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A technicality

"The third requirement, that f⁻¹ be continuous, is essential. Consider for instance the function $f:[0,2\pi )\to S^{1}$ (the unit circle in $\mathbb {R} ^{2})$ defined by $f(\phi )=(cos(\phi ),sin(\phi ))$ . This function is bijective and continuous, but not a homeomorphism ( $S^{1}$ is compact but $[0,2\pi )$ is not). The function f^-1 is not continuous at the point $(1,0)$ , because although f^-1 maps $(1,0)$ to $0$ , any neighbourhood of this point also includes points that the function maps close to $2\pi$ , but the points it maps to numbers in between lie outside the neighbourhood.[2]"

I quote the "Notes" of the article. It should probably note that this is only true if the neighborhood is not the entire set. 174.124.170.209 (talk) 03:10, 27 February 2015 (UTC)[reply]

Unintelligible to newcomers

It wasn't until now that I even knew what Homeomorphism was, the problem is that upon reading the article, I still don't know what Homeomorphism is other then some thing about how homeomorphism is a continuous function between two topological spaces that has a continuous inverse function. Homeomorphisms are the isomorphisms in the category of topological spaces—that is, they are the mappings that preserve all the topological properties of a given space. Two spaces with a homeomorphism between them are called homeomorphic, and from a topological viewpoint they are the same. Excuse my ignorance, but it seems to me like some one who knew already what Homeomorphism was would understand this Article perfectly, but a Luddite like me has absolutely no idea what this article is even beginning to speak of. — Preceding unsigned comment added by Robotwalla (talk • contribs) 19:28, 16 August 2011 (UTC)[reply]

Im wrong or this shodl be linked with Homomorphisms>Topological isomorphisms

I feel like right now, the article is quite approachable for newcomers. Should this section be removed? LudwikSzymonJaniuk (talk) 09:27, 2 April 2019 (UTC)[reply]

Knot example is not an example

The trefoil knot being deformed into the torus in 4-space is not an example of a homeomorphism, such a deformation is referred to as an ambient isotopy. Any torus embedded in 3-space so as to appear knotted is still a torus, and so is indeed "homeomorphic to a torus" in some tautological sense of the word.

Being knotted is only a property of the embedding, and not a property of the object itself. See, e.g. Knots and Links by Dale Rolfsen. 137.82.36.10 01:57, 19 September 2007 (UTC)[reply]

It might be ancient history, but the above posting is nonsense. For a start, the example involved a one-dimensional knot and a circle; the caption explicitly uses the word circle and says that the knot has been "thickened" to make the diagram more comprehensible. The statement "the trefoil knot being deformed into the torus...is not an example of a homeomorphism" is contradicted in the next sentence where the knot/torus is described as "indeed 'homeomorphic to a torus' in some tautological sense of the word". As far as I know, there is only one sense of the word homoeomorphic, and it refers to an equivalence relation; so obviously any object is homoeomorphic to itself.

The example is good, but the caption could be improved. The question of continuous deformation is really neither here nor there. That is the important point made by the example. Rdbenham (talk) 14:32, 6 April 2009 (UTC)[reply]

Although the first comment is not entirely correct, I found the above two comments, taken together, to be enlightening. I updated the description. It appears to me that an ambient isotopy of the 3D embedding space cannot continuously deform a trefoil into a circle. Can it be unknotted in 4D? Kbk (talk) 03:42, 24 November 2013 (UTC)[reply]

Correct spelling?

Isn't the correct spelling homoeomorphism rather than homeomorphism? --fil

Not in everyday use.

Charles Matthews 15:20, 9 Feb 2004 (UTC)

Ok I don't use this term every day, and I also thought homeomorphic is the correct spelling; but my spell checker (ispell-emacs) suggests homoeomorphic instead of homeomorphic. Maybe one should point out the alternative spelling. --fil

On the whole, I'd stick with the way humans spell it.

Charles Matthews 16:28, 9 Feb 2004 (UTC)

(Sorry to re-open an old thread.) Some people also leave off the first "o" in oedema and oesophagus. Might this be a English vs American thing? (People outside the US are humans too...) Alternatively, I notice that "homoeomorphism" is defined by several online dictinaries along the lines of "A near similarity of crystalline forms between unlike chemical compounds". If there are indeed two distinct words, that would explain the spell checker's response. LachlanA (talk) 02:49, 7 May 2008 (UTC)[reply]

homeomorphism makes perfect sense when speaking only of metric spaces, with no reference to topology. The article needs to be edited. I'm not sure how to word the definition to reflect this. --HellFire 13:43, 18 July 2006 (UTC)[reply]

Yeah, but any metric space is automatically a topological space. Oleg Alexandrov (talk) 15:30, 18 July 2006 (UTC)[reply]

oh thts fine in tht case.know a little about metric spaces, dont know anything about topology :-) --HellFire 10:56, 19 July 2006 (UTC)[reply]

Well (sorry to be OT), a metric space becomes a topological space if you define an "open set" as an arbitrary union of open balls. Rdbenham (talk) 14:35, 6 April 2009 (UTC)[reply]

Animation

The animation is good example of homotopy not of homeomorphism. I suggest it to be placed there, and find another image for exemplifying homeomorphism. SurDin 12:50, 7 January 2007 (UTC)[reply]

I changed the caption a bit to make it clear that homeomorphic != homotopy. Maybe we need a counter example of a homeomorphism which is not a homotopy, perhaps two linked ring and two disjoint rings? --Salix alba (talk) 22:20, 12 March 2007 (UTC)[reply]

Every homeomorphism is a homotopy, but the converse is false. Two continuous maps

f,g:X\to Y

are called homotopic if there exists a continuous map

H:X\times [0,1]\to Y

such that

H(x,0)=f(x)

and

H(x,1)=g(x)

for all

x\in X

. Now, two spaces X and Y are called homotopic if there exist continuous maps

f:X\to Y

and

g:Y\to X

such that the maps

g\circ f:X\to X

and

f\circ g:Y\to Y

are homotopic to the identity maps. Well, assume that f and g are homeomorphisms, and in fact assume that

g=f^{-1}

. Then

g\circ f=f^{-1}\circ f={\text{id}}_{X}

and

f\circ g=f\circ f^{-1}={\text{id}}_{Y}

. These composites aren't just homotopic to the identity maps, they are the identity maps! Thus two homeomorphic spaces are automatically homotopic. For a reference see pages 5 - 6 of Howard Osborn's Vector Bundles, Foundations and Stiefel-Whitney Classes, Volume 1, Academic Press, 1982. Dharma6662000 (talk) 17:19, 25 August 2008 (UTC)[reply]

Confusion

So two objects are homeomorphic if a one-to-one, continuous,invertible function (as opposed to a continuous deformation) mapping one to the other exists. Does this mean that turning an object inside out is an acceptable homeomorphism? What exactly are the restrictions on cutting and gluing to preserve topological properties (it seems a cut+glue pair, done correctly is acceptable)? Thanks.

It seems that only if orientability is preserved. Take for example a cylinder

S^{1}\times I

. Fix a point z in the circle and then cut the cylinder along

J=z\times I

. If you paste along J by identity you re-get the cylinder, but if you paste using a reflexion thru

z\times {\frac {1}{2}}

, you will get the mobius strip which is not homeomorphic to the cylinder.--kiddo 17:08, 8 April 2007 (UTC)[reply]

This is almost off-topic, but I think the sentence "Homeomorphisms are the isomorphisms in the category of topological spaces" from the NOTES section should be moved to the top of the page. That was the most informative and (for anyone who's taken high-school algebra) clear description in the entire article. —Preceding unsigned comment added by 71.198.178.194 (talk) 18:50, 11 May 2008 (UTC)[reply]

No! A mapping $f:X\to Y$ is a homeomorphism if (and only if) it and its inverse are continuous. A map is called continuous if (and only if) the preimage of open sets are open. I put "and only if" in brackets because some people find the use of the phraes "if and only if" pedantic in definitions. There exist continuous invertable maps whose inverses aren't continuous. Dharma6662000 (talk) 04:14, 25 August 2008 (UTC)[reply]

Natural continuous invertible functions which are not homeomorphisms are a bit hard to come by, e.g. no such function exists with a locally compact domain (such as a real interval) and a Hausdorff codomain. Nevertheless, one can take for example any bijection from N onto Q (both with their natural topologies inherited from the reals). — Emil J. (formerly EJ) 14:30, 25 August 2008 (UTC)[reply]

If you want a continuous bijection which is not a homeomorphism. You can take the map that wraps a half-open interval round the circle. (If the domain is compact and the image Hausdorff then a continuous bijection is a homeomorphism). Dharma6662000 (talk) 19:10, 25 August 2008 (UTC)[reply]

For your information, real mathematicians never use if and only if in definitions.--kmath (talk) 19:45, 25 August 2008 (UTC)[reply]

You seem to have a slight attitude problem Juan. Please read the Wikipedia rules about posts made on discussion pages. One of them says that you should be nice. For your information some mathematicians do you "if and only if" in definitions. The Japanese are an example (I worked at the University of Hokkaido for two years and it was common practise, are you suggesting that Japanese mathematicians are not real mathematicians?). It is technically correct to write "if and only if" but it is sometimes seen as being pedantic. You have made two posts (one on this page and one on the homeomorphism discussions page) that are totally pointless. One of them is just an insult (which after further inspection backfires and leaves you looking rather foolish), and the other is to repeat an earlier post with less detail and less clarity. Please do not waste my time, or other users' time with such puerile posts. Te pediría que fuerás más respetuoso en el futuro. Dharma6662000 (talk) 20:54, 25 August 2008 (UTC)[reply]

Ok, No es pa/tanto, amigo :). I don't mean to hurt anyone and after all i was only giving information about how pro-math works... It is clear that iff is used only when you are going to state an equivalence (with proof)... If you feel harm about my style: i am sorry, you already expressed your own opinion and that's your right. About Japaneses' comments maybe math isn't a problem but english.--kmath (talk) 02:02, 26 August 2008 (UTC)[reply]

I'm pleased by your reply. You do seem to be a nice guy after all. My personal style is to use "if" and not "if and only if", but I included it so that someone not familiar with the convention would be able to understand. To read the statement "an integer is called even if it is divisible by two" might cause some people a problem. Anyone new to mathematics might try to interpret the statement literally and get confussed. That's why I put "and only if" in brackets. I doubt that the Japanese have a problem with their English. Dharma6662000 (talk) 02:38, 26 August 2008 (UTC)[reply]

Well, there is [1] about this... and i accept your apologies... but now i regret for i wrote in your talk-page, sorry!.

You see, my point is that mathematicians have the duty of being honest always no matter what accordingly with us, mathematicians. Maybe you can understand better my position with the example in the discussion of area of a circle.--kmath (talk) 02:49, 26 August 2008 (UTC)[reply]

You're right! All that I ask is that when you point out such mistakes you do it in a nice way, and you add some kind of explanation, and not some kind of comment like "For your information, real mathematicians never use if and only if in definitions". Notice that in your reference it says "however, the word "if" is normally used in definitions", i.e. NOT ALWAYS. So given that "if" is normally used in definitions, we see that the statment "real mathematicians never use if and only if in definitions" is false! That's why I put "if and only if" in brackets and said that some people find its use pedantic.Dharma6662000 (talk) 02:56, 26 August 2008 (UTC)[reply]

ok, let's take all this as a friendly inter-language skirmish, bye--kmath (talk) 03:11, 26 August 2008 (UTC)[reply]

Let's! Buen idea mi amigo, buen idea. Dharma6662000 (talk) 03:15, 26 August 2008 (UTC)[reply]

Still confused

So I think I get that the trefoil-to-ring distortion is a homeomorphism, but is it or is it not a homotopy? Could the article state this explicitly? — Cheers, Steelpillow (Talk) 15:59, 12 January 2011 (UTC)[reply]

WTF is a "primage"?

I am a bit concerned by the following sentence:

The primage of certain sets which are actual open in the relative topology of the half-open interval are not open in the more natural topology of the circle (they are half-open intervals).

Is primage some kind of technical term in topology? I can't find a definition for it other than an additional payment made to the master of a freight vessel. It kind of looks like maybe a typo for pre-image, except that my reading of the context suggests they're just images (given that the function goes from the half-open interval to the circle, and not vice versa). In any case, it should be plural to match the are later in the sentence (I mean in are not open).

While I am here, is actual open an actual technical term, or just a German-inspired use of an adjective as an adverb?

Rdbenham (talk) 14:03, 6 April 2009 (UTC)[reply]

Primage is a typo for preimage. Note that the sentence refers to the inverse of the original mapping, that's why the domain and range are ostensibly reversed. You are also right about the plural and "actual open". — Emil J. 14:24, 6 April 2009 (UTC)[reply]

Poincare Quote

The Poincare quote in the introduction is really unnecessary and immature writing. If we include it, we should also include a "Webster's dictionary defines a homeomorphism as..." line. —Preceding unsigned comment added by 76.226.148.207 (talk) 17:50, 4 October 2009 (UTC)[reply]

Definition

I recommend defining f to be a homeomorphism if it has a continuous inverse. Look at this: f is a group isomorphism if either it's 1:1 homomorphism or it has a homomorphic inverse. But 1:1 is not sufficient for homeomorphisms. It's not clear why 1:1 suffices for homomorphisms but not continuous maps. The general definition that a morphism is an isomorphism is if it has a inverse morphism. But of course in proving stuff we use the 1:1 continuous open map def. My point is the inverse morphism def is much more universal and captures the concept of 'isomorphism'. Money is tight (talk) 13:05, 19 February 2010 (UTC)[reply]

Could you clarify what exactly do you want to change, and why? This article currently defines homeomorphism as a continuous function between two topological spaces that has a continuous inverse function.—Emil J. 13:22, 19 February 2010 (UTC)[reply]

Yes that what it says in the introduction. I'm suggesting we put emphasis in the Definition section. Sorry if I missed that the first time. Money is tight (talk) 21:59, 19 February 2010 (UTC)[reply]

But the definition in the Definition section is the same, it demands that the inverse function f⁻¹ is continuous. The only difference is that there is an added parenthetical explanation that this condition is equivalent to f being an open mapping. Are you suggesting to remove it? That's definitely not a good idea, the remark conveys a useful piece of information.—Emil J. 11:22, 22 February 2010 (UTC)[reply]

2022Jul26 text here defines a homeomorphism for as a function only, but this definition needs to also address correspondences. For example in the James Munkres’ second edition of Topology, page 103 reads: “An isomorphism is a bijective correspondence that preserves the algebraic structure involved. The analogous concept in topology is that of a homeomorphism; it is a bijective correspondence that preserves the topological structure involved.” Can someone please add a formal definition that addresses correspondences? (I am not a mathematician; otherwise, I would try.) For accessibility, the definition that addresses functions should remain. But this page would be greatly enhanced if it also addressed correspondences. — Preceding unsigned comment added by 136.54.56.233 (talk) 02:19, 27 July 2022 (UTC)[reply]

“Bijective correspondence” is a uncommon terminology for either a bijective function or the pair formed by a bijective function and its inverse function. In the case of algebraic structures, if a bijective function preserves the structure, the same is true for the inverse function, and the above ambiguity does not matter. Here, the ambiguity is confusing. Therefore the older terminology of correspondences must be avoided. D.Lazard (talk) 15:32, 27 July 2022 (UTC)[reply]

By correspondence I meant: a mapping (say f) from set A to set B such that the image f(a) can include multiple elements of B for each a in A. A function is a special case of a correspondence.

While many real analysis and topology texts do not address correspondences (and this seems justified in my opinion), there is such a thing as a bijective correspondence… For example, in Efe Ok’s book “Real Analysis with Economic Applications,” page 287 presents a definition of a bijective correspondence (surjectivity of a correspondence is straightforward; injectivity is more tricky, but this is defined: correspondence f is injective if for any distinct a, a’ in A, the intersection of f(a) and f(a’) equals the null set).

From my perspective, this is not about “older terminology.” The 2022Jul26 definition here is addressing only a special case of a correspondence —- the special case being a function. It seems that something as important as a wikipedia entry should address more than special cases, even though clearly a function is a very central special case. 136.54.56.233 (talk) 02:11, 28 July 2022 (UTC)[reply]

What you call a correspondence is what is usually called a binary relation. This article is about ´´´homeomorphisms´´´, which are bijective functions. As bijective correspondences and bijective functions are the same, I do not see anything to add to this article D.Lazard (talk) 06:48, 28 July 2022 (UTC)[reply]

Can the Doughnut/Mug Morph Be Turned Off?

The morph is very illustrative. But also very distracting! Can somebody make it so that it can be turned off or hidden? After watching it ten times I think I get the idea, and I want to concentrate on the text! It is beginning to make me hate coffee! CountMacula (talk) 10:51, 15 May 2011 (UTC)[reply]

Can we add some non-examples?

Objects which seem like they are homeomorphic but aren't would improve this article. For example, what are some continuous mappings that are not bicontinuous? Does it have to do with cadlag? Crasshopper (talk) 18:40, 30 August 2011 (UTC)[reply]

Continuous but not Bicontinuous

Can we get an example of a function that is continuous but not bicontinuous? Crasshopper (talk) 15:59, 30 September 2011 (UTC)[reply]

Animated example with donut

I thing that intermediate stages in animation of going from mug to doughnut is not a correct represantaition of continuous deformation, therefore are misleading. Compaire the Mugs 1. with closed top and 2. with open top. Are they same? — Preceding unsigned comment added by Lufnuf (talk • contribs) 09:55, 26 November 2014‎

Please, sign your posts in talk page with four tildes (~~~~).

The two mugs are not the same, but the first one is deformed to the second one by moving up the interior surface of the mug and expanding accordingly the matter of the mug. D.Lazard (talk) 10:24, 26 November 2014 (UTC)[reply]

Notation for homeomorphism

I am surprised that the notation $X\cong Y$ for two topological spaces being homeomorphic is not included anywhere in the article. Unless anyone has a good reason otherwise, I suggest it be included. The appropriate place would be to append the sentence "If such a function exists, we say X and Y are homeomorphic" with ", and we denote this by $X\cong Y$ ". Joel Brennan (talk) 13:06, 5 April 2019 (UTC)[reply]

This would need a source asserting that this is a standard notation in topology; that is that this notation is used in mathematical texts without being defined each time. D.Lazard (talk) 15:07, 5 April 2019 (UTC)[reply]

Animation of a continuous homeomorphism between $S^1\times S^1$ and a mug

This animation is misleading. The mug initially is filled in the animation and is pushed downwards vertically. Pick any arbitrary small neighbourhood that contains a point of the side of the mug and interior of the mug that is originally filled to see why this map is not a homotopy.

94.253.181.229 (talk) 15:32, 28 May 2019 (UTC)[reply]

I cannot understand your argument. What do you mean by "originally filled"? What is "the side of the mug" (the surface of the solid body, or its cylindrical part limited by two cylindrical surfaces)? What is the "interior" (the interior of the material of the mug or the interior of the region where one can put some liquid)?

The mug must be viewed as a solid limited by a surface. As such the animation is a deformation of the solid as well as of its surface. The first step of the deformation does not consists of filling the mug with something. It consists of a deformation of the "interior" surface (the part of the surface that touches the content of the mug when it is filled). So the animation show a homotopy, not only of the mug, but also of all the space containing it. D.Lazard (talk) 16:08, 28 May 2019 (UTC)[reply]

I refer to the homotopy where we deform the torus to the mug (lets just denote this topological subspace embedded in $R^{3}$ by $M$ . (I am not sure how to code here so I shall just use standard LaTeX) Let $H:S^{1}\times S^{1}\times I\rightarrow M$ be the homotopy represented in the animation. At some particular $t\in I$ we have $\operatorname {Im} H_{t}$ , the image of the function $H_{t}(s)=H(s,t)$ being the filled mug (to state this precisely let $C$ be the base of the mug, without loss of generality let this mug be based on the xy plane. Then $C\times I$ where we suppose the mug represented has unit length) is the solid I am referring to). Restricting $H$ to $[t,1]$ together with the appropriate linear transformation we yield a homotopy $H^{*}:C\times I\times I\rightarrow C\times I$ which by assumption that the animation represents a homotopy is continuous. Let us further suppose that $C$ is in fact a closed disk $D^{2}$ of unit radius centred at the origin, and hence $(1,0,1)$ is now a point of $C\times I$ . The animation seems to suggest, at least to me, that $H^{*}$ is a strong deformation retraction relative to $S^{1}\times I\cup D^{2}\times 0$ where $S^{1}=\partial D^{2}$ . However assuming the standard topology is endowed, which is metrisable, and we may hence unashamedly use epsilon delta methods to test continuity, pick any sufficiently large non zero $t\in I$ and the point $(1,0,1)$ . Then in every neighbourhood centred at $k=(1,0,1,t)$ we can then easily pick an epsilon, for which no small enough delta exists, such that $d(k,k^{*})<\delta$ implies $d(H^{*}(k^{*}-H^{*}(k))<\epsilon$ . Instead it appears to me that using maps similar to those constructed by Spanier in his book of Algebraic Topology in a proof in Chapter 3 of simplicial complexes seems to be a better illustration. – 94.253.181.229 (talk) 15:37, 29 May 2019 (UTC)[reply]

I cannot understand your reasoning: The deformation shown by the animation may be viewed as an homotopy between surfaces, namely the surface of the mug and the torus

T=S^{1}\times S^{1}.

It may also be viewed as an homotopy between solids (the interior volumes of these surfaces). Considering surface homotopy,

H

must be a map

H:T\times I\to \mathbb {R} ^{3}

such that the images of

H(\cdot ,0)

and

H(\cdot ,1)

are respectively

M

and

T.

This is the standard definition of an homotopy, and I cannot understand how your definition of

H

is related to it. However your definition of

H

suggests that you are considering a surface homotopy. Later in your text you consider the base of the mug, without specifying if it is a solid (with some thickness) or a surface (the lower surface of the basis). In any case, starting from an homotopy between surfaces

H

, and restricting it to a subinterval, you get an homotopy

H^{*}

between solids. At this point, I am totally confused, and I cannot read further. D.Lazard (talk) 20:19, 29 May 2019 (UTC)[reply]

The point that I made is now posted on Math Stackexchange. https://math.stackexchange.com/questions/3245164/is-this-animation-a-homotopy

I will comment again if someone paraphrases my point in a manner in which you might understand. Thank you.

213.149.61.21 (talk) 14:51, 31 May 2019 (UTC)[reply]

Giving an example of a map that is NOT a homotopy does not mean no homotopy exists. You may be thinking that points that are sent to the bottom of the mug get pushed straight down while their neighbors in the side of the mug stay stationary; instead they are gradually painted along the sides as we deform the top downwards. You can think of the points that end up on the bottom of the empty mug as being the top disk of radius ≤1/2 when it is "full", and all the points between radius 1/2 and radius 1 - ε get smeared down the inside walls of the mug (and all points between 1 - ε and 1 stay where they are). 50.207.106.34 (talk) 18:53, 9 February 2024 (UTC)[reply]

Why should differentiability matter?

The graph of a differentiable function is homeomorphic to the domain of the function.
A differentiable parametrization of a curve is a homeomorphism between the domain of the parametrization and the curve.

In both cases it should suffice to have just continuity rather than differentiability.—회기-로 (talk) 20:03, 25 May 2021 (UTC)[reply]

"Homeo" listed at Redirects for discussion

A discussion is taking place to address the redirect Homeo. The discussion will occur at Wikipedia:Redirects for discussion/Log/2021 September 5#Homeo until a consensus is reached, and readers of this page are welcome to contribute to the discussion. _signed,Rosguill ^talk 16:59, 5 September 2021 (UTC)[reply]

Delete "Homeo" is not a common abbreviation of homeomorphism. I believe that it not a common abbreviation of homeostasis, homeopathy or homeobox either. As it is ridiculous to create a dab page for disambiguating between uncommon abbreviations, deleting is the best solution. D.Lazard (talk) 17:15, 5 September 2021 (UTC)[reply]
D.Lazard, you're going to want to add that comment to the actual discussion at Wikipedia:Redirects for discussion/Log/2021 September 5#Homeo _signed,Rosguill ^talk 17:38, 5 September 2021 (UTC)[reply]

Coffee mug example misleading/Animation

The animation is an example of an isotopy, and isotopy equivalent is stronger than homeomorphic. The previous discussion headed "Animation" suggests that at some point someone edited the caption of the animation to make this clear, but the caption as it currently stands is misleading. You could change the last few words to "a coffee cup and a donut are not only homeomorphic but also isotopy equivalent." Or you could remove the animation and find something that is a better illustration of a homeomorphism. 50.207.106.34 (talk) 18:32, 9 February 2024 (UTC)[reply]

Also please notice that this same diagram on the "Homotopy" page is labeled as an isotopy. 50.207.106.34 (talk) 21:28, 9 February 2024 (UTC)[reply]

End of intro section misleading

I recommend editing the end of the intro section by changing the terminology for "a homeomorphism that is a continuous deformation" from homotopy to isotopy, because while "f is a homeomorphism that is a continuous deformation" does imply "f is a homotopy", the other direction is not always true; "f is a homotopy" does not imply "f is a homeomorphism that is a continuous deformation". Homotopies are often continuous deformations where there is no homeomorphism between f₀ and f₁. (Consider any homotopy of a nullhomotopic loop to a point.)

This statement on the page reads as a definition, and definitions should always be if AND only if, so using the term "isotopy" would be better. 50.207.106.34 (talk) 18:37, 9 February 2024 (UTC)[reply]

I have already answered at user talk:D.Lazard. In short, an isotopy requires an ambient space for being defined, and there is no ambient space here. Also, the last sentence of the lead must not be understood as definition of an homotopy, since the definition of an homotopy is given in the linked article. Also, such a definition would be out of scope here. So, this last sentence must be read as a property. D.Lazard (talk) 19:08, 9 February 2024 (UTC)[reply]

A homotopy also requires an ambient space, although we are not always explicit about identifying it, as the notion of *manifolds* being homotopy equivalent is extended from the definition of *functions* being homotopic by considering this embedding into the ambient space as the function. See Hatcher's Algebraic Topology page 3: "It is true in general that two spaces X and Y are homotopy equivalent if and only if there exists a third space Z containing both X and Y as deformation retracts."

The fact that this can be read as a property is why I described it as misleading rather than incorrect, but it would still be better described as isotopy. 50.207.106.34 (talk) 21:27, 9 February 2024 (UTC)[reply]

Again, an isotopy requires an embedding space, and is not defined for stand-alone spaces. D.Lazard (talk) 21:57, 9 February 2024 (UTC)[reply]

Then perhaps remove the term homotopy entirely, which also requires an ambient space/embedding space. 2001:5A8:4140:B900:C8E9:7B03:55AC:3D1F (talk) 22:45, 9 February 2024 (UTC)[reply]

Homotopy does not requires an ambient space. However, the lead was confusing in another way, by saying thet a homomorphism is (often) a continuous deformation, while the correct formulation is that it results often from a continuous deformation. I have fixed this. D.Lazard (talk) 12:37, 10 February 2024 (UTC)[reply]