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ATT

More precisely, let $\gamma :S^1\rightarrow R^2$ be a smooth immersed loop, the question is whether it is true that there is a point $p\in R^2-\gamma(S^1)$ such that $\gamma$ is not homotopic to constant map.

Actually I'm not sure whether I choose the right tag. Tell me if I choose wrongly.

I hope it won't turn out to be trivial.

(Does the tex turn out all right? I don't seem to have the plug-in to display it.)

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    $\begingroup$ The TeX is fine, and you may also use \mathbb{S} and \mathbb{R}. There are tags for differential topology or homotopy theory too. As to the question, for any $p\in A:=\mathbb{R}^2\setminus\gamma(\mathbb{S}^1)$ the loop $\gamma$ is nullhomotopic in $\mathbb{R}^2\setminus (p)$ if and only if $\mathrm{ind}(\gamma,p)=0$, and any point in the unbounded connected component of $A$ is such. Is this what you want? $\endgroup$
    – Pietro Majer
    Commented Mar 11, 2011 at 10:17
  • $\begingroup$ Ahh I got some ambiguity in my statement, I mean whether p can always be found. But the counterexample below resolves this question. Probably I'm too careless. $\endgroup$
    – Honglu
    Commented Mar 11, 2011 at 14:01

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You can construct an immersion $\gamma$ which remains null-homotopic after removing any point $p$ not lying in its image. It suffices to let $\gamma$ travel along a graph in such a way that it runs along every edge the same number of times in each direction. As an example, you can take a train track with one central 4-valent switch and two arcs: it looks like an "8" but the 4-valent vertex is flattened, so that each of the two circles has a cusp:

The train track
(source: unipi.it)

You can let a train travel along this train track so that it runs on each of the two arcs twice in opposite directions.

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  • $\begingroup$ Good counterexample! I didn't think of this. $\endgroup$
    – Honglu
    Commented Mar 11, 2011 at 13:58
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For a different way of looking at the same issue see http://en.wikipedia.org/wiki/Pochhammer_contour . The contour is set up to have winding number 0 around any point.

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  • $\begingroup$ I believe the correct claim is that the contour has winding number zero around the two distinguished points where the plane is punctured. Based on the picture, it certainly appears to have winding number 1 around any point in the lower right bounded region. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 12, 2011 at 16:35

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