Let $S^1=\{(x,y) \in \Bbb R^2 : x^2+y^2=1\}$. I'm trying to prove that there is no 1-1 continuous mapping between $S^1$ and the real line. The map is not necessarily onto.
Proof so far:
Suppose such a function $f$ existed. Since $S^1$ is compact and $f$ is continuous and 1-1, $f$ inverse exists and is also continuous. Hence the range of $f$ is closed.
Not sure where to proceed next.
Best Answer
Suppose $f:S^1\to\mathbb R$ is a continuous map. Write $g(x)=f(x)-f(-x)$, then $g$ is also continuous. As $$g(-x) = f(-x) - f(-(-x)) = f(-x) - f(x) = -(f(x) - f(-x)) = -g(x), $$ $g$ is an odd function. If $g(x)=0$, then $f(x)=f(-x)$. If not, then $g(x)>0$ and $g(-x)<0$ or $g(x)<0$ and $g(-x)>0$. In either case, as $S^1$ is a connected subspace of $\mathbb R^2$, by the intermediate value theorem there exists $c$ between $x$ and $-x$ such that $g(c)=0$. Thus $f(c)-f(-c)=0$, so that $f(c)=f(-c)$. It follows that $f$ cannot be one-to-one.