Dear Antonio Pasini ,

I don't understand where is the difficulty in your question.

In elementary direct approach,

( If both X and Y are homeomorphic ( ≡ ) to Z, then X is homeomorphic to Y)

If X ≡ Y in any topological space V, then V - X ≡ V - Y.

Best regards

Dear Issam Kaddoura,

your statement is not true for the case when V is a topological space. For example, let V be the gluing of two exemplars of the segment (0,1) in the point 1/2. If X=\{1/2\}, then V-X has four connected components. If Y is a point different from 1/2, then V-Y has two connected components. So, V-X and V-Y are not homeomorphic.

With best regards, Mark

Dear Mark Pankov ,

It is a good (geometrical) attempt to construct a counterexample. Unfortunately, Your example for X shape is not accepted as ( topological manifold) where the neighborhood at the intersection point does not exist. ( It is not homeomorphic to any open interval in R) and X shape is neither a topological nor a differential manifold. Best wishes

Yes, for manifolds it does not work. It is only a counterexample for your statement "If X ≡ Y in any topological space V, then V - X ≡ V - Y. " nothing more :)

Hi Antonio,

As far as I remember, every non-oriented closed surface without boundary is identified by its Euler characteristic. Hence the join of a torus with k holes and a projective plane is homeomorphic to a join of a suitable number of projective planes (is it 2k+1 of them?). Note that removing the obvious Moebius band from the former leaves an orientable surface, and removing any one of the obvious Moebius bands from the latter leaves a non-orientable surface. Hence the answer is: not one. The exact number depends on the classification of closed surfaces with one boundary piece.

@Issam Kaddoura. Dear Issam, I am afraid that your claim is false. For instance, let X and Y be two simple closed curves on a surface V. Then obviously X and Y are homeomeorphic, but in general V-X and V-Y are not. For instance, it can happen that V-X is connected and V-Y is not. The same can happen if we replace curves with ordinary (two-sided, namely orientable) bands. If they are thin enough, they do the same job as curves.

@Sergey Shpectorov. Hi Sergey, very nice to hear (well, read) from you. Thank for your answer. I will think on it. However, I have a question about it: is it true that the Euler characteristic is enough to classify non-orientable compact surfaces (with no border)? Isn't the genus relevant too? Maybe I am wrong, but I think that non-orientable compact surfaces exist that have the same characteristic but different genus. For instance, let T be a torus an T' a 'double torus' (sphere with two handles). If we replace three disjoint disks of T with Moebius bands, we get a surface S with characteristic -3 ( = 0-3) and genus 4 (= 1+3). On the other hand, if we replace one disk of T' with a Moebius strip the resulting surface, say S', has characteristic -3 (= -2-1, the same as S) but genus 3 ( = 2+1). However, I am not so sure that nothing is wrong in this example. Anyway, you are right: the answer to my question sits in the classification of non-orientable compact surfaces.

@Sergey. Sorry, the example of my previous answer must be wrong. I believe you are right: genus and characteristic go together in the non orientable case too.

Dear Antonio Pasini ,

The following statement, " we want to replace a Mobius strip with a disk in a given surface." is not clear. Do you want to replace it via surgery techniques? Can you give an example to show how this can happen? It is well known that the boundary of the Mobius strip is (homeomorphic) to a simple curve that cannot be embedded in a plane. On the other hand, the boundary of the usual disc can be embedded in the plane.

@Issam. Yes, surgery techniques, of course. Examples: the real projective plane can be obtained from the sphere by replacing a disk with a Moebius band; which is the same as pasting a disk D with a Moebius band M along their borders. Next, if you replace the above disk D with another Moebius band you get the Klein bottle. Of course, this plastic surgery can be performed only in a 4-dimensional operating theatre, but this is not a problem; dimensions can be rented for free. As fo simple curves: by definition, every closed simple curve is homeomorphic to the unit circle S^1, which lives in the plane (as well as in any manifold of any dimension greater than 1).

Are the complements of any two pairs of homeomorphic compact, connected subsets of Rn homeomorphic (n = 1, 2, 3)?

In R^1:

The compact connected set is one of the following ( equipped with the usual topology):

[a,b] a closed bounded interval.

So, if [a,b] ≡ [c,d], then [a,b]^c ≡ [c,d]^c ≡ RUR.

and similarly for the case X = {singelton}, X^c ≡ RUR.

In R^2:

1. A compact connected set is a closed disc with n holes ( n may be zero), the boundary of the holes are included.

2. Simple and non-simple closed paths.

And the previous cases in R^1.

Also, we consider their combinations whose components have different dimensions (but connected)

So, the complements of these sets are also homeomorphic.

In R^3: It is not easy to discuss all the details.

For compact connected surfaces, we have:

A sphere with r handles attached (if S is orientable.)

A sphere with k crosscaps attached, (if S is non-orientable.)

In addition to the closed knots, beside all previous cases.

In general, the complements of the trivial and nontrivial knots

in R^3 are nonhomeomorphic.

Dear friends, the question I have asked has been answered by Sergey: if the surface (which is assumed to be compact, non orientable and with no borders) has even characteristic, then there is just one way to replace a Moebius band with a disk; otherwise, jus two ways. However, in view of interesting variations of my original question which have been proposed in the meanwhile, having got the above answer is not a sufficient reason to stop this discussion.

In this way X=Y and they are two closed curve on surface, if V-X=Y-V then this is property of topologic