Proving $S^1$, the unit circle in $\mathbb{R}^2$, is path-connected

Question

I am trying to prove that $S^1 = \{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\}$ is path connected, so I need to, for any $(a,b), (c,d) \in \mathbb{R}^2$, find a function $f: [0,1] \to \mathbb{R}^2$ that is continuous, that $f(t) \in S^1$ for all $t$, and $f(0) = (a,b)$, $f(1) = (c,d)$.

I can't figure out how to get started. The function should be a straight line of some sort joining the points, but that doesn't lie entirely on the boundary of the circle. I could take an arc, which surely would lie on the circle, but I don't know how to write out its equation. Polar coordinates is the first thing that comes to mind.

Any help on the formalization would be appreciated.

Answer 1

Using polar coordinates, we can write $S^1=\{(\cos \theta,\sin\theta):\theta \in [0,2\pi]\}$. Thus we can define a map $f:[0,2\pi] \rightarrow S^1$ by $\theta \mapsto (\cos \theta,\sin\theta)$. This map is continuous surjective. Hence $S^1$ is the continuous image of an interval, which is path-connected.

Answer 2

Although i go with the above answer, I am writing this with a guess that this is what you are looking for:

Let us write the points on the circle in polar coordinates. Let $ z_1= e^{i\theta_1}$ and $z_2 = e^{i\theta_2}$ be two points. Without loss of generality assume that $\theta_1<\theta_2$. Hence $0 \leq \theta_1 < \theta_2 \leq 2\pi$ (I am assuming that $\theta_1 \neq \theta_2$.

You can deal it in two ways:

1. Define explicitly a map $f$ from $[0,1]$ as follows: $$f(t) = e^{i\left(\theta_2t+(1-t)\theta_1\right)}$$. This is a continuous map (as it is a composition of two continuous functions). Image lies on unit circle. And the initial and final points are $z_1$ and $z_2$ respectively.

2. You may define a map $f$ from $[\theta_1, \theta_2]$, as $$f(t)= e^{it}.$$ $f(t)$ lies on unit circle and the initial and final points are $z_1$ and $z_2$ respectively. Since you can define a continuous Map $g$ from $[0_1]$ to $[\theta_1, \theta 2]$, consider $f$ composition $g$. You will be done.

A path need not always be from $[0,1]$. Since any two closed intervals are equivalent, a path may be from any closed interval interval $[a,b]$.