Show that if \(A\) is a retract of \(B^2\), then every continuous map \(f: A \to A\) has a fixed point.
Solution
1. Retraction and map extension:
1.1. Since \(A\) is a retract of \(B^2\), there exists a continuous map \(r: B^2 \to A\) such that:\[ r(a) = a \quad \text{for all } a \in A. \]Let \(i: A \hookrightarrow B^2\) denote the inclusion map.
1.2. Let \(f: A \to A\) be an arbitrary continuous map.Define the map \(g: B^2 \to B^2\) by composition:\[ g = i \circ f \circ r : B^2 \to B^2. \]Because \(r, f\), and \(i\) are continuous, \(g\) is a continuous map from the closed unit disk \(B^2\) to itself.
2. Existence of a fixed point via Brouwer’s Fixed Point Theorem:
2.1. By the Brouwer Fixed Point Theorem in dimension 2, every continuous map from \(B^2\) to \(B^2\) has at least one fixed point.Thus there exists a point \(x_0 \in B^2\) such that:\[ g(x_0) = x_0. \]
3. Proof that \(x_0 \in A\) and \(f(x_0) = x_0\):
3.1. By definition of \(g\), the image of \(g\) is contained in \(A\):\[ x_0 = g(x_0) = f(r(x_0)) \in A. \]Thus the fixed point \(x_0\) necessarily belongs to the subspace \(A\).
3.2. Since \(x_0 \in A\) and \(r\) is the identity on \(A\), we have \(r(x_0) = x_0\).
3.3. Substituting \(r(x_0) = x_0\) into the equation \(x_0 = f(r(x_0))\) gives:\[ f(x_0) = x_0. \]Thus \(x_0 \in A\) is a fixed point of \(f\).
4. Conclusion:Every continuous map \(f: A \to A\) has a fixed point. Q.E.D.