Let $S_1$ be the unit circle in $\mathbb{R}^2$. Then there is a continuous one-one onto function from $S_1\setminus\{(0,1)\}$ to $\mathbb{R}^2$, which I can show by shifting the circle along the positive direction of the $y$-axis by one unit, so that it has the center $(0,1)$, and then get the image of any point on the circumference (except the topmost point $(0,2)$), as the point of intersection of the straight line passing through $(0,2)$ and the given point, with the $x$-axis.
Now I want to know if $f:S_1\rightarrow \mathbb{R}$ is any continuous function, then are there countably or uncountably many points $x,y\in S_1$ such that $f(x)=f(y)$?
Best Answer
Yes, if $f: S^1\to\mathbb R$ is continuous, then there are uncountably many pairs of points $x\neq y\in S^1$ such that $f(x)=f(y)$.
Proof: If $f$ is a constant function, then we're done. Otherwise there are $a,b\in S^1$ such that $f(a)<f(b)$. Let $c$ be any number in the interval $(f(a), f(b))$. Then there exists an $x$ on the clockwise arc from $a$ to $b$, and a $y$ on the counterclockwise arc from $a$ to $b$, such that $f(x)=f(y)=c$. Now note that there are uncountably many choices for $c$.
(This works because both arcs in $S^1$ are connected. By contrast, in $S_1\setminus\{(0,1)\}$, one of the arcs is not connected. That is why you were able to find a one-to-one map.)
By the way, to answer your question in the comments: $f: S^1\to\mathbb R$ cannot be onto, because the image of a compact space under a continuous function is compact.
(This works because $S^1$ is compact. By contrast, $S_1\setminus\{(0,1)\}$ isn't compact. That is why you were able to find an onto map.)