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Consider plane $\mathbb{R}^2\setminus \{ x_1, x_2 \}$ without two points, and such closed path on this plane: enter image description here (points on this picture are deleted points $x_1$ and $x_2$)

Question: how to prove that this path isn't homotopic to zero?

Appendix. As I see, we may fix some point $\alpha\in\mathbb{R}^2\setminus \{ x_1, x_2 \}$ and consider loops $a$ and $b$, which "walk round" point $x_1$ and $x_2$ respectively. Then my path will be homotopic to $$b^{-1}a^{-1}ba$$ as the element of fundamental group ${\pi}_{1} (\mathbb{R}^2\setminus \{ x_1, x_2 \},~\alpha)$.

Thanks a lot!

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This figure is homotopic to the "8" with a hole in each part! – checkmath Mar 14 '12 at 11:06
The fundamental group of your space is equal to free group with two generators. There are no non-trivial relations in this group, so the element you wrote down does not equal the identity. – Thomas Rot Mar 14 '12 at 11:08
@ThomasRot: why give an answer in the comments? – Martin Argerami Mar 14 '12 at 14:55
well, there are some statements that need to be proven there right? – Thomas Rot Mar 14 '12 at 16:16
up vote 3 down vote accepted

Here is what happens when you try to draw a pull back of your loop on a rather simple (but still non abelian) covering of your space :

You can pull back an homotopy on your initial space starting from a null path into a loop into on an homotopy on this covering starting from a null path into a pull back of the loop.

So if your loop were homotopic to 0, its pullback should be a loop homotopic to 0 on this covering. But it is not even a loop because the end points are pulled into two different parts of the covering. Therefore it can't be homotopic to the null path in the first place.

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+1 "nice" $ $ $ $ – draks ... Apr 8 '12 at 19:44

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