2.07 Homeomorphisms

Below the video you will find accompanying notes and some pre-class questions.



(0.15) A continuous map \(F\colon X\to Y\) is a homeomorphism if it is bijective and its inverse \(F^{-1}\) is also continuous. If two topological spaces admit a homeomorphism between them, we say they are homeomorphic: they are essentially the same topological space.

(1.13) Consider the half-open interval \([0,2\pi)\) and the continuous map \(F\colon [0,2\pi)\to S^1\) defined by \(F(\theta)=e^{i\theta}\). This is continuous and bijective. However, it is not a homeomorphism! We will see that the circle has fundamental group \(\mathbf{Z}\) and the interval is simply-connected, so they cannot be homeomorphic.

Criterion for a map to be a homeomorphism

(3.33) Let \(X\) be a compact space and let \(Y\) be a Hausdorff space. Then any continuous bijection \(F\colon X\to Y\) is a homeomorphism.
(5.00) We need to show that \(F^{-1}\) is continuous, i.e. that for all open sets \(U\subset X\) the preimage \((F^{-1})^{-1}(U)\) is open in \(Y\). But \((F^{-1})^{-1}(U)=F(U)\), so we need to show that images of open sets are open. It suffices to show that complement of \(F(U)\) is closed.

(6.23) \(U\subset X\) is open, so \(X\setminus U\) is closed, and since \(X\) is compact, this means \(X\setminus U\) is closed (closed subsets of compact spaces are compact). The image of a compact set is also compact, so \(F(X\setminus U)\) is compact. A compact subset of a Hausdorff space is closed, so \(F(X\setminus U)\) is closed, so \(F(U)=Y\setminus F(X\setminus U)\) is open, as required.


(8.46) In the session on the subspace topology, we saw that the 2-torus \(T^2\) can be thought of as a subset of \(\mathbf{R}^3\) and also as a subset of \(\mathbf{R}^4\). More precisely, I will show that they are both homeomorphic to \(S^1\times S^1\).

(10.22) We need a map \(F\colon S^1\times S^1\to T'\) which will be \((e^{i\theta},e^{i\phi})\mapsto (\cos\theta,\sin\theta,\cos\phi,\sin\phi)\). It is a continuous map (we saw that \(\cos\) and \(\sin\) are continuous functions on the circle) and it is bijective. The circle is a closed and bounded set in \(\mathbf{R}^2\), so it is compact; the product \(S^1\times S^1\) is compact by Tychonoff's theorem. The image \(T'\) is a subspace of a Hausdorff space, hence Hausdorff. Therefore \(F\) is a homeomorphism.

(11.53) We need to do the same for \(T\), and the same argument will apply provided I can give you a continuous bijection \(G\colon S^1\times S^1\to T\). I claim that the following map will do: \[G(e^{i\theta},e^{i\phi})= \left(\begin{array}{ccc}\cos\phi&-\sin\phi&0\\ \sin\phi&\cos\phi&0\\ 0&0&1\end{array}\right) \left(\begin{array}{c}0\\2+\cos\theta\\\sin\theta\end{array}\right).\] In other words, I am taking the unit circle in the \(yz\)-plane centred at \((0,2,0)\) (angle coordinate \(\theta\)) and rotating it by an angle \(\phi\) around the \(z\)-axis.

Pre-class questions

  1. Let \(X\) be the set \(\{0,1\}\) equipped with the discrete topology and let \(Y\) be the set \(\{0,1\}\) equipped with the indiscrete topology. Write down a continuous bijection \(X\to Y\). Are these spaces homeomorphic? If not, why does the theorem from the video fail to apply in this case?


Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.