Proof that there is no continuous 1-1 map from the unit circle in $\Bbb R^2$ to $\Bbb R$. 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. 
 A: 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.
A: let f: $S^1 $ $\rightarrow$ R is a 1-1 continuous map. Observe that Img(f) is a compact and connected subset of R (i.e a compact interval [a,b] for some real numbers a and b)and f is a bijective continuous map from $S^1$ to Img(f)...there is a theorem in point set topology that if F: X $\rightarrow$ Y is a continuous bijection map where X is compact and Y is hausdorff then F is a homeomorphism...
but you can prove that $S^1$ is not homeomorphic with Img(f).
(one way to show this you can compare there fundamental groups otherwise just take a point out from img(f) correspondingly take out of the inverse image of the same point then you can get one side a connected set on the other hand a disconnected set but restriction map is still a homeomorphism so either both should be connected or disconnected. That will give you an contradiction)
