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Is there a continuous surjective map from $S^1$ to $[0,1]$?

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up vote 2 down vote accepted

HINT: Start with a projection map and then modify it slightly.

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More concretely, $f : S^1 = \{ e^{it} | 0 \le t < 2\pi \} \rightarrow [0,1]$ is given by $f(e^{it}) = |sin(t)|$ – HK Lee Oct 12 '12 at 8:17
@user37116: If I had wanted to give a concrete map, I’d have done so. I can’t keep you from posting a complete solution as an answer, but I don’t appreciate having one appended as a comment to what was obviously intended not to be a complete solution. – Brian M. Scott Oct 12 '12 at 8:22
And it would be nice if the downvoter would explain what’s wrong, though I’m not going to hold my breath. – Brian M. Scott Oct 12 '12 at 9:56

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