Show that for a space \(X\), the following three conditions are equivalent:
Every map \(S^1 \to X\) is homotopic to a constant map, with image a point.
Every map \(S^1 \to X\) extends to a map \(D^2 \to X\).
\(\pi_1(X, x_0) = 0\) for all \(x_0 \in X\).
Deduce that a space \(X\) is simply-connected iff all maps \(S^1 \to X\) are homotopic. [In this problem, ‘homotopic’ means ‘homotopic without regard to basepoints’.]
Solution
1. If two loops \(f_0,f_1:S^1\to X\) are freely homotopic, then after choosing a point \(s_0\in S^1\) their based homotopy classes are conjugate by the path traced by the basepoint during the homotopy.
Let \[ H:S^1\times I\to X \] be a homotopy from \(f_0\) to \(f_1\), and define \[ h(t)=H(s_0,t). \] Thus \(h\) is a path from \(x_0=f_0(s_0)\) to \(x_1=f_1(s_0)\). The square obtained by cutting \(S^1\) at \(s_0\) has boundary path \[ f_0\cdot h\cdot\overline{f_1}\cdot\bar h, \] which is null-homotopic because it is the image of the boundary of the homotopy square under \(H\). Hence, in path-homotopy classes, \[ [f_0]=[h\cdot f_1\cdot\bar h]=\beta_h([f_1]). \] In particular, a based loop freely homotopic to a constant loop represents the identity element of its fundamental group.
2. Condition (a) implies condition (b).
Let \(f:S^1\to X\). By (a), there is a homotopy \[ H:S^1\times I\to X \] from \(f\) to a constant map with value \(p\in X\). Since \(H\) is constant on the whole top circle \(S^1\times\{1\}\), it factors through the quotient \[ (S^1\times I)/(S^1\times\{1\}), \] which is the cone on \(S^1\) and is homeomorphic to \(D^2\). The descended map \(D^2\to X\) restricts on the boundary circle \(S^1\times\{0\}\) to \(f\). Thus \(f\) extends over \(D^2\).
3. Condition (b) implies condition (a).
Let \(f:S^1\to X\) extend to \[ F:D^2\to X. \] The disk is contractible to its center by \[ r_t(z)=(1-t)z. \] Hence \[ H(z,t)=F(r_t(z)), \qquad z\in S^1, \] is a homotopy from \(f=F|_{S^1}\) to the constant map with value \(F(0)\). Thus (a) holds.
4. Condition (a) implies condition (c).
Fix \(x_0\in X\) and a loop \(f:S^1\to X\) based at \(x_0\). By (a), \(f\) is freely homotopic to a constant map. By 1, the based class \([f]\in\pi_1(X,x_0)\) is conjugate, via the basepoint track, to the class of a constant loop. The latter is the identity, and conjugating the identity again gives the identity. Thus every element of \(\pi_1(X,x_0)\) is trivial, so \[ \pi_1(X,x_0)=0. \] Since \(x_0\) was arbitrary, (c) holds.
5. Condition (c) implies condition (a).
Let \(f:S^1\to X\), choose \(s_0\in S^1\), and put \(x_0=f(s_0)\). By (c), \[ [f]=1\in\pi_1(X,x_0). \] Thus \(f\) is homotopic relative to the basepoint to the constant loop at \(x_0\). Forgetting the basepoint condition gives an ordinary homotopy to a constant map, which is (a).
6. Conditions (a), (b), and (c) are equivalent.
The equivalence of (a) and (b) is <1>2–<1>3, while <1>4–<1>5 give the equivalence of (a) and (c).
7. If \(X\) is simply connected, then every two maps \(S^1\to X\) are homotopic.
6.Thus every map \(S^1\to X\) is homotopic to a constant map.If \(x,y\in X\), choose a path \(h:I\to X\) from \(x\) to \(y\).Then\[ H(s,t)=h(t) \]is a homotopy from the constant map at \(x\) to the constant map at \(y\).Consequently any two maps \(S^1\to X\) are homotopic through their respective constant maps.
8. Conversely, if all maps \(S^1\to X\) are homotopic, then \(X\) is simply connected.
First, for any \(x,y\in X\), the constant maps \(c_x,c_y:S^1\to X\) are homotopic. If \(H:S^1\times I\to X\) is such a homotopy and \(s_0\in S^1\), then \[ t\longmapsto H(s_0,t) \] is a path from \(x\) to \(y\). Hence \(X\) is path connected.
Second, every map \(S^1\to X\) is homotopic to every constant map, so condition (a) holds. By 6, condition (c) holds as well. Thus \(X\) is path connected and has trivial fundamental group, hence is simply connected.