Proving Topological Equivalence without finding a function

Question

My topology book defines topological equivalence as

Two metric spaces $(A,d_A)$ and $(B,d_B)$ are topologically equivalent if there are inverse functions $f:A\to B$ and $g:B\to A$ such that $f$ and $g$ are both continuous.

I've been using this definition to prove topological equivalence, but it get very tedious hunting for a continuous function and its inverse, and then proving the continuity of both.

How can I prove the existence of such functions without finding them? Does anybody know of an example of a proof that does this (perhaps even a proof here on math SE that shows topological equivalence without actually finding the functions)?

Answer 1

Your question can be generalized: how do you prove the existence of anything in mathematics, without finding it?

I have no answer to this question, I don't know how to do it.

There are of course many, many existence proofs in mathematics, and in particular many in topology. Often you can prove existence of a homeomorphism just by quoting one of those theorems, however that simply means that the construction which proves existence is hidden in the proof of the theorem.

A good example of this process is the classification of surfaces:

• If $S$ is a compact connected surface there exists a homeomorphism from $S$ to one of the surfaces in the following list: the sphere; a connected sum of $k$ tori; or a connected sum of $k$ projective planes (for some $k \ge 1$).

When you work through the proof of that theorem, you will see that it's a lot of hard work to construct the desired homeomorphisms. There's no getting around that.

But, I can then go on to prove existence corollaries without having to construct a homeomorphism, for example:

• If $S$ is a compact, connected surface of Euler characteristic 1 then there exists a homeomorphism from $S$ to the projective plane.

Proof: None of the other surfaces in the above list have Euler characteristic $1$, and Euler characteristic is invariant under homeomorphism. QED.

However, this does not really allow you to avoid the hard work of constructing a homeomorphism. Instead, this is just hiding the construction of the homeomorphism in the proof of the classification theorem.

But to me, there is a more important point: you are asking to take all the fun out of mathematics. I think of an existence proof as a hunt for an elusive quarry deep in a dark dank forest, or as digging deep into the earth for a rare, mythical, sparkly gem. There's something beautiful to be found, and I'm going to find it!

Answer 2

A short answer to your question is: In general, it is very hard to determine if two metric (or topological) spaces are homeomorphic.

Instead, there are ways to conclude when these spaces are NOT homeomorphic.

Let $X$ and $Y$ be topological spaces and suppose there exists a homeomorphism $\varphi\colon X\to Y$ (that is, $\varphi$ is a continuous bijection and $\varphi^{-1}$ is continuous).

One knows, for example, that continuous functions preserves connectedness and compactness, thus if $X$ (or $Y$) is compact (or connected) then $Y$ (or $X$) MUST be compact (or connected).

So, here's a way to conclude that two topological spaces are not homeomorphic: if one of them is compact (or connected) and the other is not, such a homeomorphism can't exist.

Answer 3

I don't think this answers your question but i believe it is a usefull thing to know and use it under circumstances.

$Theorem:$ Let $X,Y$ be topological spaces and $f : X \rightarrow Y$ be a bijective continuous function. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

If you find such a function and prove that it has a continuous inverse then ok.

If this is difficult then you can prove that the first space is compact and the last space is Hausdorf.

Some times it is easier to prove the second claim.It depends on how complicated are the topological spaces you study at the moment you want to prove their topological equivalence.

Besides that i don't know a way to prove the topological equivalence of two topological spaces without finding a homeomorphism.

Answer 4

The work of finding homeomorphisms can be hidden in a theorem whose proof gives a (often recursive) construction of a homeomorphism without explicitly giving one. This then can give a list of criteria to check to see that a space is homeomorphic to a standard one. E.g.

If $X$ is compact metric and has no isolated points then $X$ is homeomorphic to the Cantor middle third set.

If $X$ is countable metric without isolated points, then $X$ is homeomorphic to $\mathbb{Q}$.

Etc. There are quite a few such theorems, often with very non-trivial proofs; The following doesn't have a completely explicit proof in the sense you can write down a formula but it uses completeness arguments to show a homeomorphism must exist.

If $X$ is a completely metrisable separable locally convex topological vector space, $X$ is homeomorphic to $\mathbb{R}^\omega$.