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. $\endgroup$ – Matthew Daly Aug 14 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 at 8:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.