[proofplan] We define the integer-valued function $s \mapsto n(\gamma_s, z_0)$ for the family of closed curves $\gamma_s := H(\cdot, s)$. We show this function is continuous in $s$ by reducing to the continuity argument for winding numbers: the compactness of the image of $H$ and the uniform distance from $z_0$ give uniform estimates on the integral difference. Since a continuous integer-valued function on the connected set $[0,1]$ must be constant, $n(\gamma_0, z_0) = n(\gamma_1, z_0)$. [/proofplan] [step:Define the family of closed curves and establish uniform separation from $z_0$] For each $s \in [0,1]$, define the closed curve $\gamma_s: [a,b] \to \mathbb{C} \setminus \{z_0\}$ by $\gamma_s(t) := H(t,s)$. This is closed because $H(a,s) = H(b,s)$ for all $s$. Since $H: [a,b] \times [0,1] \to \mathbb{C} \setminus \{z_0\}$ is continuous on the compact set $[a,b] \times [0,1]$, the image $\operatorname{im}(H)$ is compact and does not contain $z_0$. Therefore \begin{align*} \delta := \inf_{(t,s) \in [a,b] \times [0,1]} |H(t,s) - z_0| = \min_{(t,s) \in [a,b] \times [0,1]} |H(t,s) - z_0| > 0. \end{align*} In particular, $|\gamma_s(t) - z_0| \geq \delta$ for all $t \in [a,b]$ and $s \in [0,1]$. [/step] [step:Show that $s \mapsto n(\gamma_s, z_0)$ is continuous on $[0,1]$] Fix $s_0 \in [0,1]$. For any $s \in [0,1]$: \begin{align*} n(\gamma_s, z_0) - n(\gamma_{s_0}, z_0) &= \frac{1}{2\pi i} \left(\oint_{\gamma_s} \frac{dz}{z - z_0} - \oint_{\gamma_{s_0}} \frac{dz}{z - z_0}\right). \end{align*} Writing these as parametric integrals over $[a,b]$, the first integral becomes $\int_a^b \frac{\gamma_s'(t)}{\gamma_s(t) - z_0}\, d\mathcal{L}^1(t)$ when $\gamma_s$ is piecewise $C^1$. However, $\gamma_s$ may only be continuous (not piecewise $C^1$) for intermediate $s$. We instead use the logarithmic characterisation. Define $g: [a,b] \times [0,1] \to \mathbb{C} \setminus \{0\}$ by $g(t,s) := H(t,s) - z_0$. Then $|g(t,s)| \geq \delta > 0$ for all $(t,s)$. Consider the map $\phi: [a,b] \times [0,1] \to S^1$ defined by $\phi(t,s) := g(t,s)/|g(t,s)|$. For each fixed $s$, $\phi(\cdot, s)$ is a continuous closed curve in $S^1$, and the winding number $n(\gamma_s, z_0)$ equals the degree of this map, i.e., the integer \begin{align*} n(\gamma_s, z_0) &= \frac{1}{2\pi} \left[\arg(\gamma_s(t) - z_0)\right]_{t=a}^{t=b} \end{align*} where $\arg(\gamma_s(t) - z_0)$ denotes a continuous branch of the argument along the curve. Since $H$ is uniformly continuous on $[a,b] \times [0,1]$ (continuous on a compact set), for any $\varepsilon > 0$ there exists $\eta > 0$ such that $|s - s_0| < \eta$ implies $|H(t,s) - H(t,s_0)| < \varepsilon$ for all $t \in [a,b]$. Choosing $\varepsilon < \delta$, we ensure $|\gamma_s(t) - \gamma_{s_0}(t)| < \delta \leq |\gamma_{s_0}(t) - z_0|$ for all $t$. Under this condition, the curve $t \mapsto (\gamma_s(t) - z_0)/(\gamma_{s_0}(t) - z_0)$ has image contained in the half-plane $\{\operatorname{Re}(w) > 0\}$ (since $|w - 1| < 1$ implies $\operatorname{Re}(w) > 0$). A closed curve in $\{\operatorname{Re}(w) > 0\}$ has winding number zero around the origin, so \begin{align*} n(\gamma_s, z_0) - n(\gamma_{s_0}, z_0) = 0 \end{align*} whenever $|s - s_0| < \eta$. Thus $s \mapsto n(\gamma_s, z_0)$ is locally constant, hence continuous. [guided] The key difficulty is that the intermediate curves $\gamma_s$ need not be piecewise $C^1$, so we cannot directly differentiate the contour integral with respect to $s$. Instead, we use a topological argument. For each $s \in [0,1]$, define $\gamma_s(t) := H(t,s)$. The winding number $n(\gamma_s, z_0)$ is well-defined because $\gamma_s$ is a continuous closed curve avoiding $z_0$ (we have $|H(t,s) - z_0| \geq \delta > 0$). The winding number can be defined for continuous (not just piecewise $C^1$) curves via the lifting characterisation: $n(\gamma_s, z_0)$ is the unique integer such that there exists a continuous function $\theta: [a,b] \to \mathbb{R}$ with $\gamma_s(t) - z_0 = |\gamma_s(t) - z_0| e^{i\theta(t)}$ and $n(\gamma_s, z_0) = (\theta(b) - \theta(a))/(2\pi)$. Now we show local constancy. Fix $s_0 \in [0,1]$. Since $H$ is continuous on the compact set $[a,b] \times [0,1]$, it is uniformly continuous: for any $\varepsilon > 0$, there exists $\eta > 0$ such that $|s - s_0| < \eta$ implies $|H(t,s) - H(t,s_0)| < \varepsilon$ for all $t \in [a,b]$. Choose $\varepsilon = \delta$ (the uniform separation from $z_0$). Then for $|s - s_0| < \eta$ and all $t \in [a,b]$: \begin{align*} |\gamma_s(t) - \gamma_{s_0}(t)| < \delta \leq |\gamma_{s_0}(t) - z_0|. \end{align*} Why does this imply equal winding numbers? Consider the quotient $w(t) := (\gamma_s(t) - z_0)/(\gamma_{s_0}(t) - z_0)$. The estimate gives $|w(t) - 1| = |\gamma_s(t) - \gamma_{s_0}(t)|/|\gamma_{s_0}(t) - z_0| < 1$. So $w(t)$ lies in the open disc $B(1,1) \subset \{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$, which is a simply connected subset of $\mathbb{C} \setminus \{0\}$. The winding number is additive: $n(\gamma_s, z_0) = n(\gamma_{s_0}, z_0) + n(w, 0)$ where $n(w, 0)$ is the winding number of the closed curve $t \mapsto w(t)$ around $0$. But $w$ has image in the simply connected set $B(1,1)$ which does not contain $0$, so $n(w, 0) = 0$. (A curve in a simply connected subset of $\mathbb{C} \setminus \{0\}$ has winding number zero around $0$ because it is contractible within that set.) Therefore $n(\gamma_s, z_0) = n(\gamma_{s_0}, z_0)$ for all $s$ with $|s - s_0| < \eta$. This shows $s \mapsto n(\gamma_s, z_0)$ is locally constant on $[0,1]$. [/guided] [/step] [step:Conclude by connectedness of $[0,1]$] The function $s \mapsto n(\gamma_s, z_0)$ is locally constant on the connected set $[0,1]$, hence constant. In particular, $n(\gamma_0, z_0) = n(\gamma_1, z_0)$. [/step]