There's no continuous injection from the unit circle to $\mathbb R$

Question

I read a proof that goes as follows:

Let $U$ be the unit circle, and let $f : U \longrightarrow \mathbb R$ be a continuous mapping. $U$ is compact and connected, so $f(U)$ is a closed, bounded interval in $\mathbb R$. Hence there are $a$ and $b$ in $\mathbb R$ such that $f(U) = [a,b]$. There are two points $x_0$ and $y_0$ in $U$ such that $a = f(x_0)$ and $b = f(y_0)$. Let $A$ and $B$ be two diametrically opposite points in $U$ distinct from $x_0$ and $y_0$. Then,

$f(\widehat {x_0 A y_0}) = [a,b]$ and $f(\widehat {x_0By_0}) = [a,b]$.

This shows that $f$ is not injective.

I don't understand that highlighted step. Can anyone explain? Thanks.

Answer 1

The arcs are also compact and connected, so their images must also be closed, bounded intervals, and they necessarily contain $\{a, b\}$.

Answer 2

This is due to the intermediate value theorem. Since $f(x_0) = a$ and $f(y_0) = b$, the image of an arc from $x_0$ to $y_0$ must be the entire interval $[a,b]$. But on the unit circle, there are exactly two such arcs, so the map is at best 2-1.

Answer 3

We can consider $f$ as a mapping from $U$ to $[a,b]$ so it is continuous and bijective; it is also closed, because $U$ is compact. Therefore $f$ is a homeomorphism.

Let $x\in U$; then $U\setminus\{x\}$ is connected. On the other hand, $[a,b]\setminus\{c\}$ is not connected if $a<c<b$. Thus either $f(x)=a$ or $f(x)=b$ and so $f(U)\subseteq\{a,b\}$. Contradiction.

Answer 4

You're quite right that the highlighted step is incorrect. Instead of choosing $A$ and $B$ to be any diametrically opposite points other than $x_0$ and $y_0$, you should choose them so that they are separated by $x_0$ and $y_0$. (They need not be diametrically opposite; that requirement is neither necessary nor sufficient for the proof to work.) In other words, $x_0$ and $y_0$ divide the circle $U$ into two arcs, each of which has $x_0$ and $y_0$ as its endpoints. Then $f$ will map each of those arcs to a connected subset of $\mathbb R$ (because the arcs are connected and $f$ is continuous) containing $a$ and $b$ (because the arcs contain $x_0$ and $y_0$). So the image of each arc will be all of $[a,b]$. Now you have that $f$ can't be one-to-one, because the interiors of the two arcs are disjoint, yet their images have (lots of) points in common.