Prove that a continuous function $f$ can't be bijective Let $S$ be the unitary circumference on $R^{2}$ and $f:S \rightarrow [0,1]$ a continuous function. Prove that $f$ can't be bijective
I guess it can't be bijective because, $f$ can't be injective for the images of $0$ and $1$, but I don't know how to prove it.
 A: Assume $f\colon S\to [0,1]$ is bijective and continuous. Let $a=f^{-1}(\frac12)\in S$. Then $f$ restricted to $S\setminus\{a\}$ is still a continuous bijection with $[0,\frac12)\cup (\frac12,1]$. On the other hand, there is an obvious continuos bijection $g\colon(0,1)\to S\setminus\{a\}$, so we obtain a continuous bijectoin $f\circ g\colon (0,1)\to [0,\frac12)\cup (\frac12,1]$. This contradicts the Intermediate Value Theorem.
A: let $f(0) = a$ and $f(1) = b$, $a$ and $b$ are points on the circle. Let $c$ and $d$ be two points on the two open intervals between $a$ and $b$, and let $x$ and $y$ be such that $f(x)=c$ and $f(y)=d$. Since $f$ is continuous, The image of the interval $(x,y)$ is a path from $c$ to $d$, which must contain either $a$ or $b$. therefore there is a $z$ in the interval such that either $f(z)=f(0)$, or $f(z)=f(1)$, so $f$ is not bijective.
A: Suppose $f$ is injective. Its image is  a  connected compact set so it is of the form $[a,b]$. Let $c=\frac {a+b}2$. The inverse of $f$ from $[a,b] \to S$ is continuous automatically since the domain and range are compact Huadorff spaces. Hence, $f$ is a homeomoprhism. If you remove $c$ from $[a,b]$ it become disconnected whereas removing a point from $S$ retains its connectedness.
