Show that for \(X\) path connected, \(\pi_1(X) = \mathbb{1} \iff \forall \text{cts.}~f: S^1 \rightarrow X\) \(f\), extends to a continuous map \(F: B^2 \rightarrow X\).
Solution
1. If every continuous map \(f:S^1\to X\) extends to \(F:B^2\to X\), then \(\pi_1(X)\) is trivial.
Let \(f:S^1\to X\) be any based loop. If \(F:B^2\to X\) extends \(f\), then contracting \(B^2\) to its center and composing with \(F\) gives a free null-homotopy of \(f\). Since \(X\) is path connected, a freely null-homotopic based loop represents the identity in \(\pi_1(X)\): if the free null-homotopy moves the basepoint along a path \(\gamma\), the original loop represents the conjugate by \(\gamma\) of the constant loop, hence is trivial. Thus every element of \(\pi_1(X)\) is the identity.
2. Conversely, assume \(X\) is path connected and \(\pi_1(X)=1\). Then every map \(f:S^1\to X\) is null-homotopic.
Choose a point \(s_0\in S^1\). The map \(f\) is a loop based at \(f(s_0)\). Since the fundamental group at that basepoint is trivial, there is a homotopy \(H:S^1\times I\to X\) from \(f\) to the constant map at \(f(s_0)\).
3. A null-homotopy of \(f\) is equivalent to an extension over \(B^2\).
Identify \[ B^2\cong S^1\times I/(S^1\times\{1\}), \] with \(S^1\times\{0\}\) as the boundary. If \(H\) is the null-homotopy from 2, then \(H\) is constant on \(S^1\times\{1\}\), so it descends through this quotient to a continuous map \(F:B^2\to X\) whose restriction to the boundary is \(f\).
4. Hence\[ \boxed{\pi_1(X)=1\iff\text{every continuous }S^1\to X\text{ extends over }B^2.} \]
Combine <1>1–<1>3.