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Let $e^{2 \pi k s} = f(s)$, $f \colon [0,1] \to S^1$ subset of Complex numbers ($S^1$ = unit circle at origin). So $f$ is a loop at the basepoint $1$ in $S^1$. Show that it is not homotopic to the constant loop $c(s) = 1$. This is obvious, but what is a more rigorous way of showing it. Oh yeah, $k$ is an an integer not equal to $0$.

Without reference to complex analysis (I'm looking for a really simple kid proof).

Thanks. -Dan

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Actually, what do you mean by this? S^1 is nullhomotopic in the plane. –  Tony Mar 19 '11 at 0:43
I mean if S^1 were treated as a topological space. What I am doing is making a really simplified proof that the fundamental group of S^1 is Z and the above is the last step. –  Dan Donnelly Mar 19 '11 at 0:47
Do you know the theory of covering spaces? Or are you looking for a proof without covering spaces? –  Tony Mar 19 '11 at 2:39
@user5237: I think if you wish to avoid complex analysis, arguments to show $\pi_1(S^1) \simeq \mathbb{Z}$ you would need to use path lifting - see for example Hatchers book, or Peter May's notes www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf or a wiki with pretty pictures (en.wikibooks.org/wiki/Topology/The_fundamental_group) –  Juan S Mar 19 '11 at 5:18
covering spaces are almost always introduced by looking at $\mathbb{R}$ over $S^1$ and proving $\pi_1(S^1)\cong\mathbb{Z}$. take a look at hatcher's <a href="math.cornell.edu/~hatcher/AT/ATpage.html">free book</a>. i would say there isnt a "simple kid proof", it takes a little work to prove this rigorously. –  yoyo Mar 19 '11 at 14:40

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