Criticize the following “proof” that \(S^2\) is simply connected. Let \(f\) be a loop in \(S^2\) based at \(x_0\). Choose a point \(p\) of \(S^2\) not lying in the image of \(f\). Since \(S^2 - \ts{p}\) is homeomorphic with \(\mathbb{R}^2\), and \(\mathbb{R}^2\) is simply connected, the loop \(f\) is path homotopic to the constant loop.
Solution
Goal: Identify and critique the fundamental flaw in the proposed proof that \(S^2\) is simply connected, and explain how the gap is rigorously resolved.
1. Identification of the logical fallacy:The flaw: The step “Choose a point \(p\) of \(S^2\) not lying in the image of \(f\)” assumes without justification that the continuous loop \(f: [0, 1] \to S^2\) cannot be surjective.
2. Existence of space-filling curves:Proof of non-triviality:
2.1. Continuous surjective paths exist: By Peano’s Theorem, there exists a continuous surjective map from the unit interval \([0, 1]\) onto the unit square \([0, 1]^2\).
2.2. By post-composing with the standard quotient map \([0, 1]^2 \to S^2\), one can construct a continuous closed loop \(f: [0, 1] \to S^2\) whose image is the entire 2-sphere, \(\operatorname{im}(f) = S^2\).
2.3. For such a space-filling loop, \(S^2 \setminus \operatorname{im}(f) = \emptyset\). No such missing point \(p\) exists, so the punctured-sphere homeomorphism \(S^2 \setminus \{p\} \cong \mathbb{R}^2\) cannot be directly applied to the loop \(f\).
3. Rigorous resolution:To prove \(\pi_1(S^2, x_0) = 0\) rigorously, one must circumvent this issue via either:1. Homotopic Perturbation / Approximation: Use the Lebesgue Covering Lemma on an open cover of small charts to approximate \(f\) by a homotopic piecewise-linear or smooth loop \(g \simeq f\) that is not surjective, and then contract \(g\) in \(S^2 \setminus \{p\} \cong \mathbb{R}^2\).2. Seifert-van Kampen Theorem: Cover \(S^2\) by two overlapping open hemispheres \(U = S^2 \setminus \{(0, 0, 1)\}\) and \(V = S^2 \setminus \{(0, 0, -1)\}\), each homeomorphic to \(\mathbb{R}^2\) (simply connected) with path-connected intersection \(U \cap V \simeq S^1\), directly concluding \(\pi_1(S^2) \cong \pi_1(U) \ast_{\pi_1(U \cap V)} \pi_1(V) = 0\). Q.E.D.