Exercise 58.9

Question

We define the \emph{	extbf{degree}} of a continuous map $h\colon S^1 	o S^1$ as follows:
\par 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 $\pi(S^1,x_0)$. The element $\gamma(x_0)$ is independent of the choice of the path $\alpha$, since the fundamental group of $S^1$ is abelian.
\par Now given $h\colon S^1 	o S^1$, choose $x_0 \in S^1$ and let $h(x_0) = x_1$. Consider the homomorphism
\begin{equation*}
  h_*\colon \pi(S^1,x_0) \longrightarrow \pi_1(S^1,x_1).
\end{equation*}
Since both groups are infinite cyclic, we have
\begin{equation}\label{3a}	ag{$\ast$}
  h_*(\gamma(x_0)) = d \cdot \gamma(x_1)
\end{equation}
for some integer $d$, if the group is written additively. The integer $d$ is called the \emph{degree} of $h$ and is denoted by $\deg h$.
\par 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 of \eqref{3a}.
\begin{enumerate}[label=(\alph*)]
\item Show that $d$ is independent of the choice of $x_0$.
\item Show that if $h,k\colon S^1 	o S^1$ are homotopic, they have the same degree.
\item Show that $\deg(h \circ k) = (\deg h) \cdot (\deg k)$.
\item Compute the degrees of the constant map, the identity map, the reflection map $ho(x_1,x_2) = (x_1,-x_2)$, and the map $h(z) = z^n$, where $z$ is a complex number.
\item Show that if $h,k\colon S^1 	o S^1$ have the same degree, they are homotopic.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$] $\DeclareMathOperator{\id}{id}$
  Let $x_0 
e y_0 \in S^1$, and let $h(y_0) = y_1$. Let $\alpha$ be a path from $x_0$ to $y_0$; then, $\beta' = h \circ \beta$ is a path from $x_1$ to $y_1$. Then, $\gamma(y_1) = [\overline{\beta}'] * \gamma(x_1) * [\beta']$, and so
  \begin{align*}
    h_*(\gamma(y_0)) &= h_*([\overline{\beta}] * \gamma(x_0) * [\beta]) = [\overline{\beta}'] * h_*(\gamma(x_0)) * [\beta']\
    &= [\overline{\beta}'] *d \cdot \gamma(x_1) * [\beta'] = d \cdot ([\overline{\beta}'] * \gamma(x_1) * [\beta']) = d \cdot \gamma(y_1).
  \end{align*}
\end{proof}
\begin{proof}[Proof of $(b)$]
  Choose $x_0 \in S^1$ and let $h(x_0) = y_0, k(x_0) = y_1$. By Lemma $58.4$, there exists a path $\alpha$ from $y_0$ to $y_1$ such that $k_* = \hat{\alpha} \circ h_*$. Then, we have
  \begin{align*}
    \deg k \cdot \gamma(y_0) = k_*(\gamma(x_0)) &= \hat{\alpha}(h^*(\gamma(x_0)))\
    &= \hat{\alpha}(\deg h\cdot\gamma(y_1)) = \deg h\cdot\hat{\alpha}(\gamma(y_1)) = \deg h\cdot\gamma(y_0),
  \end{align*}
  and so $\deg k = \deg h$.
\end{proof}
\begin{proof}[Proof of $(c)$]
  Choose $x_0 \in S^1$ and let $k(x_0) = y_0, h(y_0) = y_1$. Then,
  \begin{align*}
    \deg (h \circ k) \cdot \gamma(y_1) &= (h \circ k)_*(\gamma(x_0)) = (h_* \circ k_*)(\gamma(x_0))\
    &= h_*(\deg k\cdot\gamma(y_0)) = \deg k\cdot h_*(\gamma(y_0)) = (\deg h)\cdot(\deg k)\gamma(y_1),
  \end{align*}
  and so $\deg(h \circ k) = (\deg h)\cdot(\deg k)$.
\end{proof}
\begin{proof}[Solution for $(d)$]
  Let $k\colon S^1 	o S^1$ be the constant map. Then, we have $k_*(\gamma(x_0)) = 0$, and so $\deg k = 0$.
  \par Let $\id\colon S^1 	o S^1$ be the constant map. Then, we have $\id_*(\gamma(x_0)) = \gamma(x_0)$, and so $\deg \id = 1$.
  \par Let $h\colon S^1 	o S^1$ such that $z \mapsto z^n$. Let $p\colon \mathbb{R} 	o S^1$ be the standard covering. $p^{-1}(b_0) = 0$, which is fixed by $h$. Let $\gamma$ be the loop $\gamma(s) = e^{2\pi is}$, which generates $\pi_1(S_1,b_0)$; note we can choose $\gamma,b_0$ in this way by the fact that degree is independent of choice of generator and $x_0$. Then, $(h \circ \gamma)(s) = e^{2\pi ins} = p(ns)$, and so the lift of $h \circ \gamma$ is $s \mapsto ns$, i.e., $h_*(\gamma(b_0)) = n\gamma(b_0)$. Thus, $\deg h = n$.
  \par The case for $ho$ follows from that of $h$, where $n = -1$.
\end{proof}
\begin{proof}[Proof of $(e)$]
  By $(a)$, we use $b_0$ as our base point, and let $h(b_0) = x_0, k(b_0) =
  y_0$. Consider $h_*\colon \pi_1(S^1,b_0) 	o \pi_1(S^1,x_0)$, $k_*\colon \pi_1(S^1,b_0) 	o \pi_1(S^1,y_0)$; by assumption, there exists $n$, such that $h_*(\gamma(b_0)) = n \cdot \gamma(x_0), k_*(\gamma(b_0)) = n \cdot \gamma(y_0)$.
  \par Now let $\gamma(s) = e^{2\pi is}$, which generates $\pi_1(S_1,b_0)$ as
  above. Then, let $\widetilde{h \circ \gamma},\widetilde{k \circ \gamma}$ be
  the lifts of $h \circ \gamma, k \circ \gamma$ through the standard covering
  $p\colon \mathbb{R} 	o S^1$, at $h_0 \in p^{-1}(x_0),k_0 \in p^{-1}(y_0)$ respectively. Let $h_1 = \widetilde{h \circ \gamma}(1), k_1 = \widetilde{k \circ \gamma}(1)$.
  \par Now we have $h_*(\gamma(b_0)) = (h_1-h_0) \gamma(x_0), k_*(\gamma(b_0)) = (k_1-k_0) \gamma(y_0)$, and so $h_1-h_0 = k_1-k_0 = n$. We want to find a path homotopy from $\widetilde{h \circ \gamma}$ to $\widetilde{k \circ \gamma}$. We first define $\widetilde{\widetilde{h \circ \gamma}} = \widetilde{h \circ \gamma} - h_0 + k_0$ so that both $\widetilde{\widetilde{h \circ \gamma}}, \widetilde{h \circ \gamma}$ have the same end points; we then see there exists a path homotopy $	ilde{F}$ between them since $\mathbb{R}$ is contractible.
  \par Now consider $	ilde{h}(x) = e^{2\pi i(k_0-h_0)}h(x)$. Then, $\widetilde{\widetilde{h \circ \gamma}}$ is the lift of $	ilde{h} \circ \gamma$ at $k_0$. Then, $p \circ 	ilde{F}$ is a path homotopy from $\widetilde{\widetilde{h \circ \gamma}}$ to $\widetilde{k \circ \gamma}$. Since this is a path homotopy, it factors through $(p,\id)$ to get a homotopy $F$ from $	ilde{h}$ to $k$.
  \par We now see that there exists a path homotopy $G$ between $	ilde{h}$ and $h$ given by $G(x,t) = h(x)e^{2\pi is(k_0-h_0)}$, and so there exists a path homotopy combining $F,G$ between $h$ and $k$.
\end{proof}

Exercise 58.9

Answers

Comments

Exercise 58.9

Answers

Comments

Add answer