Define \(f: S^1 \times I \longrightarrow S^1 \times I\) by \(f(\theta, s) = (\theta + 2\pi s, s)\), so \(f\) restricts to the identity on the two boundary circles of \(S^1 \times I\). Show that \(f\) is homotopic to the identity by a homotopy \(f_t\) that is stationary on one of the boundary circles, but not by any homotopy \(f_t\) that is stationary on both boundary circles. [Consider what \(f\) does to the path \(s \mapsto (\theta_0, s)\) for fixed \(\theta_0 \in S^1\).]
Solution
Write points of \(S^1\) by angles modulo \(2\pi\).
1. The maps\[ f_t(\theta,s)=(\theta+2\pi ts,s), \qquad 0\le t\le1, \]give a homotopy from the identity to \(f\) that is stationary on the boundary circle \(S^1\times\{0\}\).
At \(t=0\), \[ f_0(\theta,s)=(\theta,s), \] while at \(t=1\) one obtains the given map \(f\). For every \(t\) and every \(\theta\), \[ f_t(\theta,0)=(\theta,0), \] so the lower boundary circle is fixed pointwise throughout the homotopy.
2. Fix \(\theta_0\in S^1\) and let\[ \alpha(s)=(\theta_0,s), \qquad 0\le s\le1. \]Then\[ (f\circ\alpha)(s)=(\theta_0+2\pi s,s). \]
This is immediate from the definition of \(f\). Both paths run from \((\theta_0,0)\) to \((\theta_0,1)\) because adding \(2\pi\) does not change the point of \(S^1\).
3. The paths \(\alpha\) and \(f\circ\alpha\) are not homotopic relative to their endpoints in \(S^1\times I\).
Let \[ r:S^1\times I\to S^1, \qquad r(\theta,s)=\theta, \] be projection to the circle. Then \(r\circ\alpha\) is the constant loop at \(\theta_0\), whereas \[ r\circ f\circ\alpha(s)=\theta_0+2\pi s \] traverses \(S^1\) once.
If \(\alpha\) and \(f\circ\alpha\) were homotopic relative to endpoints, composing such a homotopy with \(r\) would give a based homotopy between these two loops in \(S^1\). But the first loop represents \(0\) and the second represents \(1\) in \[ \pi_1(S^1,\theta_0)\cong\mathbb Z. \] They therefore cannot be based-homotopic.
4. There is no homotopy from the identity to \(f\) that is stationary on both boundary circles.
Suppose \[ F_t:S^1\times I\to S^1\times I \] were such a homotopy, with \(F_0=\operatorname{id}\) and \(F_1=f\), and with both boundary circles fixed pointwise for every \(t\). Define \[ H(s,t)=F_t(\alpha(s)). \] Then \[ H(s,0)=\alpha(s), \qquad H(s,1)=f\circ\alpha(s). \] Because \(\alpha(0)\) and \(\alpha(1)\) lie on the two boundary circles and those circles are stationary, \[ H(0,t)=\alpha(0), \qquad H(1,t)=\alpha(1) \] for all \(t\). Thus \(H\) would be a homotopy relative to endpoints between \(\alpha\) and \(f\circ\alpha\), contradicting 3.
5. Hence \(f\) is homotopic to the identity relative to either chosen single boundary circle, but not relative to both boundary circles simultaneously.
1.For the upper boundary circle one can instead use\[ g_t(\theta,s)=(\theta+2\pi t(s-1),s). \]At \(t=0\) this is the identity, at \(t=1\) it agrees with \(f\) because angles differing by \(2\pi\) define the same point of \(S^1\), and at \(s=1\) it fixes \((\theta,1)\) for every \(t\).The impossibility of fixing both is
4.