(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 $\cos$ 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.