Endpoint-fixed homotopic paths have equal holomorphic line integrals

By [L1], the image of the homotopy square lies in $\mathrm{\Omega }$. For each point $x$ of $[0,1{]}^2$, openness of $\mathrm{\Omega }$ gives an open disc $D_x\subseteq \mathrm{\Omega }$ centered at $H(x)$, and the sets $H^{-1}(D_x)$ form an open cover of the compact square $[0,1{]}^2$. By [L2], there is $\delta >0$ such that every subset of the square of diameter less than $\delta$ lies in some $H^{-1}(D_x)$. Choose $N$ with $\sqrt{2}\mathrm{/}N<\delta$.

For $1\le j,k\le N$, write $Q_{jk}=[\frac{j-1}{N},\frac{j}{N}]\times [\frac{k-1}{N},\frac{k}{N}]\mathrm{.}$ Each cell has diameter $\sqrt{2}\mathrm{/}N<\delta$, so step 1.1 places $H[Q_{jk}]$ inside an open disc $D_{jk}\subseteq \mathrm{\Omega }$. Put $a_{jk}=H(\frac{j-1}{N},\frac{k-1}{N}),b_{jk}=H(\frac{j}{N},\frac{k-1}{N}),c_{jk}=H(\frac{j}{N},\frac{k}{N}),d_{jk}=H(\frac{j-1}{N},\frac{k}{N})\mathrm{.}$ Since every disc is convex, the straight segments from $a_{jk}$ to $b_{jk}$, from $b_{jk}$ to $c_{jk}$, from $c_{jk}$ to $d_{jk}$, and from $d_{jk}$ to $a_{jk}$ all lie in $D_{jk}$. Let $P_{jk}$ be the closed polygonal contour obtained by traversing those four segments in that order.

Because $D_{jk}$ is star-shaped, [L5] gives ${\int }_{P_{jk}}f(z)dz=0.$ Also [L3] gives a primitive $F_{jk}$ of $f$ on $D_{jk}$.

Summing the zero integrals from step 3.1 over all cells, every interior polygon edge appears once in each orientation, so [L6] and [L7] cancel all interior contributions. The surviving outer boundary is the bottom polygonal path $P_0$ built from the straight segments joining ${\gamma }_0((j-1)\mathrm{/}N)$ to ${\gamma }_0(j\mathrm{/}N)$, the top polygonal path $P_1$ built in the forward direction from the straight segments joining ${\gamma }_1((j-1)\mathrm{/}N)$ to ${\gamma }_1(j\mathrm{/}N)$ but occurring in the outer boundary with reverse orientation, and the two side edges. By [L1] both side edges are constant, and for a constant path $s$ one has $s\ast s=s$, so [L6] gives ${\int }_{s}fdz={\int }_{s\ast s}fdz=2{\int }_{s}fdz$, hence ${\int }_{s}fdz=0$. Therefore ${\int }_{P_0}f(z)dz={\int }_{P_1}f(z)dz\mathrm{.}$

For each $1\le j\le N$, the bottom subpath ${\gamma }_0{\mathrm{\mid }}_{[(j-1)\mathrm{/}N,j\mathrm{/}N]}$ and the chord segment from ${\gamma }_0((j-1)\mathrm{/}N)$ to ${\gamma }_0(j\mathrm{/}N)$ both lie in $D_{j1}$. Since $F_{j1}$ is a primitive of $f$ on $D_{j1}$, [L4] gives the same endpoint increment for both, so their integrals are equal. Summing over $j$ and using [L6] yields ${\int }_{{\gamma }_0}f(z)dz={\int }_{P_0}f(z)dz\mathrm{.}$ The same argument with the top-row discs $D_{jN}$ gives ${\int }_{{\gamma }_1}f(z)dz={\int }_{P_1}f(z)dz\mathrm{.}$

Combining steps 4.1 and 4.2 gives ${\int }_{{\gamma }_0}f(z)dz={\int }_{{\gamma }_1}f(z)dz,$ as required.