We can regard \(\pi_1(X, x_0)\) as the set of basepoint-preserving homotopy classes of maps \((S^1, s_0) \to (X, x_0)\). Let \([S^1, X]\) be the set of homotopy classes of maps \(S^1 \to X\), with no conditions on basepoints. Thus there is a natural map \(\Phi: \pi_1(X, x_0) \longrightarrow [S^1, X]\) obtained by ignoring basepoints. Show that \(\Phi\) is onto if \(X\) is path-connected, and that \(\Phi([f]) = \Phi([g])\) iff \([f]\) and \([g]\) are conjugate in \(\pi_1(X, x_0)\). Hence \(\Phi\) induces a one-to-one correspondence between \([S^1, X]\) and the set of conjugacy classes in \(\pi_1(X)\), when \(X\) is path-connected.
Solution
1. Sliding the basepoint of a loop along a path changes its based class by conjugation and does not change its free homotopy class.
Let \(p:I\to X\) be a path from \(x\) to \(y\), and let \(f\) be a loop based at \(y\). The loop \[ p\cdot f\cdot\bar p \] is based at \(x\). There is a free homotopy from this loop to \(f\) obtained by moving the basepoint from \(x\) to \(y\) along \(p\) while shortening the initial and terminal copies of \(p\) by the same amount.
More explicitly, for \(t\in I\) let \[ p_t(s)=p\bigl(t+(1-t)s\bigr), \qquad 0\le s\le1, \] the terminal segment of \(p\) from \(p(t)\) to \(y\). Then the loop \[ L_t=p_t\cdot f\cdot\bar p_t \] is based at \(p(t)\). Using the standard three-piece parametrization of this concatenation gives a continuous map \[ S^1\times I\to X. \] At \(t=0\) it is \(p\cdot f\cdot\bar p\), while at \(t=1\) the two path pieces are constant and the loop is homotopic to \(f\) by reparametrization. Thus the two loops are freely homotopic.
2. If \(X\) is path connected, then \(\Phi\) is surjective.
Let \(f:S^1\to X\) be arbitrary and put \[ y=f(s_0). \] Choose a path \(p\) from \(x_0\) to \(y\). Regard \(f\) as a loop based at \(y\) and define \[ g=p\cdot f\cdot\bar p, \] a loop based at \(x_0\). By 1, \(g\) and \(f\) are freely homotopic. Hence \[ \Phi([g])=[f]\in[S^1,X]. \] Since \([f]\) was arbitrary, \(\Phi\) is onto.
3. If \([f]\) and \([g]\) are conjugate in \(\pi_1(X,x_0)\), then\[ \Phi([f])=\Phi([g]). \]
Suppose \[ [g]=[h]\,[f]\,[h]^{-1} \] for a loop \(h\) based at \(x_0\). Then \(g\) is based-homotopic to \[ h\cdot f\cdot\bar h. \] By 1, the latter loop is freely homotopic to \(f\). Therefore \(g\) and \(f\) define the same unbased homotopy class, so \[ \Phi([g])=\Phi([f]). \]
4. If\[ \Phi([f])=\Phi([g]), \]then \([f]\) and \([g]\) are conjugate in \(\pi_1(X,x_0)\).
The equality of their images means that there is a free homotopy \[ H:S^1\times I\to X \] from \(f\) to \(g\). Track the chosen point \(s_0\in S^1\) through the homotopy: \[ h(t)=H(s_0,t). \] Since both \(f\) and \(g\) are based at \(x_0\), the path \(h\) is a loop at \(x_0\).
Cutting the homotopy cylinder along \(\{s_0\}\times I\) produces a square whose boundary maps to \[ f\cdot h\cdot\bar g\cdot\bar h. \] This boundary loop is null-homotopic because it bounds the mapped square. Consequently \[ [f]=[h]\,[g]\,[h]^{-1}. \] Thus \([f]\) and \([g]\) are conjugate.
5. The fibers of \(\Phi\) are exactly the conjugacy classes in \(\pi_1(X,x_0)\).
The forward and reverse implications are 3 and 4.
6. If \(X\) is path connected, \(\Phi\) induces a bijection\[ \{\text{conjugacy classes in }\pi_1(X,x_0)\} \longrightarrow [S^1,X]. \]
By 5 the map is constant exactly on conjugacy classes, so it factors through the set of conjugacy classes and the induced map is injective. By 2 it is surjective.