Let \(f\) be the map of \(S^1 \times [0, 1]\) to itself defined by \[ f(e^{i\theta},s)=(e^{i(\theta+2\pi s)},s), \] so that \(f\) restricts to the identity on the two boundary circles of \(S^1 \times [0, 1]\).
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 that is stationary on both boundary circles.
Hint: Consider what \(f\) does to the path \(s \mapsto (e^{i\theta_0} , s)\) for fixed \(e^{i\theta_0} \in S^1\).
Solution
Let \[ A=S^1\times[0,1]. \]
1. The map \(f\) is homotopic to the identity through a homotopy stationary on the boundary circle \(S^1\times\{0\}\).
Define \[ H:A\times[0,1]\longrightarrow A \] by \[ H_t(e^{i\theta},s) = \left(e^{i(\theta+2\pi(1-t)s)},s\right). \] At \(t=0\) this is \(f\), while at \(t=1\) it is the identity.
If \(s=0\), then for every \(t\), \[ H_t(e^{i\theta},0) = (e^{i\theta},0). \] Thus the homotopy is stationary on \(S^1\times\{0\}\).
2. Let\[ \alpha:[0,1]\longrightarrow A, \qquad \alpha(s)=(1,s). \]Then \(\alpha\) and \(f\circ\alpha\) are not homotopic relative to their endpoints.
Both paths run from \[ p=(1,0) \qquad\text{to}\qquad q=(1,1). \] Let \[ \operatorname{pr}_1:A\longrightarrow S^1 \] be projection onto the first factor. Then \[ (\operatorname{pr}_1\circ\alpha)(s)=1, \] so \(\operatorname{pr}_1\circ\alpha\) is the constant loop at \(1\), whereas \[ (\operatorname{pr}_1\circ f\circ\alpha)(s) = e^{2\pi i s}, \] which traverses \(S^1\) once.
If \(\alpha\) and \(f\circ\alpha\) were homotopic relative to endpoints in \(A\), composing such a homotopy with \(\operatorname{pr}_1\) would give a based homotopy in \(S^1\) from the constant loop to the once-around loop. But these represent respectively \[ 0 \qquad\text{and}\qquad 1 \] in \[ \pi_1(S^1,1)\cong\ZZ. \] Hence no such endpoint-fixed homotopy exists.
3. There is no homotopy from \(f\) to the identity that is stationary on both boundary circles.
Suppose, for contradiction, that \[ K:A\times[0,1]\longrightarrow A \] is a homotopy from \(f\) to \(\operatorname{id}_A\) such that \[ K_t(x)=x \] for every \[ x\in S^1\times\{0,1\} \] and every \(t\in[0,1]\).
2. The map\[ F:[0,1]\times[0,1]\longrightarrow A, \qquad F(s,t)=K_t(\alpha(s)), \]is a homotopy from\[ F(s,0)=f(\alpha(s)) \]to\[ F(s,1)=\alpha(s). \]Because \(\alpha(0)=p\) and \(\alpha(1)=q\) lie on the two boundary circles and \(K\) is stationary there,\[ F(0,t)=p, \qquad F(1,t)=q \]for every \(t\).Thus \(F\) is a homotopy relative to endpoints from \(f\circ\alpha\) to \(\alpha\), contradicting
2.
Therefore no homotopy from \(f\) to the identity can be stationary on both boundary circles.
Hence \(f\) is homotopic to the identity relative to one boundary circle, but not relative to the full boundary \(S^1\times\{0,1\}\).