There's a question in my complex analysis book:

Let $G$ be a region and let $\gamma_0$ and $\gamma_1$ be two closed smooth curves in $G$. Suppose $\gamma_0\sim\gamma_1$ and $\Gamma$ is a homotopy between them. Also suppose that $\gamma_t(s)=\Gamma(s,t)$ is smooth for each $t$. If $w\in \mathbb{C}-G$, define $h(t)=n(\gamma_t;w)$ and show that $h:[0,1]\rightarrow\mathbb{Z}$ is continuous.

This is an exercise right after the section that proves the following version of Cauchy's theorem:

If $\gamma_0$ and $\gamma_1$ are two closed rectifiable curves in $G$ and $\gamma_0 \sim \gamma_1$ then $$\int_{\gamma_{0}}f=\int_{\gamma_1} f$$ for every function $f$ analytic on $G.$

I must be missing something. It seems like you can just say the following:

'Proof:' Since $\frac{1}{z-w}$ is analytic in $G$ ($w$ isn't in $G$), and $\gamma_0\sim\gamma_t\ \forall t\in[0,1]$ (consider a rescaled version of $\Gamma$), then by Cauchy's theorem, $n(\gamma_t;w)=n(\gamma_0;w)=n(\gamma_1;w)$. Thus, the winding number is constant along $t$ and thus certainly continuous.

My problem with that is that this doesn't require the assumption that $\Gamma$ is a smooth homotopy. I think the author had meant for the reader to work harder for this result. What is the proof 'supposed' to look like?


2 Answers 2


This appears to be exercise 9 in Conway's Functions of One Complex Variable I section IV.6. I'll use the second edition numbering. In the section he mentions exercise 9 as a good alternative proof of 6.7 Cauchy's Theorem (Third Version) for use in a course intended for scientists and engineers, so I agree with the OP that he probably does not intend the reader to just ignore the smoothness assumption and invoke theorem 6.7.

Given the smoothness condition exactly as he states it, I think the answer given here from Rudin is hard to beat: show that $h\left( t\right)= \eta \left( \gamma_t;w \right) $ is continuous.

The Rudin proof uses only the condition that for fixed t, $\gamma_t(s)$ is smooth (defined in Conway to mean having a continuous first derivative), and shows directly that the winding number is constant, therefore continuous.

That said, I can't help but think this is also not what the author was thinking of. In the section just above the proof of 6.7 he describes an alternative strategy to obtain Cauchy's theorem where you show the winding number is continuous, then use the fact that it is integer valued to see that it is constant. He says exercise 9 will make this argument rigorous. The Rudin proof shows that the winding number is constant directly, so it just seems unlikely that this is the proof he intended. It seems more likely he means a standard $\delta - \epsilon$ continuity proof.

I'm going to offer what I can do along these lines, but in order to bound the quantity $|\gamma'_{t}(s) - \gamma'_{t_p}(s)|$, I have to add a continuity assumption to one of the homotopy's partials that is not assumed in the problem statement.

So suppose we have instead: $\gamma_0$ and $\gamma_1$ are smooth curves with $\gamma_0 \sim \gamma_1$ where $\Gamma(s,t)$ is a homotopy between them. For reference, in the text that means that $\Gamma : I^2 \rightarrow \mathbb{C}$ ($I = [0,1]$) is continuous and: $$\Gamma(s,0) = \gamma_0(s) \\ \Gamma(s,1) = \gamma_1(s) \\ \Gamma(0,t) = \Gamma(1,t)$$ Rather than the smoothness condition stated, assume that $\partial_s \Gamma$ exists and is continuous as a function of both variables.

Also note that the winding number is given: $$ n(\gamma;w) = \frac{1}{2\pi i} \int_\gamma (z-w)^{-1} dz $$

Denote $\gamma_t(s) = \Gamma(s,t)$. To show continuity at a point $t_p \in I$, consider: $$ n(\gamma_t;w) - n(\gamma_{t_p};w) = \frac{1}{2\pi i} \int_0^1 \frac{\gamma'_t(s)}{\gamma_t(s) - w} - \frac{\gamma'_{t_p}(s)}{\gamma_{t_p}(s) - w} ds \\ = \frac{1}{2\pi i} \int_0^1 \frac{\gamma'_t(s)(\gamma_{t_p}(s) - w) - \gamma'_{t_p}(s)(\gamma_t(s) - w)}{(\gamma_t(s) - w)(\gamma_{t_p}(s) - w)} ds $$

Now note that $\Gamma(I^2)$ is a compact set in $\mathbb{C}-\{w\}$ so its distance from w can be bounded. There exist positive m and M such that: $$0 \lt m \lt |\gamma_t(s) - w| \lt M$$ for arbitrary s and t.

Then we have, $$ |n(\gamma_t;w) - n(\gamma_{t_p};w)| \lt \frac{1}{2\pi m^2} \int_0^1 \left| \gamma'_t(s)(\gamma_{t_p}(s) - w) - \gamma'_{t_p}(s)(\gamma_t(s) - w)\right| ds $$

Adding and subtracting the cross term $\gamma'_{t_p}(\gamma_{t_p} - w)$ gives, $$ |n(\gamma_t;w) - n(\gamma_{t_p};w)| \lt \frac{1}{2\pi m^2} \int_0^1 \left| (\gamma'_t - \gamma'_{t_p})(\gamma_{t_p} - w) - \gamma'_{t_p}\left((\gamma_t - w) - (\gamma_{t_p} - w) \right) \right| ds \\ \le \frac{1}{2\pi m^2} \int_0^1 \left| (\gamma'_t - \gamma'_{t_p})(\gamma_{t_p} - w)\right| + \left| \gamma'_{t_p}\left(\gamma_t - \gamma_{t_p}\right) \right| ds $$

Since $\gamma'_{t_p}(s)$ is continuous, it's bounded above by a constant, $M_1$. Let $M' = max\{M,M_1\}$ (note this step could be done with the original smoothness assumption): $$ |n(\gamma_t;w) - n(\gamma_{t_p};w)| \lt \frac{M'}{2\pi m^2} \int_0^1 \left| \gamma'_t - \gamma'_{t_p} \right| + \left| \gamma_t - \gamma_{t_p} \right| ds $$

Since $\Gamma$ is continuous on a compact set, it is uniformly continuous. For any $\epsilon \gt 0$, there must then exist $\delta_1$ such that, $$|(s',t') - (s,t)| \lt \delta_1 \Rightarrow |\Gamma(s',t') - \Gamma(s,t)| \lt \frac{\epsilon \pi m^2}{M'}$$

This will hold in particular when $s' = s$, so when $|t - t_p| \lt \delta_1$: $$ |n(\gamma_t;w) - n(\gamma_{t_p};w)| \lt \frac{M'}{2\pi m^2} \int_0^1 \left| \gamma'_t - \gamma'_{t_p} \right| ds + \frac{\epsilon}{2} $$

Finally, using that $\partial_s \Gamma$ is continuous on the compact set $I^2$ we again get uniform continuity. So, for any $\epsilon \gt 0$, there must exist $\delta_2 \gt 0$ such that, $$|(s',t') - (s,t)| \lt \delta_2 \Rightarrow |\partial_s\Gamma(s',t') - \partial_s\Gamma(s,t)| \lt \frac{\epsilon \pi m^2}{M'}$$

So in particular when $s' = s$, $$|t - t_p| \lt \delta_2 \Rightarrow |\gamma'_t(s) - \gamma'_{t_p}(s)| \lt \frac{\epsilon \pi m^2}{M'}$$

Letting $\delta = min\{\delta_1,\delta_2\}$ whenever $|t - t_p| \lt \delta$, $$ |n(\gamma_t;w) - n(\gamma_{t_p};w)| \lt \epsilon $$

So the winding number is uniformly continuous, therefore it's continuous.

If someone sees a reason that:

  1. $\Gamma$ being continuous
  2. $\partial_s \Gamma$ existing
  3. For fixed $t$, $s \rightarrow \partial_s \Gamma(s,t)$ is continuous

Together imply that $\partial_s \Gamma(s,t)$ is continuous in both variables then I think this would directly answer the question. My hunch is that this is not true. Is there another reason we should be able to bound $|\partial_s \Gamma(s,t) - \partial_s \Gamma(s,t_p)|$ given just those 3 assumptions?


I had the same problem. I believe your proof is correct. It sounded weird to me too, but then I find the following results:


It seems, a function $g:[a,b]\to\mathbb{Z}$ is continuous if and only if $g$ is constant. So it makes sense you could prove $n(\gamma_t,w)$ is constant.

  • 1
    $\begingroup$ The link is broken. Is the hypothesis of smoothness irrelevant? $\endgroup$ Oct 9, 2019 at 0:51
  • 1
    $\begingroup$ @AlejandroTolcachier The homotopy invariance theorem for contour integrals applies in greater generality to continuous homotopies with piecewise $C^1$ sample paths. That is, for each $s$ the sample contour $t\mapsto\Gamma(s,t)$ is piecewise $C^1$ but $\Gamma:[0,1]^2\to U$ is only continuous $\endgroup$
    – FShrike
    May 9 at 18:15
  • $\begingroup$ The link is available here: web.archive.org/web/20151123172903/http://people.duke.edu/… I think the idea in this section of Conway is that with stronger smoothness assumptions you get easier proofs (he includes a third proof where continuity of the homotopy's second partials gives the winding number as differentiable [beginning of 6.7 set f=1/(z-w)]), but if you take the long road and prove the theorem in full generality you get a new method that can be reused in other situations. $\endgroup$
    – mmcfa
    May 9 at 19:27

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