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How can one prove that the fundamental group of the topologists sine curve is trivial? I haven't been able to make any progress on this. A hint in the right direction preferred over a complete solution.

qwert4321
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1 Answers1

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Supposing you have the parametrization of the curve given here, try to show that any curve with base point $(0,0)$ has to be constant.

Then show that any curve with basepoint not at $(0,0)$ cannot pass through $(0,0)$, and use the fact that the topologist sine curve without $(0,0)$ is homeomorphic to the real line.

Patrick Da Silva
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  • So I have managed with the first part, and I believe with the second part - I showed the topologist sine curve without (0,0) is homeomorphic to [ 1, ∞ ), which is contractible and therefore has a trivial fundamental group. Is this correct? (You stated that that the topologist sine curve without (0,0) is homeomorphic to the real line, which I couldn't see -- but showing is homeomorphic to [ 1, ∞ ) seemed to do the trick). Your hint was very helpful, thank you. – qwert4321 Jan 11 '15 at 12:01
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    @qwert4321 : I don't see how you've shown this, because it's not what you want. The homeomorphism is simply given by $(x, \sin(1/x)) \mapsto x$, so that the topologist sine curve is homeomorphic to $]0,\infty[$, which is homeomorphic to the real line (but if I told you it was homeomorphic to $]0,\infty[$, it would have been kind of a bigger hint!). But the contractible part and the rest of your argument is fine! – Patrick Da Silva Jan 11 '15 at 19:54