Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
"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
add comment

1 Answer

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.

share|improve this answer
Thanks, that seems much shorter. Why would anyone do something complicated like in the link above? –  Matt N. Jun 30 '11 at 7:00
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.