We define the degree of a continuous map \(h: S^1 \to S^1\) as follows.
Let \(b_0\) be the point \((1, 0)\) of \(S^1\); choose a generator \(\gamma\) for the infinite cyclic group \(\pi_1(S^1, b_0)\). If \(x_0\) is any point of \(S^1\), choose a path \(\alpha\) in \(S^1\) from \(b_0\) to \(x_0\), and define \(\gamma(x_0) = \hat{\alpha}(\gamma)\). Then \(\gamma(x_0)\) generates \(\pi_1(S^1, x_0)\). The element \(\gamma(x_0)\) is independent of the choice of the path \(\alpha\), since the fundamental group of \(S^1\) is abelian.
Now given \(h: S^1 \to S^1\), choose \(x_0 \in S^1\) and let \(h(x_0) = x_1\). Consider the homomorphism
\[ h_*: \pi_1(S^1, x_0) \to \pi_1(S^1, x_1). \]
Since both groups are infinite cyclic, we have
\[ h_*(\gamma(x_0)) = d \cdot \gamma(x_1) \]
for some integer \(d\), if the group is written additively. The integer \(d\) is called the degree of \(h\) and is denoted by \(\deg h\).
The degree of \(h\) is independent of the choice of the generator \(\gamma\); choosing the other generator would merely change the sign of both sides.
Show that \(d\) is independent of the choice of \(x_0\).
Show that if \(h, k: S^1 \to S^1\) are homotopic, they have the same degree.
Show that \(\deg(h \circ k) = (\deg h) \cdot (\deg k)\).
Compute the degrees of the constant map, the identity map, the reflection map \(\rho(x_1, x_2) = (x_1, -x_2)\), and the map \(h(z) = z^n\), where \(z\) is a complex number.
Show that if \(h, k: S^1 \to S^1\) have the same degree, they are homotopic.
Solution
Let \(\gamma(x)\) denote the preferred generator described in the exercise.
Let \(x_0,x_0'\in S^1\), and choose a path \(\alpha:x_0\to x_0'\). Put \(x_1=h(x_0)\), \(x_1'=h(x_0')\), and \(\beta=h\circ\alpha\). Naturality of change of basepoint gives \[ \widehat\beta\circ h_{*,x_0}=h_{*,x_0'}\circ\widehat\alpha. \] By the definition of the preferred generators, \[ \widehat\alpha(\gamma(x_0))=\gamma(x_0'),\qquad \widehat\beta(\gamma(x_1))=\gamma(x_1'). \] If \(h_{*,x_0}(\gamma(x_0))=d\gamma(x_1)\), applying the displayed naturality equation gives \[ h_{*,x_0'}(\gamma(x_0'))=d\gamma(x_1'). \] Thus \(d\) is independent of \(x_0\).
If \(h\simeq k\), choose a path \(\eta(t)=H(x_0,t)\) traced by the basepoint under a homotopy \(H\). The standard homotopy/basepoint-change theorem gives \[ \widehat\eta\circ h_*=k_*. \] Since \(\widehat\eta\) carries the preferred generator at \(h(x_0)\) to that at \(k(x_0)\), the integer multiplying the generator is unchanged. Hence \(\deg h=\deg k\).
If \(k(x_0)=x_1\) and \(h(x_1)=x_2\), then \[ k_*(\gamma(x_0))=(\deg k)\gamma(x_1), \] so \[ (h\circ k)_*(\gamma(x_0)) =h_*k_*(\gamma(x_0)) =(\deg k)(\deg h)\gamma(x_2). \] Thus \[ \deg(h\circ k)=(\deg h)(\deg k). \]
A constant map induces the zero homomorphism, so its degree is \(0\). The identity induces the identity homomorphism, so its degree is \(1\). The reflection \(\rho(e^{2\pi it})=e^{-2\pi it}\) reverses the standard generator, so \(\deg\rho=-1\). Finally, \(z\mapsto z^n\) lifts under \(t\mapsto e^{2\pi it}\) to \(t\mapsto nt\), so it carries the generator to \(n\) times itself and has degree \(n\).
Suppose \(\deg h=\deg k=d\). Choose lifts of \(h\) and \(k\) to the universal covering \(q:\mathbb R\to S^1\): \[ h(e^{2\pi it})=e^{2\pi iH(t)},\qquad k(e^{2\pi it})=e^{2\pi iK(t)}. \] The degree condition is exactly \[ H(t+1)=H(t)+d,\qquad K(t+1)=K(t)+d. \] Hence for \(0\le s\le1\), \[ L_s(t)=(1-s)H(t)+sK(t) \] also satisfies \(L_s(t+1)=L_s(t)+d\). Therefore \[ F(e^{2\pi it},s)=e^{2\pi iL_s(t)} \] is well defined and continuous on \(S^1\times I\), with \(F(-,0)=h\) and \(F(-,1)=k\). Thus \(h\simeq k\).