A plane domain with trivial fundamental group is homologically simply connected

Let $\mathrm{\Gamma }={\sum }_{j<r}{m}_j{\gamma }_j$ be a cycle with trace in $\mathrm{\Omega }$, and let $f$ be holomorphic on $\mathrm{\Omega }$. Choose a basepoint $z_0\in \mathrm{\Omega }$. Let $E:=\{{\gamma }_j(a_j):j<r,m_j\mathrm{\ne }0\}\cup \{{\gamma }_j(b_j):j<r,m_j\mathrm{\ne }0\}\mathrm{.}$ For each $q\in E$, [L5] gives a polygonal path ${\lambda }_q$ in $\mathrm{\Omega }$ from $z_0$ to $q$; by [L6] each ${\lambda }_q$ is a rectifiable contour.

Fix $j<r$ with $m_j\mathrm{\ne }0$, and write $a_j={\gamma }_j(a_j)$ and $b_j={\gamma }_j(b_j)$. The contour ${\delta }_j:={\lambda }_{a_j}\ast {\gamma }_j\ast {\lambda }_{b_j}^{-}$ is a based loop at $z_0$. By the triviality hypothesis, its loop class is the identity, so [L1] makes ${\delta }_j$ path-homotopic relative to the endpoints to the constant loop at $z_0$. Applying [L2] and then [L6] gives $0={\int }_{{\delta }_j}f(z)dz={\int }_{{\lambda }_{a_j}}f(z)dz+{\int }_{{\gamma }_j}f(z)dz-{\int }_{{\lambda }_{b_j}}f(z)dz,$ hence ${\int }_{{\gamma }_j}f(z)dz={\int }_{{\lambda }_{b_j}}f(z)dz-{\int }_{{\lambda }_{a_j}}f(z)dz\mathrm{.}$

By [L4] and step 1.2, ${\int }_{\mathrm{\Gamma }}f(z)dz={\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\gamma }_j}f(z)dz={\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\lambda }_{b_j}}f(z)dz-{\sum }_{\begin{array}{c}j<r\\ m_j\mathrm{\ne }0\end{array}}{m}_j{\int }_{{\lambda }_{a_j}}f(z)dz\mathrm{.}$ Grouping the two finite sums by endpoint $q\in E$ and using the boundary formula from [L4], this becomes ${\int }_{\mathrm{\Gamma }}f(z)dz={\sum }_{q\in E}\mathrm{\partial }\mathrm{\Gamma }(q){\int }_{{\lambda }_q}f(z)dz=0,$ because $\mathrm{\Gamma }$ is a cycle. Thus ${\int }_{\mathrm{\Gamma }}f(z)dz=0$ for every holomorphic $f$ and every cycle $\mathrm{\Gamma }$ in $\mathrm{\Omega }$.

The criterion in [L3] now shows that $\mathrm{\Omega }$ is homologically simply connected.