# Talk:Homeomorphism

WikiProject Mathematics (Rated B-class, High-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 B Class
 High Importance
Field: Topology

## A technicality

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

## Unintelligible to newcomers

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

## 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)

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)

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)

## 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)

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)

Yeah, but any metric space is automatically a topological space. Oleg Alexandrov (talk) 15:30, 18 July 2006 (UTC)
oh thts fine in tht case.know a little about metric spaces, dont know anything about topology :-) --HellFire 10:56, 19 July 2006 (UTC)

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)

## 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)

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)
Every homeomorphism is a homotopy, but the converse is false. Two continuous maps ${\displaystyle f,g:X\to Y}$ are called homotopic if there exists a continuous map ${\displaystyle H:X\times [0,1]\to Y}$ such that ${\displaystyle H(x,0)=f(x)}$ and ${\displaystyle H(x,1)=g(x)}$ for all ${\displaystyle x\in X}$. Now, two spaces X and Y are called homotopic if there exist continuous maps ${\displaystyle f:X\to Y}$ and ${\displaystyle g:Y\to X}$ such that the maps ${\displaystyle g\circ f:X\to X}$ and ${\displaystyle f\circ g:Y\to Y}$ are homotopic to the identity maps. Well, assume that f and g are homeomorphisms, and in fact assume that ${\displaystyle g=f^{-1}}$. Then ${\displaystyle g\circ f=f^{-1}\circ f={\text{id}}_{X}}$ and ${\displaystyle f\circ g=f\circ f^{-1}={\text{id}}_{Y}}$. 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)

## 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 ${\displaystyle S^{1}\times I}$. Fix a point z in the circle and then cut the cylinder along ${\displaystyle J=z\times I}$. If you paste along J by identity you re-get the cylinder, but if you paste using a reflexion thru ${\displaystyle z\times {\frac {1}{2}}}$, you will get the mobius strip which is not homeomorphic to the cylinder.--kiddo 17:08, 8 April 2007 (UTC)
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)

No! A mapping ${\displaystyle f:X\to Y}$ 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)

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)
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)
For your information, real mathematicians never use if and only if in definitions.--kmath (talk) 19:45, 25 August 2008 (UTC)
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)
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)
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)
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)
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)
ok, let's take all this as a friendly inter-language skirmish, bye--kmath (talk) 03:11, 26 August 2008 (UTC)
Let's! Buen idea mi amigo, buen idea. Dharma6662000 (talk) 03:15, 26 August 2008 (UTC)
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)

## 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)

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)

## 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)

## 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)

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)
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)
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)

## 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)

## 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)

## 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)

## 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 (talkcontribs) 09:55, 26 November 2014‎