# What is an efficient way to show two spaces are homeomorphic?

Okay here is the setup to a problem I have. We have $\mathbb{R}^3$ along with the standard topology and then two subspaces that are given the subspace topology. I forget the explicit definitions of the two subspaces $A,B$ but it is essentially a 3d parabola and a kind of cone shape. I can see looking at them that they would be homeomorphic I think but I don't know how to prove it.

So right now I have $A=\{(x,y,z) \in \mathbb{R^3}: \text{Something}\}$ and $B=\{(x,y,z) \in \mathbb{R^3}: \text{Something}\}$ and I want to show they are homeomorphic so what is a good way to do this?

Also I don't really see the significance of $A,B$ be given the subspace topology what difference does that make?

In general is it better to explicitly give such a homeomorphism or is there something else I can do?

• Giving $A$ and $B$ the subspace topology is just the normal way to turn them into topological spaces. If you don't give them a topology, then they're just sets. Also, if you give $A$ two different topologies, then the two copies are no longer homeomorphic, at least not via the identity map. So it makes a difference as to what $A$ will be homeomorphic to. Feb 20 '17 at 18:28
• As for how to show that $A$ and $B$ are homeomorphic, depending on their specific characteristics, you may be able to either write down a specific homeomorphism between them, or give a more abstract argument along these lines: math.stackexchange.com/questions/165629/… That specific result may not be useful to you, though. Feb 20 '17 at 18:31
• There is no efficient way to show that two spaces are homeomorphic. This task is extremely difficult and in most cases you just construct a homeomorphism directly or indirectly as a composition of homeomorphisms. It is alot easier to show that two spaces are not homeomorphic by looking at topological invariants. Feb 20 '17 at 18:58
• do not vandalise your posts... You may edit your question to clarify it but not delete it.
– Surb
Mar 4 '17 at 22:37

Every subset of $\mathbb R^3$ is its own special snowflake, and there is no algorithm for deciding when two of them are homeomorphic. It's not simply that no algorithm is known; no algorithm for solving this can possibly exist because there are too many subspaces.