# A question regarding the proof of $\pi_1(\mathbb{R}^2 \backslash \{ 0 \}, 1) \cong \mathbb{Z}$

The following proof is the first hit for me on Google to find a proof that the fundamental group of the punctured plane is isomorphic to $\mathbb{Z}$: link

I've worked through it until the bottom of page 2 where it says:

"To prove this statement we consider the plane $C$ cut along $L$. The points of the cut have the parametrisation $R_+ \times [0, 2\pi]$ in polar coordinates."

Here $C$ is the complex numbers and $L$ is an arbitrary ray starting at $0$. My question is: how do they get $R_+ \times [0, 2\pi]$ for the points of the cut. If I cut along a ray I think the points should all be in $R_+ \times \phi_0$ for $\phi_0$ fixed.

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"The cut" in this text refers to the result, I believe –  Grigory M Jun 29 '11 at 14:07
@Grigory: That would make more sense but if that was the case, shouldn't it be $R_+ \times (0, 2\pi]$ to exclude the points where I cut? (assuming I cut along the $x$-axis) –  Matt N. Jun 29 '11 at 14:10
What does "should" mean here? One can consider either $\mathbb R_+\times(0,2\pi]$ or $\mathbb R_+\times[0,2\pi]$ -- both correspond to cutting the plane in some sense. –  Grigory M Jun 29 '11 at 14:14
I think I understand. It's all not so well-explained in there but they're not really interested in the cut plane but rather in proving that everything outside the ray is simply-connected. So it doesn't matter whether one includes the points in the cut ($[0,2\pi]$) or whether one excludes them ($(0,2\pi)$). –  Matt N. Jun 30 '11 at 6:58

that it is homotopic to $S^1$. Informally, just draw radial lines out from the origin,
embedded in $\mathbb R^2$and use convexity.