I'm trying to show that the unit circle is path-connected. I have a sketch of a proof, and was hoping someone could tell me if this is on the right track.
Given points $(a,b), (m,n)$ on the circle, I need to demonstrate that there exists a path from one to the other, a continuous function $\gamma: [0,1] \to \mathbb{R}^2$ that lies entirely on the unit circle and has the property that $\gamma(0) = (a,b)$ and $\gamma(1) = (m,n)$. Intuitively, I want to take the circular arc joining these two points, so I just need to, formally, express this arc as the continuous image of $[0,1]$. Using polar coordinates, I can write $$ (a,b) = (\cos \theta, \sin \theta), \; (m,n) = (\cos \psi, \sin \psi) $$ for unique $\theta, \psi \in [0,2\pi)$. I can identify a point on this arc with a unique angle in $[0, 2\pi)$ using polar coordinates, i.e., via the map $$ g: \phi \mapsto (\cos \phi, \sin \phi). $$ So I first thought to write down a bijection between $[0,1]$ and $[\theta, \psi]$ by $$ f(t) = (1 - t)\theta + t \psi. $$ This is a linear bijection between $[0,1]$ and $[\theta, \psi]$ with $f(1) = \psi$ and $f(0) = \theta$. Now I just compose maps to get a map from $[0,1]$ to $\mathbb{R}^k$: $$ g(f(t)) = g((1 - t)\theta + t \psi) = (\cos((1 - t)\theta + t \psi), \sin((1 - t)\theta + t \psi). $$ This map is a composition of continuous functions and therefore bijective, and $g(f(0)) = (a,b)$, $g(f(1)) = (m,n)$. By construction, we get a point on the unit circle, so all of the requirements for a path are satisfied.
How does this look? Two things I'm concerned with are:
I don't require the map from $[0,1] \to \mathbb{R}^2$ to be bijective (and if I'm just parametrizing an arc, it surely won't be surjective), but it seems of particular importance that the map from $[0,1] \to [\theta, \psi]$ is bijective. Is this correct? Or rather, does it only matter that the map $g$ that sends a point to the unit circle is surjective onto the arc? I'm trying to piece together the crucial pieces of this proof and understand where this function came about.
The function $f$ defined above is just a degree-one polynomial and surely continuous, but I didn't prove that $g$ is continuous. Is it a well-known fact that trigonometric functions are continuous, and therefore a two-component function with trig functions are continuous? This exercise came from Artin's Algebra book, and I'm trying to figure out the level of detail he was expecting when he wrote this problem. That polynomials are continuous seems well-known enough that I can take it for granted.