Let \(H^2\) be the subspace of \(\mathbb{R}^2\) consisting of all points \((x_1, x_2)\) with \(x_2 \geq 0\). A 2-manifold with boundary (or surface with boundary) is a Hausdorff space \(X\) with a countable basis such that each point \(x\) of \(X\) has a neighborhood homeomorphic with an open set of \(\mathbb{R}^2\) or \(H^2\). The boundary of \(X\) (denoted \(\partial X\)) consists of those points \(x\) such that \(x\) has no neighborhood homeomorphic with an open set of \(\mathbb{R}^2\).
Show that no point of \(H^2\) of the form \((x_1, 0)\) has a neighborhood (in \(H^2\)) that is homeomorphic to an open set of \(\mathbb{R}^2\).
Show that \(x \in \partial X\) if and only if there is a homeomorphism \(h\) mapping a neighborhood of \(x\) onto an open set of \(H^2\) such that \(h(x) \in \mathbb{R} \times 0\).
Show that \(\partial X\) is a 1-manifold.
Solution
- Let \(q=(q_1,0)\in\partial H^2\). We use local homology. For a sufficiently small half-disk \(D_+\) about \(q\), both \(D_+\) and \(D_+-\{q\}\) are contractible: the punctured half-disk deformation retracts onto its semicircular boundary arc. Hence the long exact sequence of the pair gives \[ H_2(D_+,D_+-\{q\})=0. \] By excision this is the local group \[ H_2(H^2,H^2-\{q\})=0. \]
By contrast, if \(z\) lies in an open subset \(V\subset\mathbb R^2\), excision to a small disk about \(z\) gives \[ H_2(V,V-\{z\})\cong H_2(D^2,D^2-\{0\})\cong\mathbb Z. \] Local homology is preserved by homeomorphisms. Therefore no neighborhood of \(q\) in \(H^2\) can be homeomorphic to an open subset of \(\mathbb R^2\).
- Suppose first that a chart \[ h:U\longrightarrow V\subset H^2 \] satisfies \(h(x)\in\mathbb R\times\{0\}\). By part (a), \(h(x)\) has no neighborhood in \(H^2\) homeomorphic to an open subset of \(\mathbb R^2\); transporting this statement through \(h\), neither does \(x\). Hence \(x\in\partial X\).
Conversely, let \(x\in\partial X\). By the definition of a manifold with boundary, \(x\) has a chart to an open subset of either \(\mathbb R^2\) or \(H^2\). The first possibility is excluded by the definition of \(\partial X\). Thus there is a half-plane chart \(h:U\to V\subset H^2\). If \(h(x)\) had positive second coordinate, a small Euclidean disk about \(h(x)\) would lie in \(H^2\), giving an ordinary \(\mathbb R^2\)-neighborhood of \(x\), again a contradiction. Therefore \(h(x)\in\mathbb R\times\{0\}\).
- Let \(x\in\partial X\) and choose a chart as in (b), with \(h(x)=(a,0)\). Shrinking the chart if necessary, take \[ V=(a-\varepsilon,a+\varepsilon)\times[0,\varepsilon) \] inside its image. By part (b), \[ h(U\cap\partial X)=V\cap(\mathbb R\times\{0\}) =(a-\varepsilon,a+\varepsilon)\times\{0\}. \] Hence \(U\cap\partial X\) is an open neighborhood of \(x\) in \(\partial X\) homeomorphic to an open interval. Since \(\partial X\) is a subspace of the Hausdorff second-countable space \(X\), it is Hausdorff and second countable. Thus \(\partial X\) is a 1-manifold.