I was asked to show that $f: \mathbb{R} \to \mathbb{R}$ given by $f: x \mapsto 2x$ induces a continuous map $\bar{f}:S^1 \to S^1$ for $S^1 = \mathbb{R}/(2\pi\mathbb{Z}) = [0, 2\pi)$. I was told to think in terms of projecting the unit circle onto the reals and arguing that $\bar{f}$ is a composition of continuous maps. I didn't fully understand this argument and in fact can't get myself to even believe this map is continuous.
If we were working over the reals, I agree it's obviously continuous, but the fact that $\bar{f}$ is defined over a clopen quotient field seems problematic for me. This is because a continuous map, intuitively in my mind, is one that guarantees that small perturbations in the input space lead to small perturbations in the output space. If I think of the unit circle however, the doubling map at input $\pi$ gives $2\pi \cong 0$ in the output space. However, at input $\pi – \epsilon$, it gives $2\pi – \epsilon$ which is very far from $0$. So by decreasing the input by an infinitesmal amount, the output "jumped" from $0$ to $2\pi – \epsilon$, so I don't see how such a map could be continuous.
I would appreciate clarity on the "composition" argument, and how that relates to my mistaken graphical intuition about "small perturbation."
Best Answer
For this, it is probably better to identify $S^1$ with $S^1 = \{z \in \mathbb{C} : |z| = 1\} = \{e^{it} : t \in [0, 2\pi)\} \subset \mathbb{C}$ equipped with the induced metric from $\mathbb{C}$. I think this identification is a homeomorphism. In this case, the doubling map $x \to 2x$ is the map $f(e^{it}) = e^{2it}$. Equivalently, $f(z) = z^2$. Since $z^2$ is continuous, $f$ is continuous.
This explains why $2\pi - \varepsilon$ is close to $0$: the point on the unit circle with angle $2\pi - \varepsilon$ is close to the point with angle $0$.