Let \(X\) be a path-connected topological space such that \(\pi_1(X) = 0\). Prove that every continuous map \(f: S^1 \to X\) can be extended to a continuous map \(F: D^2 \to X\) (where \(S^1 = \partial D^2\)).
Solution
Goal: Prove that if \(\pi_1(X) = 0\), every continuous map \(f: S^1 \to X\) extends continuously to \(F: D^2 \to X\).
1. Realization of the 2-disk \(D^2\) as a quotient cone of the cylinder \(S^1 \times [0, 1]\):Proof:
1.1. Define the continuous map \(q: S^1 \times [0, 1] \to D^2\) by\[q(z, t) = (1 - t)z \quad \text{for } z \in S^1 \subset \mathbb{C}, \, t \in [0, 1].\]
1.2. The map \(q\) is surjective: for any point \(w \in D^2\), if \(w = 0\), \(q(z, 1) = 0\) for any \(z \in S^1\); if \(w \neq 0\), \(w = r e^{i\theta} = (1 - (1 - r)) e^{i\theta} = q(e^{i\theta}, 1 - r)\).
1.3. The fibers of \(q\) are singletons \(\{(z, t)\}\) for \(t < 1\), and the top circle \(S^1 \times \{1\}\) is mapped to the single point \(0 \in D^2\).
1.4. Since the cylinder \(S^1 \times [0, 1]\) is compact and \(D^2\) is Hausdorff, the continuous surjection \(q\) is a closed map, hence a quotient map.
1.5. Therefore \(q\) induces a homeomorphism \((S^1 \times [0, 1]) / (S^1 \times \{1\}) \cong D^2\).
2. Existence of a null-homotopy from the triviality of \(\pi_1(X)\):Proof:
2.1. Fix a basepoint \(s_0 \in S^1\) and let \(x_0 = f(s_0) \in X\).
2.2. The map \(f: S^1 \to X\) is a loop based at \(x_0\), representing a homotopy class \([f] \in \pi_1(X, x_0)\).
2.3. By hypothesis, \(\pi_1(X, x_0) = 0\), so \([f] = 0\).
2.4. Thus \(f\) is homotopic relative to \(s_0\) (and freely homotopic) to the constant loop \(c_{x_0}(z) = x_0\): there exists a continuous map \(H: S^1 \times [0, 1] \to X\) such that\[H(z, 0) = f(z) \quad \text{and} \quad H(z, 1) = x_0 \quad \text{for all } z \in S^1.\]
3. Factoring the homotopy through the quotient to define the extension \(F\):Proof:
3.1. On the collapsed subspace \(S^1 \times \{1\}\), the homotopy \(H\) satisfies \(H(z, 1) = x_0\) for all \(z \in S^1\), so \(H\) is constant on \(S^1 \times \{1\}\).
3.2. By the universal property of quotient maps, since \(H\) is constant on the fibers of the quotient map \(q: S^1 \times [0, 1] \to D^2\), there exists a unique continuous map \(F: D^2 \to X\) making the diagram commute:\[F(q(z, t)) = H(z, t) \quad \text{for all } (z, t) \in S^1 \times [0, 1].\]
3.3. On the boundary circle \(\partial D^2 = S^1\), points are represented by \(w = z = q(z, 0)\) for \(z \in S^1\).
3.4. Evaluating \(F\) on the boundary:\[F(z) = F(q(z, 0)) = H(z, 0) = f(z) \quad \text{for all } z \in S^1.\]
3.5. Thus \(F: D^2 \to X\) is a continuous extension of \(f\).
4. Conclusion:Proof:Every continuous map \(f: S^1 \to X\) extends to a continuous map \(F: D^2 \to X\).