Let $S^1$ denote the unit circle in the plane $\Bbb R^2$. Pick out the true statement(s):

(a) There exists $f : S^1 \to\Bbb R$ which is continuous and one-one.

(b) For every continuous function $f : S^1 \to\Bbb R$, there exist uncountably many pairs of distinct points $x$ and $y$ in $S^1$ such that $f(x) = f(y)$.

(c) There exists $f : S^1 \to\Bbb R$ which is continuous and one-one and onto.

I have only idea that for (c) $f$ is nearly homeomorphism as we cant say that $f$ inverse is continuous or not and the sets are not homeomorphic. No idea about (a) and (b).

  • $\begingroup$ Hint: (is it homework?) What can you say about the continuous image of a compact set? $\endgroup$
    – hardmath
    Sep 6 '12 at 4:18


(c) $S^1$ is compact, and continuous functions preserve compactness.

(a) After you’ve done (c), you should know what kind of subset of $\Bbb R$ the continuous image of $S^1$ must be, unless it’s just a single point. It’s a very nice kind of set: it’s connected (why?), but if you remove almost any of its points, what’s left is not connected. Show that if $x$ is one of these so-called cut points whose removal disconnects $f[S^1]$, there must be at least two distinct points $p,q\in S^1$ such that $f(p)=f(q)=x$.

(b) If you do (a) using the hint above, this one will come almost for free.

  • $\begingroup$ @hardmath: Thanks: I did misread it. $\endgroup$ Sep 6 '12 at 4:25
  • $\begingroup$ M Scott, please elaborate or make me understand the $b$ $\endgroup$
    – Marso
    Jan 7 '13 at 10:16

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