Recall that \(S^n\) is defined to be \(\{(x_1,x_2,\ldots,x_n,x_{n+1})\in\mathbb R^{n+1}\mid x_1^2+x_2^2+\cdots+x_n^2+x_{n+1}^2=1\}\). We shall be interested in the values \(n=1,2\). Then define \(f,g:S^1\to S^2\) by \(f(\cos(s),\sin(s))=(\cos(s),\sin(s),0)\) and \(g(\cos(s),\sin(s))=(\cos(s),-\sin(s),0)\). Show that \(f\) is homotopic to \(g\).
Solution
Define \[ H:S^1\times I\longrightarrow S^2 \] by \[ H((x,y),t) =\bigl(x,\,y\cos(\pi t),\,y\sin(\pi t)\bigr). \]
1. The formula for \(H\) takes values in \(S^2\).
If \((x,y)\in S^1\), then \[ x^2+y^2=1. \] Therefore \[ \begin{aligned} \|H((x,y),t)\|^2 &=x^2+y^2\cos^2(\pi t)+y^2\sin^2(\pi t)\\ &=x^2+y^2\bigl(\cos^2(\pi t)+\sin^2(\pi t)\bigr)\\ &=x^2+y^2\\ &=1. \end{aligned} \] Hence \(H((x,y),t)\in S^2\) for every \(((x,y),t)\in S^1\times I\).
2. The map \(H:S^1\times I\to S^2\) is continuous.
The coordinate functions \[ (x,y,t)\longmapsto x, \qquad (x,y,t)\longmapsto y\cos(\pi t), \qquad (x,y,t)\longmapsto y\sin(\pi t) \] are continuous on \(S^1\times I\) as restrictions and products of continuous real-valued functions. Thus the corresponding map into \(\mathbb R^3\) is continuous. By 1 its image lies in the subspace \(S^2\), so \(H\) is continuous as a map to \(S^2\).
3. At \(t=0\), the map \(H\) equals \(f\).
For \((x,y)\in S^1\), \[ H((x,y),0) =\bigl(x,y\cos0,y\sin0\bigr) =(x,y,0). \] Writing \((x,y)=(\cos s,\sin s)\) gives \[ H((\cos s,\sin s),0) =(\cos s,\sin s,0) =f(\cos s,\sin s). \]
4. At \(t=1\), the map \(H\) equals \(g\).
For \((x,y)\in S^1\), \[ H((x,y),1) =\bigl(x,y\cos\pi,y\sin\pi\bigr) =(x,-y,0). \] Thus \[ H((\cos s,\sin s),1) =(\cos s,-\sin s,0) =g(\cos s,\sin s). \]
5. Therefore \(f\simeq g\).
By 2, \(H\) is continuous, and <1>3–<1>4 give \[ H(-,0)=f, \qquad H(-,1)=g. \] Hence \(H\) is a homotopy from \(f\) to \(g\).