In the Complex Analysis text by Ahlfors, he says that we can define the winding number $n(\gamma,a)$ for any continuous, closed curve $\gamma$ which doesn't pass through the point $a$ (differentiability is not required).
Divide $\gamma$ into subarcs $\gamma_1,\dots,\gamma_n$, such that each subarc lies in disk which doesn't include $a$.
Form the line segments $\sigma_1,\dots,\sigma_n$ between the endpoints (with direction as the curve's).
Lastly, define $n(\gamma,a)$ to be $n(\sigma,a)$ where $\sigma=\sigma_1+\dots+\sigma_n$.
I want to show that this number is independent of the choice of the subarcs. That is if $\delta_1,\dots,\delta_m$ is another admissible partition, with line segments $\tau_1,\dots,\tau_m$ than $n(\sigma,a)=n(\tau,a)$.
I've been thinking about this for a couple of days now, and any help will be appreciated.
Hint: If you have two partitions $(\sigma_k)$ and $(\tau_l)$, you can always find a common refined partition $(\rho_m)$, that is a partition such that each $\sigma_k$ and each $\tau_l$ is a $\rho_m$. Now if you prove that $n(\sigma,a)=n(\rho,a)$ and $n(\tau,a)=n(\rho,a)$, you're done; in other words, you only need to consider the case where one of the two partitions is a refinement of the other, and I believe that case should be much easier.