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On page 185 of Saff, Fundamentals of Complex Analysis, the author states that in the punctured plane ($\mathbb{C}- \{0\}$), the unit circle traversed once in the positive direction is NOT continuously deformable to the unit circle traversed twice in the positive direction.

I can't understand why the deformation $z(s,t):[0,1]^{2} \rightarrow \mathbb{C} - \{0\}$ given by $$z(s,t) = e^{2\pi t(1+s)i}$$ doesn't do the trick. For $s=1$ don't we get twice the speed?

BlueBuck
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I think you misunderstand the statement. Any path in $\mathbb C \smallsetminus \{0\}$ can be homotoped to any other path if the endpoints may be moved, since $\mathbb C\smallsetminus\{0\}$ is connected. For the statement about circles to be interesting, it is important to interpret it in terms of circles, or, equivalently, in terms of continuously deforming paths without moving either endpoint.

So the question is to prove that the function $S^1 = \mathbb R / \mathbb Z \to \mathbb C$ given by sending $\theta \mapsto \exp(2\pi i\theta )$ is not homotopic to $\theta \mapsto \exp(4\pi i \theta)$. There are many proofs of this fact. I will give one that I think is quite informative, but without a lot more work applies only to smooth deformations between smooth paths. (By approximating other paths, this proof can be used to conclude the statement about continuous deformations of continuous paths, but I will not do so.) The reason I like this proof is that it uses the complex numbers.

The basic fact to be used is that the contour integral of a meromorphic 1-form $\alpha = f(z)\mathrm d z$ along a curve that avoids all poles of $f$ is invariant under endpoint-fixing deformations of the curve; if the endpoints move, then the integral changes by something depending only on the endpoints. In particular, for any curve $\gamma : S^1 \to \mathbb C$ avoiding the poles of $f$, $\oint_\gamma f(z)\mathrm dz$ is invariant under small changes to $\gamma$. It thus suffices to find a 1-form (holomoprhic on $\mathbb C\smallsetminus \{0\}$, with a pole at $0$) that distinguishes the two curves in question.

The function $f(z)\mathrm dz = z^{-1}\mathrm d z$ does the trick. Its integral around the pole is just $2\pi i$, whereas if you go twice around you get $4\pi i$.

Theo Johnson-Freyd
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  • Thank you for pointing out that the winding number function distinguishes the curves! I guess I accept that they aren't homotopic, but I am still struggling with why the deformation I've listed doesn't work based on definition 5 given in Saff on page 182. i.e z(s,t) paratemtrizes a loop for each s, and z(0,t), z(1,t) parametrize the loops described above respectively. – BlueBuck Mar 10 '14 at 01:55
  • @BlueBuck: Dear Blue, The point is that the right-hand endpoint is not fixed in your homotopy. This is why Theo makes a point of saying that the statement is only true (and sensible) if you fix the endpoints of your loops. Regards, – Matt E Mar 10 '14 at 02:38
  • @BlueBuck As Matt E says, in your homotopy, the $t=1$ endpoint follows the path $s\mapsto\exp(2\pi i(1+s))=\exp(2\pi i s)$, which starts at $1$ and walks once around the circle to return to $1$. Yours is absolutely a homotopy between the two paths, but it is not a homotopy "rel boundary". For this you would need a map $z:[0,1]^2\to\mathbb C$ which not only has the correct restrictions at $s=0$ and $s=1$ but also restricts along $t=0$ and $t=1$ to constant paths sitting at the endpoints, i.e. in this case satisfying $z(s,0)=z(s,1)=1$ for all $s$ (and $z(0,t)=\dots$ and $z(1,t)=\dots$). – Theo Johnson-Freyd Mar 10 '14 at 02:46
  • Thanks for the further explanation, I'll have to read up on this some more from other sources. Reading through Saff I was unable to find any mention of the requirement that the deformations be endpoint preserving. In the mean time though I have accepted your answer! Thank you. – BlueBuck Mar 10 '14 at 04:04