Is the following statement is true/false ?

There exist $f : S^1 \rightarrow \mathbb{R} $ which is continuous and onto.

My thinking: yes, because for every function $f : S^1 \rightarrow \mathbb{R}$ there exist uncountably many pairs of distinct points $ x $ and $y$ $\in S^1$ such that $f(x) = f(y)$.

  • 1
    $\begingroup$ That's not what "onto" means. $\endgroup$
    – user694818
    Aug 14 '19 at 7:13

The statement is wrong. The image of a compact set under a continuous mapping is compact.

  • $\begingroup$ In other words: $f$ has a minimum $m$ and a maximum $M$ on $S^1$, so that $f(S^1) \subset [m, M]$. $\endgroup$
    – Martin R
    Aug 14 '19 at 8:00

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