Is there a continuous function $f:S^1 \to \mathbb R$ which is one-one? Is there a continuous function $f:S^1 \to \mathbb R$ which is one-one?

Suppose such a function exists. Let $u : [0,1] \to S^1$ be a suitable path that traces around the circle, and consider $g = fu$. This $g : [0,1] \to \mathbb R$ is one-to-one except that $g(0) = g(1)$.
Consider $y = g(1/2)$. It must be either greater than or less than $g(0)$. Pick some value $z$ between $g(0)$ and $y$. Apply the intermediate value theorem on both the intervals $[0,1/2]$ and $[1/2,1]$ and show that there must be some $c$ and $d$ in $(0,1/2)$ and $(1/2,1)$ respectively with $g(c) = g(d) = z$, contradicting $g$ being one-to-one.
In simple terms: the function must go either up or down from where it starts, but then it needs to go back through territory it's already visited to get back to the first point.
