Let \(X\) be a 2-manifold; let \(U_1, \ldots, U_k\) be a collection of disjoint open sets in \(X\); and suppose that for each \(i\), there is a homeomorphism \(h_i\) of the open unit ball \(B^2\) with \(U_i\). Let \(\epsilon = 1/2\) and let \(B_\epsilon\) be the open ball of radius \(\epsilon\). Show that the space \(Y = X - \bigcup h_i(B_\epsilon)\) is a 2-manifold with boundary, and that \(\partial Y\) has \(k\) components. The space \(Y\) is called “X-with-\(k\)-holes.”
Solution
For each \(i\), write \[ D_i=h_i(B_{1/2})\subset U_i. \] The sets \(D_i\) are pairwise disjoint open disks. Put \[ Y=X-\bigcup_{i=1}^k D_i. \] As a closed subspace of the Hausdorff second-countable surface \(X\), the space \(Y\) is Hausdorff and second countable.
Away from the circles \[ C_i=h_i(\{z:\|z\|=1/2\}), \] points of \(Y\) have the same ordinary surface charts they had in \(X\). At a point of \(C_i\), use polar coordinates in the annulus \[ h_i\bigl(\{1/2\le\|z\|<3/4\}\bigr). \] Locally the inequality \(r\ge1/2\) becomes a half-plane inequality, so each such point has a chart onto an open subset of \(H^2\) taking \(C_i\) to the boundary line. Hence \(Y\) is a 2-manifold with boundary.
By the intrinsic-boundary criterion of E-RKSXA, its boundary is exactly \[ \partial Y=C_1\sqcup\cdots\sqcup C_k. \] Each \(C_i\) is homeomorphic to \(S^1\), and the circles are pairwise disjoint because the original \(U_i\)’s are disjoint. Thus \(\partial Y\) has exactly \(k\) connected components.