## Homeomorphisms in Topology

One question I've had lately in my independent study of topology is the problem of how to show two sets are homeomorphic to each other. I am not sure how I would go about doing this in a general, or even specific case. One problem that wants me to demonstrate this is in Mendelson:

Prove that an open interval (a,b) considered as a subspace of the real line is homeomorphic to the real line.

Now, I know that a homeomorphism is a map where both it and its inverse are continuous, so I'm wondering if all that is needed to show two sets are homeomorphic is to simply define a mapping between the two. However, I don't know if this is right. Can someone please clarify?

Thanks.

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age

Blog Entries: 1
Recognitions:
Gold Member
 Quote by philosophking I'm wondering if all that is needed to show two sets are homeomorphic is to simply define a mapping between the two.
Yes, that is pretty much it. The mapping has to be one-to-one so that an inverse can exist. The mapping has to be continuous and the inverse has to be continuous. So here the task is to define a mapping from the interval (a,b) to the entire real line $(-\infty,+\infty)$.

 How would you do that? I saw someone once parametrize to make it work, but I guess I just don't know how one would go about doing that.

Mentor

## Homeomorphisms in Topology

What's with the LaTeX preview feature? It never seems to work for me.

Back to the matter at hand.

Hint: think tan(x) restricted to the domain (-pi/2 , pi/2).

Regards,
George

 Mentor If you don't want to muck around mapping (a , b) onto (-pi/2 , pi/2), an easier example that I think works is f: (a , b) -> R given by f(x) =1/(x -a) + 1/(x - b). Regards, George
 Great, thanks a lot you guys. That was a big help. I guess it just depends on being creative in your map-defining.
 How about showing that two spaces are not homeomorphic. To do that you would have to show that one has a topological property that the other does not right? By topological property I guess I mean any property of a topological space that is always preserved by homeomorphism... Maybe its even harder to prove that any propety is always preserved by homeomorphisms... (I don't immediately see how to prove such a statement). maybe I've just confused myself. I guess a better way to prove two spaces are not homeomorphic would be to just assume they are homeomorphic and show that the existence of a homeomorphism between them leads to a contradiction. Am I right? or does that first method work after all?

Recognitions:
Homework Help
 Quote by Cincinnatus How about showing that two spaces are not homeomorphic. To do that you would have to show that one has a topological property that the other does not right?
well, trivially, the answer is "yes" since "is a member of some equivalence class of topological spaces modulo homeomorphism" is a topological property. That is to say determining homeomorphism type and showing they are not equivalent would obviously do.

 By topological property I guess I mean any property of a topological space that is always preserved by homeomorphism... Maybe its even harder to prove that any propety is always preserved by homeomorphisms... (I don't immediately see how to prove such a statement).
no, it is usually easy to prove certain properties are always preserved by homeomorphism. What is hard is generating a complete set of data that uniquely characterizes the space and is not as difficult as working out every possible homeomorphic image.

 maybe I've just confused myself. I guess a better way to prove two spaces are not homeomorphic would be to just assume they are homeomorphic and show that the existence of a homeomorphism between them leads to a contradiction. Am I right? or does that first method work after all?
your "two methods" appear to be the same: show that they have some different invariants under homeomorphism (that would be the contradiction). there are some easy to calculate invariants: homotopy groups, homology groups etc.

In the case of smooth (connected compact) 2-folds then the genus and orientability determine the space up to homeomorphism. That is every compact connected surface is homeomorphic to either a torus with n holes (n in N) or a torus with n holes and a mobius cap stitched in. (equivalently the fundamental group determines it).

in the case of 3-folds (things that look locally like R^3) then we do not know what data is sufficient to characterize them. see eg the poincare and geometrization conjectures.