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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.

Thanks for your help!

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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) – Rudy the Reindeer 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 ($(0,2\pi)$). – Rudy the Reindeer Jun 30 '11 at 6:58
up vote 2 down vote accepted

A different approach: use the fact that the punctured plane retracts to a circle, so

that it is homotopic to $S^1$. Informally, just draw radial lines out from the origin,

and move part of the plane inwards to get your circle. Or imagine the circle standardly

embedded in $\mathbb R^2$and use convexity.

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Thanks, that seems much shorter. Why would anyone do something complicated like in the link above? – Rudy the Reindeer Jun 30 '11 at 7:00

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