# Winding numbers are continuous: The proof was too easy

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?

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.