show that $h\left( t\right)= \eta \left( \gamma_t;w \right) $ is continuous. 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 $\Gamma: I² \rightarrow G$ is a continuous map such that:
    $$ \left\{ \begin{array}{rclll}
 \Gamma\left(s,0 \right)=\gamma_0\left( s\right)   & \Gamma\left(s,1 \right)=\gamma_1\left( s\right)  &  & 0\leqslant s \leqslant 1  \\
 \Gamma\left( 0,t\right) =\Gamma\left( 1,t\right)  &
 & &0\leqslant t \leqslant 1
 \end{array}
 \right. \ $$ 
    Also suppose that $\gamma_t=\Gamma\left(s,t \right) $ is smooth for each t. If $w \in \mathbb{C}-G$ define $h\left( t\right)= \eta \left( \gamma_t;w \right) $ and show that $h:\left[ 0,1\right] \rightarrow \mathbb{Z} $ is continuous.
my aproach is the next:
If $t_0,t_1 \in \left[ 0,1\right] $ then 
\begin{eqnarray*}
  \left| h\left(t_0 \right)-h\left( t_1\right)  \right| &=& \left| \eta \left( \gamma_{t_0};w \right)-\eta \left( \gamma_{t_1};w \right)\right| \\
  &=& 2\pi \left| \int_{\gamma_{t_0}}^{}\frac{1}{z-w}dz-\int_{\gamma_{t_1}}^{}\frac{1}{z-w}dz\right| \\
  &\leqslant& 2\pi \int_{0}^{1} \left| \left( \frac{\gamma_{t_0}'\left( s\right) }{\gamma_{t_0}\left( s\right) -w}-\frac{\gamma_{t_1}'\left( s\right) }{\gamma_{t_1}\left( s\right) -w}\right)\right|  \left| ds\right|  
 \end{eqnarray*}
 A: Here is how Rudin does it: since $\Gamma (I^2)$ is compact, there is a $\delta>0$ such that $d(w,\Gamma(s,t))>\delta$ for all $(s,t)\in I^2.$ Fix $t_0\in I.$ Since $\Gamma$ is uniformly continuous $|\Gamma(s, t) - \Gamma(s', t')|<\epsilon$ whenever $|s-s'| + |t-t'|$ is small enough.
The foregoing implies that  
$\tag 1|\gamma_t(s)-\gamma_{t_0}(s)|<|w-\gamma_{t_0}(s)| \ \text{if} \ |t-t_0|  \ \text{is small enough}.$ 
Noting that  $\gamma_{t_0}(s)-w\neq 0$, define $\gamma(s)=\frac{\gamma_t(s)-w}{\gamma_{t_0}(s)-w}.$ The smoothness hypothesis implies that 
$\tag2 \frac{\gamma'}{\gamma}=\frac{\gamma_t'}{\gamma_t-w}-\frac{\gamma_{t_0}'}{\gamma_{t_0}-w}.$ 
And it is easy to check using $(1)$ that $|1 - \gamma(s) | <1.$ Thus, the curve defined by $\gamma $ is contained within the the disk of radius $1$ centered at $(1,0)$ and so $\int\frac{\gamma'}{\gamma}=\int\frac{\gamma'}{\gamma-0}=0$, so that on integrating $(2)$, we get 
$\tag3 n(\gamma_t,w)=\int\frac{\gamma_t'}{\gamma_t-w}=\int\frac{\gamma_{t_0}'}{\gamma_{t_0}-w}=n(\gamma_{t_0},w).$ 
The continuity of $h$ at $t_0$ now follows immediately from $(3).$
