Give an example to show that the conclusion of the Borsuk lemma need not hold if \(f\) is not injective.
Solution
Goal: Provide an explicit counterexample demonstrating that the conclusion of the Borsuk Lemma (that an image of a non-separating compact planar set does not separate the plane) fails when the continuous map \(f\) is not injective.
1. Statement of the Borsuk Lemma:The Borsuk Lemma (Munkres §62 Lemma 62.4) states that if \(A \subset \mathbb{R}^2\) is a compact subspace that does not separate \(\mathbb{R}^2\) (i.e. \(\mathbb{R}^2 \setminus A\) is connected), and \(f: A \to \mathbb{R}^2\) is an injective continuous map (an imbedding), then \(f(A)\) does not separate \(\mathbb{R}^2\) (i.e. \(\mathbb{R}^2 \setminus f(A)\) is connected).
2. Construction of non-injective counterexample:Proof:
2.1. Let \(A = [0, 1] \times \{0\} \subset \mathbb{R}^2\) be the unit interval on the \(x\)-axis.
2.2. The set \(A\) is compact, convex, and homeomorphic to the interval \([0, 1]\).
2.3. Its complement \(\mathbb{R}^2 \setminus A\) is connected (path-connected), so \(A\) does not separate the plane.
2.4. Define the continuous map \(f: A \to \mathbb{R}^2\) by:\[f(t, 0) = (\cos 2\pi t, \sin 2\pi t).\]
2.5. The map \(f\) is continuous, but fails to be injective since \(f(0, 0) = f(1, 0) = (1, 0)\).
3. Failure of the Borsuk conclusion for the image:Proof:
3.1. The image \(f(A)\) is the unit circle \(S^1 = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\}\).
3.2. By the Jordan Curve Theorem, the complement \(\mathbb{R}^2 \setminus S^1\) consists of exactly two connected components: the bounded open disk \(B_1(\mathbf{0})\) and the unbounded region \(\mathbb{R}^2 \setminus \overline{B_1(\mathbf{0})}\).
3.3. Thus \(f(A)\) separates the plane into 2 components, whereas \(\mathbb{R}^2 \setminus A\) had only 1 component.
4. Conclusion:The conclusion of the Borsuk lemma fails without the hypothesis of injectivity. Q.E.D.