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 $f$ exists. $f(S^1)$ must be connected and compact. $f$ is one-one so it cannot be constant. It follows that $f(S^1) = [a, b]$ for $a \ne b$.
Pick a point $s \in S^1$ such that $f(s) \not\in \{a, b\}$. $S^1 - \{s\}$ is connected, but $[a, b] - \{f(s)\}$ is not. We conclude that such a function $f$ cannot exist.

Alternatively, if you're familiar with the concept of the fundamental group: The restriction $f : S^1 \to f(S^1)$ must be a homeomorphism (being a continuous bijection from a compact space onto a Hausdorff space). But we know that $\pi_1(S^1) = \mathbb Z$ whereas any connected subset of $\mathbb R$ is contractible; hence it has a trivial fundamental group.
