# Geometric difference between a homotopy equivalance and a homeomorphism

1. Jul 19, 2012

### Flying_Goat

Geometrically, what is the difference between saying 'X is homotopic equivalent to Y' and 'X is homeomorphic to Y'? I know that a homeomorphism is a homotopy equivalence, but I can't seem to visualise the difference between them. It seems to me that both of these terms are about deforming spaces continuously and I don't see why(intuitively) homotopy equivalance is a weaker notion than a homeomorphism.

Quoting from wikipedia, "A solid disk is not homeomorphic to a single point, although the disk and the point are homotopy equivalent". There does not exist a continuous bijection between a point and a solid disk which stops them from being homeomorphic. Does that mean you can't 'continously shrink' a solid disk into a point? If not, then geometrically what does the fact that those two are homotopic equivalent tell us?

I am in a situation where I know the definitions, but can't see the 'picture'.

Any help would be appreciated.

2. Jul 20, 2012

### Bacle2

Actually, the fact that there is not an isomorphism between a solid disk and a

point--by cardinality reasons, for one-- precisely shows that the two are not

homeomorphic. The fact that the solid disk (embedded in R^2, I assume) is

contractible precisely means that it _can_ be shrunk to a point within the space.

Homotopy equivalent here (within R^2) does have a nice geometric interpretation:

Draw a cylinder ; fill the bottom circle into a disk, and draw a single point {p} in the

top circle. Then you can see how the solid disk can be "continuously massaged" into

the point {p}; say by shrinking the disk gradually into a point. You can think

maybe of a movie where you start with the disk and end with the point {p}, where

there are no surprises in-between.

A homotopy within an abstract topological space does not have as nice of a

geometric interpretation that I know off.

3. Jul 20, 2012

### lavinia

Here are two surfaces that are homotopy equivalent but not homeomorphic.

The Mobius band and the cylinder. Each can be smoothly shrunk by a homotopy to their equatorial circles. Yet one is an orientable surface and the other is not.

Similarly a solid Klein bottle is homotopy equivalent to a solid torus.

Here is a wilder example as an exercise.

Remove the z-axis and the circle of radius 1 in the xy-plane from Euclidean 3 space. Show that this is homotopy equivalent to a torus.

4. Jul 20, 2012

### mathwonk

cute example. just visualizing it mentally, it appears that the given subspace of R^3 is homeomorphic to the normal bundle to the torus, "hence" homotopy equivalent to the torus.

note homotopy allows squashing, while homeomorphisms are bijective.

the simpler examples are also vector bundles, but on the circle. It seems any vector bundle on a space is homotopy equivalent to that space. This is just an extension of the fact that R^n for any n, is homotopy equivalent to a point.

in particular homotopy equivalent spaces need not have the same dimension, but homeomorphic ones do.

(I am just tossing this off the top of my head, and not checking the definition of homotopy carefully, but it seems ok.)

Last edited: Jul 20, 2012
5. Jul 20, 2012

### Flying_Goat

Thanks for your replies. That made alot of sense. So a homotopy equivalence allows collapsing a bunch of points into one, such as that of a disk to a point. Homeomorphism is 'stronger' in the sense that the deformation is 'bijective'.

6. Jul 20, 2012

### homeomorphic

Deformation retractions are easier to visualize. There's a theorem that says that two spaces are homotopy equivalent if and only if they are both deformation retracts of some other space (possibly, maybe some extra hypotheses are needed here).

I think it's not too hard to see this for CW complexes if I remember right, maybe using the homotopy extension property for CW pairs or some such thing, but I would have to think a bit to remember how the argument goes. Too lazy for the moment.

Also, an interesting theme in more advanced topology is that often homotopy information tells you something about homeomorphism type for certain very special spaces. For example, for certain nice enough 3-manifolds, the homeomorphism type is determined by the homotopy type. For my qualifying exam, I outlined the proof of that fact. It's pretty involved.

Some other amazing examples of this happen in high dimensional topology, as with the h-cobordism theorem and the s-cobordism theorem (not exactly saying homotopy equivalent is equivalent to homeomorphic in this case, but the fact that certain maps are homotopy equivalences implies that two manifolds are homeomorphic).

7. Jul 21, 2012

### lavinia

Interesting homeomorphic

A couple questions

Can you give an example of two homotopy equivalent compact manifolds without boundary that are not homeomorphic?

Are there fundamental groups that determine the homeomorphism type of a manifold?

8. Jul 21, 2012

### mathwonk

I believe two simply connected oriented differentiable 4 manifolds (compact no boundary) are homeomorphic if even the cup product forms on second cohomology are isomorphic.

I do not expect the fundamental group to determine the homeomorphism type of a manifold since I believe all smooth complex algebraic surfaces (real 4 manifolds) in CP^3 are simply connected.

Of course the homeomorphism type of compact orientable 2 manifolds with no bndry are determined by very little data, even the 1st homology group hence also the fundamental group.

9. Jul 21, 2012

### homeomorphic

I'm assuming you mean of the same dimension, otherwise, we could take any ℝ^n for two different values of n. In dimension 2, of course, among compact 3-manifolds without boundary, there's no difference between homeomorphism type and homotopy type. I should know an example for compact manifolds of the same dimension that are homotopy equivalent, but not homeomorphic, but I can't think of any concrete examples. However, the following tells me they exist.

That's roughly what Freedman's theorem says, but you need a Kirby-Siebenmann invariant to pin down the homeomorphism type (which is a Z mod 2 invariant, so it just splits it into two cases). And in this case, the intersection forms determine the homotopy type (an older theorem of Whitehead). So, the guys with the same intersection forms, but different Kirby-Siebenmann invariants are evidently homotopy equivalent, but not homeomorphic (because Freedmans theorem is an if and only if statement).

I don't think so. However, the fundamental group controls Whitehead torsion, which determines which homotopy equivalences are simple-homotopy equivalences. By playing around with handles and cobordisms and Whitehead torsion, you get the s-cobordism theorem. So, in these cases, the fundamental groups and induced maps between them will tell you that some relevant Whitehead torsion vanishes, which is one of the hypotheses you need to conclude that two ends of a suitable cobordism between the manifolds are homeomorphic.

Actually, the theorem whose proof I outlined for my qual actually says that if you have a continuous map that induces an isomorphism on the fundamental group, it is homotopic to a homeomorphism for Haken 3-manifolds, I guess (the book I used called it compact, P^2-irreducible, sufficiently large 3-manifolds). It's generally not true that just having isomorphic fundamental groups will give you a homeomorphism, although that is true for aspherical manifolds because, in that case, any isomorphism of the fundamental groups will be induced by a continuous map. A big theme in 3-manifolds is to study how the fundamental group influences the topology of the 3-manifold. For example, if the fundamental group decomposes as a free product, then, under suitable conditions, the manifold decomposes as a connected sum with each summand having the free factors as its fundamental group.

10. Jul 21, 2012

### Bacle2

Re the intersection forms of 2n-manifolds:

If Manifolds are homeomorphic* , then their Q_M and Q_M' are equal;
homeomorphism would preserve the respective 2-homologies; a homotopy-
equivalence would be enough, actually.

But we cannot say: "if Q_M and Q_M' are equivalent, then M and M'
are diffeomorphic". : we can keep the intersections (therefore the equivalence of the Q_M's ) . Take , e.g.,
M' = M \/ (some handles with no effect on middle-dimensional homology).
Then map X' to X by collapsing those handles.

Say M = CP^2, M' = CP^2 # S^1 x S^3, the connected sum,
and map M' ---> M , and then collpase the S^1 x S^3 to a point. Then, the 2nd homology
and coh. respectively, are preserved, and so are the intersections. But, clearly, there is no homeomorphism **

* I hope poster homeomorphic did not trademark his name.

11. Jul 21, 2012

### lavinia

These are not compact

12. Jul 21, 2012

### homeomorphic

Oh, sorry, I forgot you wanted compact. So, you'd just need two closed manifolds of different dimensions that have the same homotopy type. It's not obvious how to do that if you don't want either to have boundary or if it's possible to do at all. But my other example works.

13. Jul 21, 2012

### homeomorphic

Freedmans theorem says the equivalence of intersection forms, plus the Kirby Siebenmann invariant implies homeomorphism. It's very important to say homeomorphism, not diffeomorphism. Actually, Freedman's theorem is a source of many exotic 4-manifolds because we have diffeomorphism invariants like Seiberg-Witten invariants that can distinguish 2-manifolds that have the same intersection form and Kirby-Siebenmann invariant.

S^1 × S^3 is not simply connected (fundamental group is Z), so Freedman's theorem doesn't apply here.

14. Jul 21, 2012

### mathwonk

I am confused. I read that ctc wall proved that simply connected differentiable oriented 4 manifolds with isomorp-hic intersection forms on second homology are h cobordant:

http://www.maths.ed.ac.uk/~aar/papers/w4.pdf

Then I thought I read in wikipedia that h cobordism implies homeomorphism in dimension 4, after freedman. What have I missed?

notice I carefully restricted my statement to homeomorphism of differentiable manifolds, hence piece - wise linear, so the kirby siebenmann invariant apparently does vanish.

Last edited: Jul 21, 2012
15. Jul 21, 2012

### homeomorphic

Yes, that's right. Doesn't look like you missed anything.

16. Jul 22, 2012

### lavinia

I think that the closed manifolds must have the same dimension since homotopy equivalence implies isomorphic homology.

17. Jul 22, 2012

### lavinia

OK. These 4 manifold theorems seem very powerful. I will read that paper by Wall.

My question was a little different. Take a group - say the fundamental group of the Klein bottle or of the two ringed torus. How many closed manifolds without boundary can you make that have these group as fundamental groups?

Is there an easy construction that shows that there are infinitely many?

Last edited: Jul 22, 2012
18. Jul 22, 2012

### homeomorphic

Right, the one with bigger dimension will have higher up homology groups, so, yeah, it's not possible to pull off that idea if you're talking about closed manifolds.

19. Jul 22, 2012

### homeomorphic

It's easy to construct a CW complex having any given fundamental group by taking a wedge of circles and then gluing on disks to kill the relations. For 4-manifolds, the same type of trick will work if you use handles instead of cells, but there's some subtlety there because for 3-manifolds, not every group can be realized as a fundamental group. So, you can show that there exists a 4-manifold with each fundamental group. And you can attach higher handles until it gets capped off, I think, so that it's closed without affecting the fundamental group. Then, just connect sum with a bunch of simply-connected guys, and that should give you infinitely many with the same fundamental group.

20. Jul 22, 2012

### lavinia

very cool. So just to get a feel for this, how do I construct a closed 4 manifold without boundary whose fundamental group is the fundamental group of the Klein bottle?

Also suppose you take a high dimensional lattice, say a 95 dimensional lattice (free abelian group). On the universal covering space of this manifold, this huge lattice must act on a simply connected 4 manifold. That would be interesting to see. Do you just add 95 copies of S^1xS^3 to the 4 sphere with 190 open balls removed?

Last edited: Jul 22, 2012