# Main Theorems/Techniques for proving Homeomorphism?

General Question: what are the most common Theorems/Methods used to prove Homeomorphism?

I encountered:
- find the map explicitly
- use the Compact-to-Hausdorff Lemma
- find cts maps $f$ and $g$ s.t. $f\circ g=g\circ f= i$ where $i$ is the identity map.

Can anyone deepen/correct/enlarge my list?

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Are you aware that the third point is the definition of (a pair of) homeomorphisms? – Dan Rust Dec 21 '13 at 18:56
More or less, what really confuses me is the necessity of checking both compositions (I thought that once $f\circ g=i\Rightarrow f^{-1}=g\Rightarrow f^{-1}$ cts and you are done) – Ton Dec 21 '13 at 19:20
It's to ensure bijectivity. Suppose $X=[0,1]$ and $Y=S^1=\{e^{i2\pi t}\mid t\in[0,1]\}$. Let $g\colon X\to Y$ be given by $g(x)=e^{i\pi x}$ and $f\colon Y\to X$ is given by $f(z)=Re(z)$. Because $g$ maps $X$ to the 'upper hemisphere' of the circle, we see that $f\circ g$ is the identity. However, $g\circ f$ maps points on the lower hemisphere to the upper hemisphere so it is not the identity. – Dan Rust Dec 21 '13 at 19:52

Suppose that $f: X \rightarrow Y$ is a bijective continuous map. In general, from this it does not follow that $f^{-1}$ is continuous, so $f$ is not necessarily a homeomorphism.

However, there are some special cases where $f: X \rightarrow Y$ bijective and continuous implies that $f$ is a homeomorphism. In any introductory course on functional analysis, one comes across to the open mapping theorem. As an easy corollary, you get the bounded inverse theorem:

Bounded Inverse Theorem: Let $X$ and $Y$ be Banach spaces. Suppose that $T: X \rightarrow Y$ is a bijective, linear and continuous map. Then $T$ is a homeomorphism.

This is a very special case, here linearity and the algebraic structure of $X$ and $Y$ is vital for the above theorem to hold. There is also the following theorem, proven in the 1960's.

Theorem: Any two separable infinite-dimensional Banach spaces are homeomorphic.

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Do you have a reference for the theorem about separable Banach spaces? +1 – Olivier Bégassat Dec 21 '13 at 22:27
@OlivierBégassat: I believe the result is due to Kadets, here his paper from 1967: link (I have not read this paper, but it seems accessible) Also, I forgot the condition that the spaces should be infinite-dimensional.. – spin Dec 21 '13 at 22:59

Probably the most useful method is to give a full classification of a certain class of spaces. A classification of spaces satisfying some property $P$ amounts to giving a set of criteria for such spaces (hopefully a finite set of criteria) such that two spaces with property $P$ are homeomorphic if and only if both spaces coincide for all criteria. For instance

• $2$-dimensional manifolds ($3$-dimensional manifolds are nearly done). These are classified by their genus (or Euler characteristic, or homology, or fundamental group) and their orientability. If you want to consider manifolds with boundary and marked-points too, it amounts to also counting boundary circles and marked points.
• Perfect totally-disconnected compact metric spaces. The classification of these spaces is actually really easy - there's only one space up to homeomorphism with these properties and it's the Cantor set. A common technique to show that a space is homeomorphic to the Cantor set is to simply check that it satisfies all of the above properties.
• Finite space. This one is an easy one. You just need to check cardinality and then all permutations of the elements and the action of this on the topology. I suppose this amounts to your first point of 'finding a map explicitly', but I mention it because we can do this in an algorithmic way, so finite spaces are particularly suited to this method - even if the computational complexity of such a check isn't feasible to be done in practice.
• Knot complements in the sphere. Because a knot complement $S^3\setminus K$ is homeomorphic to $S^3\setminus K'$ if and only if $K$ and $K'$ are isotopic knots, this amounts to then classifying knots. This is a hard problem but (as far as I'm aware) is expected to be solvable and already has a vast array of machinery that has been developed for this purpose.
• Denjoy continua. These are spaces which are embedded in the torus $T^2$ which, without getting technical, can be thought of as removing a disk from the torus and then 'stretching' that hole around the torus along some geodesic with irrational slope (we take the limit of this stretching process). If the 'stretching line' of a Denjoy continuum $D$ has slope $\alpha$ with continued fraction expansion $[a_0:a_1,\ldots]$ and another Denjoy continuum $D'$ has stretching line with slope $\beta$ and continued fraction expansion $[b_0:b_1,\ldots]$, then $D$ is homeomorphic to $D'$ if and only if there exist $k$ and $k'$ such that $(a_k,a_{k+1},a_{k+2},\ldots)=(b_{k'},b_{k'+1},b_{k'+2},\ldots)$.
• Solenoids. These are inverse limits of maps from the $n$-torus to itself. They can be fully classified by just considering the degrees of the maps appearing in the inverse limit (disregarding order) up to a finite deletion of those maps appearing in the limit. This is another result attributed to Fokkink in his PhD thesis, as with Denjoy continua.

There are, of course, many examples I've either skipped, forgotten, don't know of, or don't know enough about to mention.

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I would add the open mapping theorem: if $T: X \rightarrow Y$ is a bijective continuous linear map between two Banach spaces, then $T$ is a homeomorphism. – spin Dec 21 '13 at 21:48
@spin That would make a great answer if you wanted to flesh it out a little (I'm not too knowledgeable of functional analysis). – Dan Rust Dec 21 '13 at 22:00
You can certainly eleminate a priori many of the bijections of a finite set as potential homeomorphisms in most cases. – Olivier Bégassat Dec 21 '13 at 22:29
@OlivierBégassat That's interesting - is there a well-developed theory? I've heard in passing that the theory of finite topological spaces is something which is seriously studied, but I know next to nothing about it. – Dan Rust Dec 21 '13 at 22:36
I don't know much about it, but I was thinking about minimal open sets : if $X$ is a finite topological space, and $x$ is a point of $X$, then you can consider the intersection of all open sets containing $x$, say $\mathcal{O}_x$. Any homeomorphism of $X$ will have to permute the elements ($x\in X_i$) that determine minimal open sets of the same cardinality ($|\mathcal{O}_x|=i$), so there is a partition of $$X=\coprod_{i=1}^r X_i$$ that all the homeomorphisms preserve. If that partition is non trivial, then $$\mathrm{Homeo}(X)\subset\prod_{i=1}^r\mathfrak{S}(X_i)$$ – Olivier Bégassat Dec 21 '13 at 22:44