The Möbius band \(M\) is not a surface, but what is called a “surface with boundary”. Show that \(M\) is homeomorphic to the space obtained by deleting an open disc from \(P^2\).
Solution
Goal: Prove that deleting an open disc from the real projective plane \(\mathbb{R}P^2\) yields a space homeomorphic to the Möbius band \(M\).
1. Quotient model of the real projective plane:Proof:
1.1. The real projective plane \(\mathbb{R}P^2\) is homeomorphic to the closed unit disc \(D^2 = \{x \in \mathbb{R}^2 \mid \|x\| \le 1\}\) modulo the antipodal identification on its boundary:\[\mathbb{R}P^2 \cong D^2 / \sim, \quad \text{where } x \sim -x \text{ for } x \in S^1 = \partial D^2.\]
1.2. Let \(q: D^2 \to \mathbb{R}P^2\) be the canonical quotient map.
2. Deleting an open disc from \(\mathbb{R}P^2\):Proof:
2.1. Let \(D_{1/2} = \{x \in \mathbb{R}^2 \mid \|x\| < 1/2\}\) be the open central disc of radius \(1/2\).
2.2. Since \(D_{1/2} \subset \operatorname{Int}(D^2)\) contains no boundary points, \(U = q(D_{1/2})\) is an open topological disc in \(\mathbb{R}P^2\).
2.3. The punctured space \(Y = \mathbb{R}P^2 \setminus U\) is homeomorphic to the quotient of the closed annulus \(A = \{x \in \mathbb{R}^2 \mid 1/2 \le \|x\| \le 1\}\) by the boundary identification \(x \sim -x\) on the outer boundary circle \(\|x\| = 1\).
3. Homeomorphism between the identified annulus \(A/\sim\) and the Möbius strip \(M\):Proof:
3.1. In polar coordinates \((r, \theta)\) on \(A\), the equivalence relation identifies \((1, \theta) \sim (1, \theta + \pi)\) for all \(\theta \in [0, \pi]\), while leaving points with \(1/2 \le r < 1\) un-identified.
3.2. Cut the annulus along the segments \(\theta = 0\) and \(\theta = \pi\).
3.3. Consider the semi-annulus \(R = [0, \pi] \times [1/2, 1]\).
3.4. The outer arc \((1, \theta)\) connects \((1, 0)\) to \((1, \pi) \sim (1, 0)\), forming a continuous strip.
3.5. The radial edges at \(\theta = 0\) and \(\theta = \pi\) are glued with a reversal of orientation on the radial parameter, exactly realizing the standard quotient presentation of the Möbius strip:\[M = ([0, 1] \times [-1, 1]) / \sim, \quad (0, t) \sim (1, -t).\]
3.6. The inner boundary circle \(r = 1/2\) becomes the single continuous boundary circle \(\partial M\).
3.7. Hence \(Y = A/\sim\) is homeomorphic to \(M\).
4. Conclusion:The space obtained by deleting an open disc from \(\mathbb{R}P^2\) is homeomorphic to the Möbius band \(M\). Q.E.D.