Show that if a path-connected, locally path-connected space \(X\) has \(\pi_1(X)\) finite, then every map \(f: X \to S^1\) is nullhomotopic.
Solution
Goal: Prove that every continuous map \(f: X \to S^1\) from a path-connected, locally path-connected space with finite \(\pi_1(X)\) is nullhomotopic, by lifting \(f\) to the universal cover \(\mathbb{R}\).
1. Triviality of the induced homomorphism on fundamental groups:Proof:
1.1. Fix a basepoint \(x_0 \in X\) and let \(s_0 = f(x_0) \in S^1\).
1.2. The induced map \(f_*: \pi_1(X, x_0) \to \pi_1(S^1, s_0) \cong \mathbb{Z}\) is a group homomorphism.
1.3. The image \(f_*(\pi_1(X, x_0))\) is a subgroup of \(\mathbb{Z}\).
1.4. By the First Isomorphism Theorem for groups, \(|f_*(\pi_1(X, x_0))|\) divides \(|\pi_1(X, x_0)| < \infty\), so the image is a finite subgroup of \(\mathbb{Z}\).
1.5. Since \(\mathbb{Z}\) is infinite cyclic and torsion-free, its only finite subgroup is the trivial subgroup \(\{0\}\).
1.6. Thus \(f_*(\pi_1(X, x_0)) = \{0\}\).
2. Lifting \(f\) to the universal cover \(\mathbb{R}\):Proof:
2.1. Let \(p: \mathbb{R} \to S^1\) be the standard universal covering map given by \(p(t) = e^{2\pi i t}\), and choose \(t_0 \in \mathbb{R}\) with \(p(t_0) = s_0\).
2.2. The fundamental group of the total space is trivial: \(p_*(\pi_1(\mathbb{R}, t_0)) = \{0\}\).
2.3. From 1, we have \(f_*(\pi_1(X, x_0)) = \{0\} \subseteq p_*(\pi_1(\mathbb{R}, t_0))\).
2.4. Since \(X\) is path-connected and locally path-connected, the Covering Space Lifting Criterion applies: there exists a unique continuous map \(\widetilde{f}: X \to \mathbb{R}\) such that\[p \circ \widetilde{f} = f \quad \text{and} \quad \widetilde{f}(x_0) = t_0.\]
3. Null-homotopy of \(f\):Proof:
3.1. The codomain \(\mathbb{R}\) is convex and contractible.
3.2. Define the straight-line homotopy \(\widetilde{H}: X \times [0, 1] \to \mathbb{R}\) by\[\widetilde{H}(x, t) = (1 - t) \widetilde{f}(x) + t \cdot 0.\]
3.3. The map \(\widetilde{H}\) is continuous, with \(\widetilde{H}(x, 0) = \widetilde{f}(x)\) and \(\widetilde{H}(x, 1) = 0\) (the constant map to \(0 \in \mathbb{R}\)).
3.4. Define \(H: X \times [0, 1] \to S^1\) by post-composing with the covering map:\[H(x, t) = p(\widetilde{H}(x, t)) = e^{2\pi i \widetilde{H}(x, t)}.\]
3.5. Since \(p\) and \(\widetilde{H}\) are continuous, \(H\) is a continuous homotopy.
3.6. Evaluating at the endpoints:\[H(x, 0) = p(\widetilde{f}(x)) = f(x), \qquad H(x, 1) = p(0) = e^0 = 1.\]
3.7. Thus \(H\) is a homotopy between \(f\) and the constant map \(c_1(x) = 1\).
4. Conclusion:Proof:Every continuous map \(f: X \to S^1\) is nullhomotopic.