Path on $\mathbb{R}^2\setminus \{x_1, x_2\}$

Question

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!

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
well, there are some statements that need to be proven there right? — Thomas Rot, Mar 14 '12 at 16:16

score 4 · Accepted Answer · answered Mar 14 '12 at 15:24

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.

Path on $\mathbb{R}^2\setminus \{x_1, x_2\}$

1 Answers1