Degree of Continuous Maps on S^n (9.58.9-10)

James Munkres Topology, chapter 9.58, exercise 9-10:

9. Let $h : S^1→S^1$ be a continuous map. Let $b_0=(1,0)$, and let $γ$ generate $π_1(S^1,b_0)$. For any given point $x∈S^1$, define $γ(x)=\hat{α}(γ)$ where $α$ is a path from $b_0$ to $x$; note that the choice of $α$ is immaterial since $π_1(S^1,b_0)$ is abelian. Since $\hat{α}$ is an isomorphism, $γ(x)$ will generate $π_1(S^1,x)$. As well, we see $h_*$ is a homomorphism between $π_1(S^1,x)$ and $π_1(S^1,h(x))$, so $$h_*(γ(x))=d_x·γ(h(x))$$ for some integer $d_x$, if the fundamental groups are considered additively. This $d_x$ is called the degree of $h$ (relative to $x$), and is independent of choice of $γ$ since choosing the other generator of $π_1(S^1,b_0)$ will change signs accordingly to result in the same $d_x$.

(a) Show that $d_x$ is independent of the choice of $x∈S^1$, so we may denote it more generally by $d$.
(b) Show that if $h, k : S^1→S^1$ are homotopic, then their degrees are the same.
(c) Show that $\text{deg}(h∘k)=(\text{deg }h)·(\text{deg }k)$.
(d) Compute the degrees of the constant map, the identity map, the reflection map $(x,y)↦(x,-y)$, and the map $z↦z^n$ when $S^1$ is considered as a subgroup of $ℂ$.
(e) Show that if $h,k : S^1→S^1$ have the same degree, they are homotopic.

10. Suppose that to every map $h : S^n→S^n$ we have assigned an integer, denoted by $\text{deg }h$ and called the degree of $h$, such that:

1. Homotopic maps have the same degree
2. $\text{deg }(h∘k) = (\text{deg }h)·(\text{deg }k)$
3. The identity map has degree $1$, any constant map degree $0$, and the reflection maps $(x_1,...,x_i,...,x_{n+1})↦(x_1,...,-x_i,...,x_{n+1})$ degree $-1$.

Prove the following:
(a) There is no retraction $r : B^{n+1}→S^n$
(b) If $h : S^n→S^n$ has degree different than $(-1)^{n+1}$, then $h$ has a fixed point.
(c) If $h : S^n→S^n$ has degree different than $(-1)^n$, then $h(x)=-x$ for some $x$.
(d) If $S^n$ has a nonvanishing tangent vector field $v$, then $n$ is odd.

Proof: Notation shall be mildly abused in the explanations that follow, in that $γ$ will refer to an element of $π_1(S^1,b_0)$ when outside brackets, and a loop about $b_0$ whose homotopy class is such $γ$ when inside brackets.

9. (a) We shall show $d_x=d_{b_0}$ for all $x∈S^1$. We have $$h_*(γ)=d_{b_0}·\hat{α}(γ)$$ where $α$ is a path from $b_0$ to $h(b_0)$. Let $β$ be a path from $b_0$ to $x$, and let $δ$ be a path from $b_0$ to $h(x)$, and observe $$h_*(γ(x))=h_*(\hat{β}(γ))=[h∘\overline{β}]*[h∘γ]*[h∘β]=\widehat{h∘β}(h_*(γ))=$$ $$\widehat{h∘β}(d_{b_0}·\hat{α}(γ))=d_{b_0}·\widehat{h∘β}∘\hat{α}(γ)=d_{b_0}·\widehat{α*(h∘β)}(γ)=$$ $$d_{b_0}·\hat{δ}(γ)=d_{b_0}·γ(h(x))$$ (b) Choose $x∈S^1$. Since $h$ and $k$ are homotopic, let $α$ be a path from $h(x)$ to $k(x)$ such that $k_*=\hat{α}∘h_*$. Then if $d$ is the degree of $h$ and $β$ is a path between $b_0$ and $h(x)$, we have $$k_*(γ(x))=\hat{α}∘h_*(γ(x))=d·\hat{α}(γ(h(x))=$$ $$d·\hat{α}∘\hat{β}(γ)=d·\widehat{β*α}(γ)=d·γ(k(x))$$ (c) We simply observe $$(h∘k)_*(γ(x))=h_*∘k_*(γ(x))=(\text{deg }k)·h_*(γ(k(x))=$$ $$[(\text{deg }k)·(\text{deg }h)]·γ((h∘k)(x))$$ (d) The constant map induces trivial homomorphisms, so $d=0$ in this case. The identity map induces identity homomorphisms, so $d=1$ in this case. Let $f$ be the reflection map; when $p : ℝ→S^1$ is the standard covering map $x↦(\text{cos }2πx,\text{sin }2πx)$ and $\tilde{γ} : I→ℝ$ is given by $x↦x$, we see $γ=p∘\tilde{γ}$ generates $π_1(S^1,b_0)$. As well, $p(-\tilde{γ}(x))=(\text{cos }(-2πx),\text{sin}(-2πx))=(\text{cos }2πx,-\text{sin}(2πx))=f∘γ(x)$ so $\widetilde{f∘y}=-\tilde{γ}$. Since $-\tilde{γ}$ is a path from $0$ to $-1$ in $ℝ$ we may observe $f_*(γ)=[f∘γ]=-γ$ so that $d=-1$. A similar appeal to covering maps beyond when $n=-1$ shows that generally $z↦z^n$ is degree $n$.

(e) Lemma: If $G : I×I→S^1$ is a homotopy between $h∘γ$ and $k∘γ$, then $h, k$ are homotopic. Proof: Consider $S^1$ as the circle group in $ℂ$. We see $p_1(t)=G(0,t)$ and $p_2(t)=G(1,t)$ are paths from $h(b_0)$ to $k(b_0)$, and let $q(x,t)=(\dfrac{p_1(t)}{p_2(t)})^x$. First we show $J: I×I→S^1$ defined by $J = G·q$ is another homotopy, but such that $J(0,t)=J(1,t)$ for all $t∈I$. As such, it is clear $$J(x,0)=G(x,0)·1^x=h∘γ(x)$$ and $$J(x,1)=G(x,1)·1^x=k∘γ(x)$$ and also $J(0,t)=G(0,t)·1=p_1(t)=p_2(t)·\dfrac{p_1(t)}{p_2(t)}=J(1,t)$.

Thus, since $γ$ is also a quotient map (if $γ$ weren't surjective, then $γ$ would map into $S^1-p≅ℝ$ for some $p∈S^1$ and thus be nulhomotopic), we may factor $J$ through $S^1×I$ via the quotient map $γ×i$ to obtain $K$. For $x∈S^1$, let $γ(v)=x$, and we see $$K(x,0)=K∘(γ×i)(v,0)=J(v,0)=h∘γ(v)=h(x)$$ $$K(x,1)=K∘(γ×i)(v,1)=J(v,1)=k∘γ(v)=k(x)$$ so that $K$ is a homotopy between $h$ and $k$.$~\square$

Let $α$ be a path from $b_0$ to $h(b_0)$, and let $β$ be a path from $h(b_0)$ to $k(b_0)$. Then $$\hat{β}∘h_*(γ)=\hat{β}(d·\hat{α}(γ))=d·\widehat{α*β}(γ)=k_*(γ)$$ so that $\hat{β}∘h_*=k_*$. As such, let $F : I×I→S^1$ be a path homotopy between $\overline{β}*(h∘γ)*β$ and $k∘γ$ (however, by path homotopy equivalence, let this former piecewise function be split such that $(\overline{β}*(h∘γ)*β)(1/3+t/3)=h∘γ(t)$). Also, let $H : I×I→I$ be a homotopy between the functions $f,g : I→I$ given by $f(t)=t$ and $g(t)=1/3+t/3$, specifically a homotopy from $g$ to $f$. Define $K : I×I→S^1$ by $K(x,y)=F(H(x,y),y)$. We claim $K$ is a homotopy between $h∘γ$ and $k∘γ$, so that the statement follows by the proceeding lemma. To wit, observe $$K(x,0)=F(H(x,0),0)=(\overline{β}*(h∘γ)*β)(1/3+x/3)=h∘γ(x)$$ $$K(x,1)=F(H(x,1),1)=k∘γ(x)$$ 10. (a) If $j : S^n→B^{n+1}$ is the inclusion map, then we see if $r$ were a retraction, then since $B^{n+1}$ is convex that $i_{S^n} \simeq r∘j \simeq 0∘j$, and so the identity map is homotopic to the constant map on $S^n$, contradicting (1) and (3).

(b) Suppose $h$ does not have a fixed point; then extend $h$ to $ℝ^{n+1}-0$ via $H(x)=h(\dfrac{x}{||x||})$. Now define a homotopy $G$ from $H$ to the circular antipodal map $λ(x)=-\dfrac{x}{||x||}$ on $ℝ^{n+1}-0$ by $G(x,t)=(1-t)·H(x)+t·λ(x)$. Note that by construction $G(x,t) \not = 0$ for any $x,t$ so the range is indeed within $ℝ^{n+1}-0$. Now when $j : S^n→ℝ^{n+1}-0$ is inclusion and $r : ℝ^{n+1}-0→S^n$ the standard retraction, we see $F : S^n×I→S^n$ given by $F(x,t)=r(G(j(x),t))$ is a homotopy between $H_{|S^n}=h$ and the antipodal map $x↦-x$ on $S^n$, and since this latter is seen to have degree $(-1)^{n+1}$ by (2) and (3), it follows by (1) that $h$ has the same degree.

(c) This is merely part (b) applied to the composite of the antipodal map with $h$.

(d) If a nonvanishing vector $v(x)$ is tangent to a point $x$ of $S^n$, then particularly $t·v(x) ≠ x$ for $t∈ℝ$. Now, when $j : S^n→ℝ^{n+1}-0$ is the inclusion map, consider the straight-line homotopy between $j$ and $v$ on $ℝ^{n+1}-0$. This homotopy is well defined by the nonvanishing and non-orthogonality of the vector field. As well, the straight-line homotopy between $v$ and $-j$ is also defined. Therefore $j$ is homotopic to $-j$ in $ℝ^{n+1}-0$, and since $S^n$ is a retract of this latter, we find the identity map is homotopic to the antipodal map in $S^n$. Since the antipodal map has degree $-1$ when $n$ is even, it follows $n$ must be odd.$~\square$