Index of the boundary of a graph-bounded plane region

($\gamma$ is a closed complex contour.) The wall segments are straight paths. Near either endpoint the Puiseux expansion of $\beta$ has the form $\beta (w+t)={\sum }_{j\ge 0}{c}_j{t}^{j\mathrm{/}m}$, or the analogous expansion in $w^{\mathrm{\prime }}-t$. Substituting $t=s^m$ parametrizes that graph segment by $s\mapsto (\sum c_j{s}^j,w+s^m)$; both coordinates are $C^1$ on a closed short interval, so the segment has finite length by [F4]. On the remaining compact subinterval the graph is $C^1$ because $\beta$ is analytic on $(w,w^{\mathrm{\prime }})$, again giving finite length by [F4]. The same argument applies to $\alpha$. Thus all four (possibly degenerate) pieces are rectifiable, their concatenation $\gamma$ is a rectifiable path, and it is closed because consecutive endpoints agree, so $\gamma$ is a closed complex contour.

(Based homotopies preserve the index.) Let ${\gamma }_0,{\gamma }_1:[0,1]\to \mathbb{C}\setminus \{q\}$ be closed rectifiable loops with ${\gamma }_0(0)={\gamma }_1(0)=q_0\mathrm{\ne }q$, and let $F:[0,1]\times [0,1]\to \mathbb{C}\setminus \{q\}$ be continuous with $F(j,u)={\gamma }_j(u)$ for $j\in \{0,1\}$ and $F(t,0)=F(t,1)=q_0$ for all $t$. Put $c=q_0-q\mathrm{\ne }0$, $\tilde{F}(t,u)=(F(t,u)-q)\mathrm{/}c$ and $\alpha (t,u)=\tilde{F}(t,u)\mathrm{/}\mathrm{\mid }\tilde{F}(t,u)\mathrm{\mid }$; then $\tilde{F}$ takes values in ${\mathbb{C}}^{\times }$, is a path homotopy between the loops ${\tilde{\gamma }}_j=({\gamma }_j-q)\mathrm{/}c$ based at $1$, and $\alpha$ is a path homotopy between the based circle loops ${\alpha }_j={\tilde{\gamma }}_j\mathrm{/}\mathrm{\mid }{\tilde{\gamma }}_j\mathrm{\mid }$; by [F5] the degrees of ${\alpha }_0$ and ${\alpha }_1$ are equal, and by [F5] and [F2] applied to the closed rectifiable loops ${\tilde{\gamma }}_j$ based at $1$, $n({\gamma }_j,q)=n({\tilde{\gamma }}_j,0)=\mathrm{deg}({\alpha }_j)$; hence $n({\gamma }_0,q)=n({\gamma }_1,q)$.

(Exterior case.) Let $q=(x_q,y_q)\in \mathbb{C}\setminus T$. If $y_q<w$, the vertical ray $\{(x_q,y):y\le y_q\}$ is disjoint from $T$ because every point of $T$ has $y\ge w>y_q$; if $y_q>w^{\mathrm{\prime }}$, the upward vertical ray $\{(x_q,y):y\ge y_q\}$ has the same property; if $w\le y_q\le w^{\mathrm{\prime }}$ and $x_q>\beta (y_q)$ (the endpoint case $y_q\in \{w,w^{\mathrm{\prime }}\}$ included, since every point of $T$ at height $y_q$ has first coordinate in $[\alpha (y_q),\beta (y_q)]$), the rightward horizontal ray $\{(x,y_q):x\ge x_q\}$ is disjoint from $T$; and if $x_q<\alpha (y_q)$ the leftward horizontal ray $\{(x,y_q):x\le x_q\}$ is disjoint from $T$. These cases exhaust $\mathbb{C}\setminus T$, since a point with $w\le y_q\le w^{\mathrm{\prime }}$ and $\alpha (y_q)\le x_q\le \beta (y_q)$ lies in $T$. Each chosen ray is a connected set containing $q$, contained in $\mathbb{C}\setminus {\gamma }^{\ast }$ (because ${\gamma }^{\ast }\subseteq T$) and meeting $\{p:\mathrm{\mid }p\mathrm{\mid }>R\}$ for $R$ as in [F3]; hence $n(\gamma ,q)=0$ by [F3].

(Interior case: reduction to a rectangle.) Let $q=(x_q,y_q)\in T^{\circ }$. Choose $M>\max \{x_q,{\sup }_{[w,w^{\mathrm{\prime }}]}\beta \}$ and $M^{\mathrm{\prime }\mathrm{\prime }}>-\min \{x_q,{\inf }_{[w,w^{\mathrm{\prime }}]}\alpha \}$, and put $q_0:=(\alpha (w),w)\mathrm{\ne }q$. For $t\in [0,2]$ let $R_t$ be the region bounded by the graphs ${\alpha }_t,{\beta }_t:[w,w^{\mathrm{\prime }}]\to \mathbb{R}$, ${\alpha }_t:=(1-s(t))\alpha -s(t)M^{\mathrm{\prime }\mathrm{\prime }}$ and ${\beta }_t:=(1-r(t))\beta +r(t)M$, where $r(t)=\min \{t,1\}$ and $s(t)=\max \{0,t-1\}$; then ${\alpha }_t\le {\beta }_t$ are continuous, real-analytic on $(w,w^{\mathrm{\prime }})$, and $q_0$ lies on the bottom wall of $R_t$ because ${\alpha }_t(w)\le \alpha (w)\le \beta (w)\le {\beta }_t(w)$. Parametrize the boundary of $R_t$ by the five arcs bottom-from-$q_0$, right graph of ${\beta }_t$, top wall, left graph of ${\alpha }_t$, bottom-to-$q_0$, the parameter on each arc being proportional to the arc parameter; every coordinate is built from $s(t),r(t)$ and the continuous functions $\alpha ,\beta$ by affine operations, so $(t,u)\mapsto P(t,u)$ is continuous, $P(t,0)=P(t,1)=q_0$, and $P(t,\cdot )$ is a positively oriented closed contour. Moreover $q\in R_t^{\circ }$ for every $t$: indeed ${\alpha }_t(y_q)\le \alpha (y_q)<x_q<\beta (y_q)\le {\beta }_t(y_q)$ in the first stage and ${\alpha }_t(y_q)\le \alpha (y_q)<x_q<M$ in the second, with $w<y_q<w^{\mathrm{\prime }}$ fixed by $q\in T^{\circ }$. Hence $(t,u)\mapsto P(t,u)$ is a based homotopy in $\mathbb{C}\setminus \{q\}$ from $P(0,\cdot )=\gamma$ to $P(2,\cdot )=$ the positively oriented boundary of the rectangle $R:=[-M^{\mathrm{\prime }\mathrm{\prime }},M]\times [w,w^{\mathrm{\prime }}]$, which is rectifiable as a finite concatenation of segments; by step 1.2, $n(\gamma ,q)=n(\mathrm{\partial }R,q)$.

(Interior case: reduction of the rectangle to a circle.) Write $q=(x_q,y_q)$ and for each direction $\theta$ let $\rho (\theta )$ be the unique positive number with $q+\rho (\theta )e^{i\theta }\in \mathrm{\partial }R$; explicitly $\rho (\theta )$ is the minimum of the positive numbers among $(M-x_q)\mathrm{/}\cos \theta$ for $\cos \theta >0$, $(-M^{\mathrm{\prime }\mathrm{\prime }}-x_q)\mathrm{/}\cos \theta$ for $\cos \theta <0$, $(w^{\mathrm{\prime }}-y_q)\mathrm{/}\sin \theta$ for $\sin \theta >0$ and $(w-y_q)\mathrm{/}\sin \theta$ for $\sin \theta <0$, so $\rho$ is continuous and positive on $[0,2\pi ]$. With ${\theta }_0$ the argument of $q_0-q$ and $r_0:=\mathrm{\mid }{q}_0-q\mathrm{\mid }=\rho ({\theta }_0)$, put $G(t,\theta ):=q+((1-t)\rho (\theta )+t{r}_0)e^{i\theta }$ for $t\in [0,1]$ and $\theta \in [{\theta }_0,{\theta }_0+2\pi ]$; then $G$ is continuous, $G(t,{\theta }_0)=q+\rho ({\theta }_0)e^{i{\theta }_0}=q_0$ for every $t$, so $G$ is a based homotopy of closed curves in $\mathbb{C}\setminus \{q\}$, with $G(0,\cdot )$ a positively oriented parametrization of $\mathrm{\partial }R$ and $G(1,\cdot )$ the circle of radius $r_0$ about $q$ traversed once positively. By step 1.2 and [F6], $n(\mathrm{\partial }R,q)=n(\text{circle},q)=1$.

(Interior case concluded.) For $q\in T^{\circ }$ steps 2.1 and 2.2 give $n(\gamma ,q)=n(\mathrm{\partial }R,q)=1$.

(Similarity invariance.) Let $\sigma (z)=cz+d$ with $c\mathrm{\ne }0$ and let $q\mathrm{\notin }{\gamma }^{\ast }$. By [F1] there is a continuous logarithm $\lambda$ of $\gamma -q$ along $\gamma$; then $\lambda +\log c$ is a continuous logarithm of $\sigma \circ \gamma -\sigma (q)=c(\gamma -q)$ along $\sigma \circ \gamma$, so by [F1] the two contours have the same index, $n(\sigma \circ \gamma ,\sigma (q))=n(\gamma ,q)$. Hence for $q^{\mathrm{\prime }}\in \sigma (T^{\circ })$, writing $q^{\mathrm{\prime }}=\sigma (q)$ with $q\in T^{\circ }$, step 3.1 gives $n(\sigma \circ \gamma ,q^{\mathrm{\prime }})=1$; and for $q^{\mathrm{\prime }}\mathrm{\notin }\sigma (T)$ the point $q={\sigma }^{-1}(q^{\mathrm{\prime }})$ lies in $\mathbb{C}\setminus T$, so step 1.3 gives $n(\gamma ,q)=0$ and therefore $n(\sigma \circ \gamma ,q^{\mathrm{\prime }})=0$. Multiplication by $c\mathrm{\ne }0$ is orientation-preserving, so $\sigma \circ \gamma$ is the positively oriented boundary contour of $\sigma (T)$, and clause 4 follows.

(Conclusion.) Steps 3.1 and 1.3 give the two index assertions. For the homology claim, define $H:[0,1{]}^2\to T,H(s,t)=((1-s)\alpha (y(t))+s\beta (y(t)),y(t)),y(t)=w+(w^{\mathrm{\prime }}-w)t\mathrm{.}$ This is continuous and its restriction to the positively oriented boundary of the square traces $\gamma$. Thus $\gamma$ is null-homotopic in $T\subseteq \mathrm{\Omega }$, and a null-homotopic loop represents zero in singular homology, so it is null-homologous in $\mathrm{\Omega }$. The construction selected only finitely many explicit numbers ($M$, $M^{\mathrm{\prime }\mathrm{\prime }}$, the parameters of the arcs), so no choice principle is used.