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Map from $[-1,1]\to S^1$ that are continuous at exactly $1$ or $0$ points?

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The following is a problem from a GRE review booklet:

Let $f$ be a function with domain $[-1,1]$, such that the coordinates of each point $(x,y)$ of its graph satisfy $x^2 + y^2 = 1$. The total number of points at which $f$ is necessarily continuous is... ?

My solution was to define the map $f$ by $$f(x)=\begin{cases}\sqrt{1-x^2}\quad\quad \textrm{if }x\in\mathbb{Q}\\ -\sqrt{1-x^2}\quad \textrm{otherwise}\end{cases}$$Which is discontinuous at all points in $(-1,1)$ but continuous at the two points $-1$ and $1$.

This leads me to believe that every map from $[-1,1]\to S^1$ must necessarily be continuous at two points. According to the answer key, this is indeed the correct solution.

I am confused, however, on why it is not possible for such a map to exist with zero or only $1$ point of continuity. I tried proving an arbitrary map from $[-1,1]\to S^1$ must be continuous at two points using contradiction, but couldn't get anywhere. Does anyone have any insight? Thank you.


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