# Can a smooth, immersed loop in R^2 become not nullhomotopic by removing a point?

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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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? –  Pietro Majer Mar 11 '11 at 10:17
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. –  Honglu Mar 11 '11 at 14:01

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: