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?

It cannot be 1-1 because of the one-dimensional version of Borsuk-Ulam: for every continuous $f: S^1 \rightarrow \mathbb{R}$ there exists an $x \in S^1$ such that $f(x) = f(-x)$. 
The proof in the one-dimensional case (it's true for all $S^n$ and $\mathbb{R}^n$ but harder to prove) is not very hard: define $g(x) = f(x) - f(-x)$. Suppose for some $p \in S^1$, $g(p) \neq 0$ (otherwise we would be done anyway), then $g(-p) = -g(p)$ by definition and so then $g$ assumes positive and negative values in $\mathbb{R}$, and so also assumes the value $0$ by the intermediate value theorem (we use the connectedness of $S^1$). And we are done for such a value: $g(x) = 0$ iff $f(x) = f(-x)$.
