Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 13E / b / ii / Solution

Use the elementary irrational-rotation density fact: if $\alpha$ is irrational, $\{n\alpha \mathrm{mod}1:n\in \mathbb{Z}\}$ is dense in the circle. One proof uses the pigeonhole principle on fractional parts of $0,\alpha ,\dots ,N\alpha$. Two lie within $1/N$; taking their difference and, if needed, its negative yields an integer multiple of $\alpha$ with fractional part $\delta \in (0,1/N)$. The multiples $0,\delta ,\dots ,\lfloor 1/\delta \rfloor \delta$ form a $1/N$-net of the circle and are themselves integer multiples of $\alpha$ modulo one. Taking $N$ arbitrarily large proves density.

For any target coordinates $(u,v)$ modulo one, points of the line with parameter $t=u+n$ have first coordinate exactly $u$ modulo one and second coordinate $\alpha u+n\alpha$ modulo one. The density fact approximates $v$ arbitrarily well. Since products of small coordinate intervals form a neighbourhood basis of the torus,Nevertheless the point $\pi (0,\alpha /2)$ does not lie in $\pi (L)$. Such an equality would require a line parameter $t=n\in \mathbb{Z}$ and $\alpha (n-1/2)\in \mathbb{Z}$, contradicting irrationality. HenceThis is the irrational winding of the torus: dense need not mean surjective.