will start working on the article. -  Subh83 (talk | contribs) 04:34, 22 November 2011 (UTC)
thread_title:  New article for homotopy equivalence 

message: Does anyone know the etymology of the word "homotopy"? I would guess it comes from the greek word "topos"="place" such as "utopia" or "distopia", but it is only a guess. If someone could check the etymology of the word and add it to the page it would be much appreciated.
:I don't know the details. But basically, Greek "homo" = same and "topos" = place. Maproom (talk) 21:29, 9 January 2014 (UTC)
thread_title:  Etymology 

message: I believe the following is incorrect:
:[F]or example, the double torus and the double torus with the rings interlinked are homotopy equivalent (since they are homeomorphic), even though the said transformation cannot be embedded in three-dimensional Euclidean space without the rings "passing through" each other.
This transformation may indeed be embedded in R^3.  Here is an illustration of exactly that.  I have edited this out for now.  If someone knows of a better example, please include it.
Peligro~enwiki (talk) 01:56, 18 November 2015 (UTC)
replies: text: Maybe a valid example is a torus and a knotted torus. Maproom (talk) 11:10, 18 November 2015 (UTC)
thread_title:  Error in Homotopy Equivalence section 

message: Substituted at 02:13, 5 May 2016 (UTC)
thread_title: Assessment comment