Exercise 58.10

Question

Suppose that to every map $h\colon S^n 	o S^n$ we have assigned an integer,
denoted by $\deg h$ and called the \emph{	extbf{degree}} of $h$, such that
\begin{enumerate}[label=(oman*)]
\item Homotopic maps have the same degree.
\item $\deg(h \circ k) = (\deg h) \cdot (\deg k)$.
\item The identity has degree $1$, any constant map has degree $0$, and the reflection map $ho(x_1,\ldots,x_{n+1}) = (x_1,\ldots,x_n,-x_{n+1})$ has degree $-1$.
\end{enumerate}
Prove the following:
\begin{enumerate}[label=(\alph*)]
\item There is no retraction $r\colon B^{n+1} 	o S^n$.
\item If $h\colon S^n 	o S^n$ has degree different from $(-1)^{n+1}$, then $h$ has a fixed point.
\item If $h\colon S^n 	o S^n$ has degree different from $1$, then $h$ maps some point $x$ to is antipode $-x$.
\item If $S^n$ has a nonvanishing tangent vector field $v$, then $n$ is odd.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  We use polar coordinates to denote $B^{n+1}$ and $S^n$, i.e., the points in $B^{n+1}$ are given by $(	heta,\phi_2,\ldots,\phi_n,ho)$ and the points in $S^n$ by $(	heta,\phi_2,\ldots,\phi_n)$, where $0 \le 	heta < 2\pi$, $0 \le \phi_i < \pi$, and $0 \le ho \le 1$. Thus, the inclusion $S^n \hookrightarrow B^{n+1}$ is given by $(	heta,\phi_2,\ldots,\phi_n) \mapsto (	heta,\phi_2,\ldots,\phi_n,1)$.
  \par Now, suppose a retraction $r\colon B^{n+1} 	o S^n$ exists. Then, define the
  homotopy $H\colon S^n 	imes [0,1] 	o S^n$ by $H(x_1,\ldots,x_n,t) = r(x_1,\ldots,x_n,t)$, which is continuous since $r$ is. Then, $H(x_1,\ldots,x_n,0)$ is constant, hence has degree $0$, whereas $H(x_1,\ldots,x_n,1)$ is the identity, hence has degree $1$, which contradicts $(i)$.
\end{proof}
\begin{proof}[Proof of $(b)$]
  We return to Cartesian coordinates. Suppose $h$ has no fixed points. We first construct a homotopy:
  \begin{equation*}
    H(x,t) = \frac{(1-t)h(x)+ta(x)}{\lVert(1-t)h(x)+ta(x)Vert}.
  \end{equation*}
  Since $H$ is continuous since it is a composition of continuous functions in both variables, it suffices to show $\lVert(1-t)h(x)+ta(x)Vert 
e 0$ for all $x,t$. So, suppose $(1-t)h(x)+ta(x) = 0$. Then, $(1-t)h(x) = tx$. Comparing norms, this is only possible when $t=1/2$, and so $h(x) = x$, which contradicts that $h$ has no fixed points.
  \par Thus, we see that $h$ is homotopic to the antipodal map. Now, since the antipodal map is the composition of $n+1$ reflections, each one reflecting each coordinate of $S^n$, we have that $\deg(h) = \deg(a) = (-1)^{n+1}$ by $(i)$, $(ii)$, and $(iii)$, a contradiction.
\end{proof}
\begin{proof}[Proof of $(c)$]
  Suppose $h$ maps no point $x$ to its antipode $-x$. We first construct a homotopy:
  \begin{equation*}
    G(x,t) = \frac{(1-t)h(x)+tx}{\lVert(1-t)h(x)+txVert}.
  \end{equation*}
  Since $G$ is continuous since it is a composition of continuous functions in both variables, it suffices to show $\lVert(1-t)h(x)+txVert 
e 0$ for all $x,t$. So, suppose $(1-t)h(x)+tx = 0$. Then, $(1-t)h(x) = -tx$. Comparing norms, this is only possible when $t=1/2$, and so $h(x) = -x$, which contradicts that $h$ maps no point $x$ to its antipode $-x$.
  \par Thus, we see that $h$ is homotopic to the identity. Since the identity has degree $1$, we have $\deg(h) = 1$ by $(i)$, a contradiction.
  %Suppose $h$ does not map any point $x$ to its antipode $-x$. Then, $g = -h$ is a map with no fixed points, and so $g \simeq a$ by the proof of $(b)$. But then, $\deg(h) = \deg(-g) = \deg(-a) = \deg(i) = 1$ by $(i)$ and $(iii)$, a contradiction.
\end{proof}
\begin{proof}[Proof of $(d)$]
  Suppose such a $v(x)$ exists. Then, let $h(x) = v(x)/\lVert v(x) Vert$; $h(x)$ then is a map $S^n 	o S^n$. Since $v(x)$ is a tangent vector field, $\braket{h(x),x} = \braket{v(x),x}/\lVert v(x) Vert = 0$, where $\braket{\cdot,\cdot}$ is the standard inner product in $\mathbb{R}^{n+1}$. Then, there are no points $h(x) = \pm x$ since $\braket{h(x),x} = 0 
e \pm1$. By the proofs of $(b)$ and $(c)$, $h(x)$ is then homotopic to the identity and the antipodal map; hence, the identity is homotopic to the antipodal map. By $(i)$, $(iii)$, and the proofs of $(b)$ and $(c)$, this then implies that $\deg 1 = \deg (-1)^{n+1}$, and so $n$ is odd.
\end{proof}

Exercise 58.10

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Exercise 58.10

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