that is what I am searching.
This depends on what constraints you put on your surfaces. Since compact surfaces are classified by their homotopy (more specifically, by their orientability and Euler characteristic), you won't find an example there.
Going to open surfaces however, an example would be the cylinder and Möbius strip : both are homotopic to a circle, but they are not homeomorphic since one is orientable and the other is not.
thanks, and where could i find the ways of calculating the homotopy groups ?
It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?
Careful where you step in : ). In general, this is a very difficult question. But have a look at http://www.math.psu.edu/katok_a/TOPOLOGY/Chapter2.pdf for instance.
You may also want to look up "classification of two dimensional manifolds", this should give you some interesting links.
The two-dimensional case is always much simpler in such questions. Also I should have mentionned this: it is very easy to find a manifold that is homotopic but not homeomorphic to a given compact 2-manifold: it just won't be two-dimensional. (Can you see how to do that?)
As to your question, have a look at https://en.wikipedia.org/wiki/Borel_conjecture.
Examples exist in higher dimensions than 3, maybe in every dimension higher than 3, not sure.
An interesting sub-case is that of homotopically equivalent compact manifolds without boundary that are not homeomorphic. There are examples of 3 manifolds.
All 2-manifolds that are compact without boundary are determined by their fundamental group.
One can also ask whether a homotopy equivalence between homeomorphic manifolds is homotopic to a homeomorphism.
The thing with classification of manifolds, for the little I remember about it, is that it is a horribly complicated subject. Dimensions 4 and 3 tend to be the trickiest : one is trivial, two is easy, and 5 and above tend to fall into general rules (not that it's a trivial topic there by any means) - but the middle ones have entire books written about them, such as http://www.springer.com/mathematics/geometry/book/978-3-540-45898-2
Take a deep breath, and start with www.math.cornell.edu/~hatcher/AT/ATch4.pdf for instance.
For homotopy 3-spheres, you deal with Poincare's conjecture (a theorem now, by
Perelman , etc.).
Actually, even for 1-manifolds , if you include ones boundary, you have the open, half-open, closed intervals that are homotopy-equivalent but not homeomorphic. But maybe the most notable examples in any dimension are the contractible spaces, which are homotopically-equivalent to a point. So you can even have spaces of different dimensions being homotopic --but trivially not homeomorphic by cardinality reasons alone.
Yes the dimensions have to be the same but i thought R and R2 for example have the same cardinality ?
Yes. That is correct.
R and R2 are not homeomorphic but this is a deep theorem.
In order to take away obvious cases such as the Moebius band and the circle, restrict to the case of compact manifolds without boundary. Such spaces are never contractible and never have the same homotopy type as compact manifolds without boundary in other dimensions.
As Wabbit pointed out, this gets extremely complicated. For three manifolds there are fundamental groups which determine the manifold up to homeomorphism and others that determine the homotopy type but not the homeomorphism type. The Poincare conjecture is true in all dimensions but the methods of proof are quite different for dimensions 5 and above than in dimension 4 and dimension3.
The homotopy groups of spheres is a whole subject by itself.
Here are some problems:
- Take ##R^3## minus both the z-axis and the unit circle in the xy-plane. Prove that this space is homotopically equivalent to a torus.
- A subspace A of a topological space W is called a strong deformation retract of W if there is a homotopy F: WxI -> W such that the following 3 conditions hold:
F(x,0) is the identity map
F(x,1) is completely contained in A
The composition A -> WxI ->A where the first arrow is the inclusions a -> (a,1) and the second is the homotopy,F, is the identity map on A.
Show that A is homotopically equivalent to W
- Show that if the unit disk has a fixed point free mapping into itself then it is homotopically equivalent to the circle. In fact the circle would be a strong deformation retract of the disk. Conclude Brouwer's Fixed Point Theorem which says that every mapping of the disk into itself has a fixed point.
Thanks for the problems. I wanted to ask a question more : for pi1 we say that we take a loop and describe it by a continuous function f : [0;1]->Manifold such that f(0)=f(1)
But what if there is a knot in the loop we contract in the space ? Is a knot contractible to a point ?
I think a loop with a knot is homotopic to a circle but not homeomorphic this could be another simple example.
One thing to note is that the non homeomorphism of ## R^n ## and ## R^m ## is actually a nice early result of homotopy theory : once you remove one point from each, they are homotopic to the corresponding (n-1)-sphere, and the calculation of the lower* homotopy groups of spheres is one of the few easy ones you can make. If you are interested in the subject you should definitely look up this calculation which is found in pretty much any introduction to higher homotopy groups.
* Of course this being homotopy, nothing is easy, which is why I had to add "lower" to this statement : the calculation of ## \pi_k(S^n) ## is easy only for ## k\leq n ## - but this turns out to be sufficient for the purpose at hand.
In homotopy theory loops are allowed to self intersect and to pass through themselves. So a knot doesn't matter. For instance, a space filling loop that completely covers the sphere is homotopic to a point.
I suppose one could restrict the homotopies to disallow crossovers but that is not what homotopy theory assumes. I imagine in Knot theory one would make this restriction.
...except I had to remove the silly statement "non homotopy equivalence of ## R^n ## and ## R^m ## " from that post, but you have now captured my blooper for all posterity to contemplate : )
I'll leave it to the opening poster to see why it's such a silly statement.
Here are some problems to give you an idea of what happens when you glue together manifolds with boundary.
- Show that the unit circle bundle of the 2 sphere is two solid tori glued together along their boundaries. Conclude that one can obtain real projective three space from two solid tori.
- Show that the 3 sphere can also be obtained from two solid tori glued together along their boundaries.
So you see that the gluing map matters. What other 3 manifolds can be obtained from gluing 2 solid tori together?
- Show that one can also obtain the three sphere by gluing two solid balls together along their boundaries,
Bloopers don't matter but I will be happy to delete if you want.
Not at all, I was just kidding, let this remain as a lesson in proof-reading before posting : )
Separate names with a comma.