2.07 Homeomorphisms

(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.

(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,

• write \(T\subset\mathbf{R}^3\) for the standard torus in \(\mathbf{R}^3\).
• define \(T'=\{(\cos\theta,\sin\theta,\cos\phi,\sin\phi)\ :\ \theta,\phi\in[0,2\pi)\}\subset\mathbf{R}^4\) to be the torus in 4-d.

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 [[./topsp03.org][\(\cos\) 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.