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:
- $\Gamma$ being continuous
- $\partial_s \Gamma$ existing
- 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?
