The Hopf Degree Theorem — Examples

By [F1] the degree of $p_k$ is the integer $k$; the explicit lift of [F3] has period increment $k$, so this increment agrees with the degree already computed by [F1].

Let $k,l\in \mathbb{Z}$. If $k=l$ then $p_k=p_l$; conversely if $p_k$ and $p_l$ are homotopic, [F2] with $m=1$ gives $k=\mathrm{deg}(p_k)=\mathrm{deg}(p_l)=l$, so the two maps are homotopic exactly when their exponents agree.

Hence the assignment $k\mapsto [p_k]$ is a bijection from $\mathbb{Z}$ to $[S^1,S^1]$, inverse to the degree, and a power map is determined up to homotopy by its exponent alone; this matches the covering-space description.

The reflection $R$ is the restriction of the invertible linear map of ${\mathbb{R}}^{m+1}$ with determinant $-1$ that preserves the unit sphere, hence restricts to a smooth diffeomorphism of $S^m$; the linear reflection reverses the ambient orientation and the outward-normal-first orientation of $S^m$ is transported from the ambient orientation, so $R$ reverses the orientation of $S^m$, and $\mathrm{deg}(R)=-1$ both by the direct reflection computation of [F1] and by the diffeomorphism criterion of [F2].

Let $f:S^m\to S^m$ have degree $-1$. By [F3] applied to $f$ and $R$, whose degrees are equal, $f$ is homotopic to $R$; conversely every map homotopic to $R$ has degree $-1$ by [F3]. By [F4] every self-map of degree $-1$ is a homotopy equivalence, and by [F3] its homotopy class is that of the reflection, so a map of degree $-1$ is a homotopy equivalence precisely when it lies in the homotopy class of a reflection.

This identifies the degree-$-1$ class, represented by a reflection, and shows the consistency of the reflection sign with the general classification; no new invariant or orientation convention is introduced beyond the outward-normal-first orientation of the sphere.

The realization lemma supplies exactly the objects described: the model $F$ with $F^{-1}(y_{-})=\{0\}$, $d{F}_0$ invertible and $F=N$ off the unit ball, and the glued map $f_k$ which equals $F$ read through the $i$-th chart near the centre $p_i$ and $N$ elsewhere; its construction and its smoothness are those verified there.

The preimage of $y_{-}$ under $f_k$ is exactly $\{p_1,\dots ,p_{\mathrm{\mid }k\mathrm{\mid }}\}$: each centre gives a preimage by $F(0)=y_{-}$, the model has no other preimage of $y_{-}$ inside the unit ball, and points outside the balls as well as points of a ball mapping outside the unit ball have value $N\mathrm{\ne }{y}_{-}$; the differential $d{f}_{k,p_i}=d{F}_0\circ d{\phi }_i$ is invertible, so $y_{-}$ is a regular value and $f_k$ is proper because $M$ is compact.

By step 1.2 and [F2], $\mathrm{deg}(f_k)=\mathrm{sgn}(d{F}_0){\sum }_{i=1}^{\mathrm{\mid }k\mathrm{\mid }}{\epsilon }_i$ with ${\epsilon }_i=\pm 1$ the orientation sign of the $i$-th chart, by [F3]; choosing every chart so that $\mathrm{sgn}(d{F}_0){\epsilon }_i=\mathrm{sgn}(k)$ makes the sum equal to $k$. For $k=0$ there is no ball and $f_0=N$ is constant, with an empty regular fibre over any point different from $N$, so $\mathrm{deg}(f_0)=0$. This realizes every prescribed degree by collapsing oriented disks with the prescribed local signs.

By [F1], and because $n$ is even, ${\mathbb{R}\mathbb{P}}^n$ is a compact connected nonorientable smooth $n$-manifold with the affine-chart structure. Parametrize its top-cell attachment by $\pi (x)=[x:\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}^2}]$ for $x\in D^n$. On the interior, the last-coordinate affine chart gives $u=x\mathrm{/}\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}^2}$, with smooth inverse $x=u\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, so $\pi$ restricts to a diffeomorphism onto the open top cell. On the boundary $\pi (x)=[x:0]$ is the antipodal attachment onto ${\mathbb{R}\mathbb{P}}^{n-1}$. In particular ${\mathbb{R}\mathbb{P}}^n$ is nonempty and has no boundary.

The continuous collapse map $q:=\bar{c}\circ Q$ is well defined and continuous, where $Q:{\mathbb{R}\mathbb{P}}^n\to {\mathbb{R}\mathbb{P}}^n\mathrm{/}{\mathbb{R}\mathbb{P}}^{n-1}$ is the quotient map and $\bar{c}$ is the homeomorphism ${\mathbb{R}\mathbb{P}}^n\mathrm{/}{\mathbb{R}\mathbb{P}}^{n-1}\to S^n$ induced by $c$ through [F2]; equivalently $q\circ \pi =c$ on $D^n$, and $\bar{c}$ is the homeomorphism of [F2], so the identification of the quotient with $S^n$ is exactly the one exhibited by the classical pinch.

The model $F$ is constant $N$ for $\mathrm{\mid }x{\mathrm{\mid }}^2\ge 3\mathrm{/}4$. Thus $\tilde{q}(\pi (x))=F(x)$ is well defined: only boundary points of $D^n$ are identified, and $F$ has the same value on them. It is smooth on the open cell because $\pi$ is a local diffeomorphism there; near the lower skeleton it is constant, since the closed smaller disk $\mathrm{\mid }x{\mathrm{\mid }}^2\le 3\mathrm{/}4$ has image disjoint from that skeleton. To exhibit a homotopy from $c$ to $F$ relative to $\mathrm{\partial }{D}^n$, write $s=\mathrm{\mid }x{\mathrm{\mid }}^2$, let $W(s)=1-\chi (s{)}^2(1-s)$ be the profile of [F3], and set $W_{\tau }(s)=(1-\tau )s+\tau W(s)$. For $x\mathrm{\ne }0$ define $C_{\tau }(x)=(2\sqrt{W_{\tau }(s)(1-W_{\tau }(s))\mathrm{/}s}x,2W_{\tau }(s)-1)$, and set $C_{\tau }(0)=y_{-}$. This is continuous jointly in $x,\tau$: near $s=0$ one has $W_{\tau }(s)=s$, and elsewhere $s>0$ the displayed square root is continuous and nonnegative. Its norm is one, $C_0=c$, $C_1=F$ because $\chi \ge 0$, and for $s=1$ it is always $N$. The map $\pi \times {\mathrm{i}\mathrm{d}}_I:D^n\times I\to {\mathbb{R}\mathbb{P}}^n\times I$ is a quotient map by [F2], since its source is compact and its target Hausdorff. Thus this family, constant on its fibres, descends continuously to a homotopy $q\simeq \tilde{q}$. No radial diffeomorphism or homeomorphism assertion about $F$ is needed.

The point $y_{-}=F(0)$ is a regular value of $\tilde{q}$ with the single preimage $\pi (0)$: $\tilde{q}(\pi (x))=F(x)=y_{-}$ forces $x=0$ by [F3], the differential satisfies $d{\tilde{q}}_{\pi (0)}=d{F}_0\circ (d{\pi }_0{)}^{-1}$ by step 1.1, hence is invertible, and points of ${\mathbb{R}\mathbb{P}}^n$ outside the image of the interior of the disk have value $N\mathrm{\ne }{y}_{-}$; therefore ${\mathrm{deg}}_2(\tilde{q})=1$ by [F4].

Since $\tilde{q}$ is homotopic to $q$, [F4] gives ${\mathrm{deg}}_2(q)={\mathrm{deg}}_2(\tilde{q})=1$; a constant map has an empty regular fibre over any other value, so its mod-two degree is $0$, and the two values of $\mathbb{Z}\mathrm{/}2$ are realized. By [F5] applied to the nonorientable manifold ${\mathbb{R}\mathbb{P}}^n$, the mod-two degree is a bijection $[{\mathbb{R}\mathbb{P}}^n,S^n]\to \mathbb{Z}\mathrm{/}2$, so the classes of $q$ and of the constant map are the two classes and no integer degree is available for ${\mathbb{R}\mathbb{P}}^n$.

(Equal total degrees.) Let $f=\mathrm{id}\bigsqcup c$ and $g=c\bigsqcup \mathrm{id}$ on $M=S^m\bigsqcup S^m$. The restrictions to the two components have degrees $1$ and $0$ in the first case and $0$ and $1$ in the second, so by [F2] both total degrees equal $1+0=1=0+1$.

(No homotopy exists.) Suppose $H:M\times I\to S^m$ were a homotopy from $f$ to $g$. Its restriction to the first copy $S^m\times I$ is a homotopy from $\mathrm{id}$ to $c$ between continuous self-maps of the sphere; by the sphere homotopy invariance recorded in [F2], $\mathrm{deg}(\mathrm{id})=\mathrm{deg}(c)$, contradicting $1\mathrm{\ne }0$ from [F1].

There is therefore a closed oriented smooth $m$-manifold, namely $S^m\bigsqcup S^m$, and two maps on it whose total degrees agree but which are not homotopic; the total degree is not a complete invariant for disconnected domains, exactly as recorded in the connectedness remark.