The Hopf Degree Theorem

Form the tangent-ray cover $O(TM)$ with points $(x,o)$, where $o$ is one of the two orientation rays of $T_{x}M$. In each tangent chart its topology and smooth structure are $U\times \{+,-\}$; transition signs are locally constant because their derivative determinants are continuous and nonzero. These charts define a two-sheeted covering $p:O(TM)\to M$. A section is exactly an orientation by [F4]: in a chart a continuous section chooses a locally constant sign, hence a smooth ray. The map $\rho :B(M)\to O(TM)$ sending a frame to its ray is locally the determinant-sign quotient and has the path-connected fibre ${\mathrm{G}\mathrm{L}}_m^{+}(\mathbb{R})$.

Both $M$ and $O(TM)$ are locally path-connected. Lift paths in $M$ to $O(TM)$ with a prescribed initial lift by Existence and uniqueness of path lifts through a covering map. Since $M$ is path-connected by [F3], each path component of the cover meets the fibre over every point. Thus there are at most two path components. If there are two, each contains exactly one point over each base point; the restricted projection is a bijective local diffeomorphism, so its inverse is a section, and $M$ is orientable. Conversely a section and its opposite have disjoint open images covering $O(TM)$, each homeomorphic to the connected $M$. Hence the cover has exactly two components precisely in the orientable case, and one otherwise.

A path in $O(TM)$ can be lifted to $B(M)$ with prescribed initial frame: by [F6], subdivide its parameter interval into finitely many pieces lying in bundle trivializations from step 1.1; on each piece keep the fibre coordinate constant, expressing the terminal frame in the next chart before continuing. This constructs a continuous lift. Join its endpoint to any prescribed frame over the same terminal ray by [F2]. Conversely every path in $B(M)$ projects under $\rho$. Thus $\rho$ induces a bijection of path components. By [F3] these are also connected components; step 2.1 gives their number, and in the oriented case their labels are the signs relative to the chosen orientation.

Given a continuous frame path $c$, first replace it by $c(\sigma (3t-1))$, constant near $0$ and $1$, where $\sigma$ is [F5]. Extend this path to $\mathbb{R}$ by its constant endpoint values. Apply [F5] to the closed set $(-\mathrm{\infty },0]\cup [1,\mathrm{\infty })$, near which the extension is smooth. Restrict the resulting smooth approximation to $[0,1]$. Its endpoints are unchanged; its image is a path in the same component, so in the oriented case the endpoint signs coincide. Local path-connectedness of each component follows from [F3]. Nonemptiness is essential: $B(\mathrm{\varnothing })$ has no components.

Choose a smooth nondecreasing $\lambda :[0,1]\to [0,1]$ given explicitly by $\lambda (t)=\sigma (3t-1)$, with $\sigma$ the standard smooth step function (The standard smooth step function) and $\epsilon =1\mathrm{/}8$, and replace $c$ by the reparametrized smooth path $c\circ \lambda$ with the same endpoints, so that $x(\lambda (t))=x_0$ and $b(\lambda (t))=b_0$ for $t\le \epsilon$ and $x(\lambda (t))=x_1$, $b(\lambda (t))=b_1$ for $t\ge 1-\epsilon$.

Define $W:=\{(x(\lambda (t)),t):t\in [0,1]\}\subseteq M\times I$. The map $t\mapsto (x(\lambda (t)),t)$ is smooth and injective (the second coordinate separates points) with derivative having second component $1\mathrm{\ne }0$, so $W$ is a compact embedded $1$-submanifold with boundary the two endpoints $(x_0,0)$ and $(x_1,1)$; it is neat in $M\times [0,1]$, and by step 1.1 its ends are exactly $W\cap (M\times [0,\epsilon ))=\{x_0\}\times [0,\epsilon )$ and $W\cap (M\times (1-\epsilon ,1])=\{x_1\}\times (1-\epsilon ,1]$.

Write $\gamma (t)=x(\lambda (t))$. At $(\gamma (t),t)$ define $Q_t:T_{\gamma (t)}M\oplus \mathbb{R}\to T_{\gamma (t)}M$ by $Q_t(v,a)=v-a\dot{\gamma }(t)$. Its kernel is precisely $\mathbb{R}(\dot{\gamma }(t),1)=T_{(\gamma (t),t)}W$, and it is surjective since $Q_t(v,0)=v$; hence it induces a smooth isomorphism of the normal quotient with $T_{\gamma (t)}M$. The framing is ${\mathrm{\Psi }}_t=b(\lambda (t){)}^{-1}\circ Q_t$ on that quotient. On each end collar $\dot{\gamma }=0$ and $b(\lambda (t))=b_i$, so $\mathrm{\Psi }$ is exactly the product pullback of ${\phi }_i=b_i^{-1}$. This supplies the required quotient map and the framing in the trivialization direction of [F2].

Therefore $(W,\epsilon ,\mathrm{\Psi })$ satisfies all the data of a framed cobordism from the framed point $(x_0,{\phi }_0)$ to $(x_1,{\phi }_1)$ in the sense of [F2], and reading the framings through the frame-bundle dictionary the two framed points $(x_0,b_0)$ and $(x_1,b_1)$ are framed cobordant. No choice beyond the inherited countable choice and the finite choice of $\lambda$ and $\epsilon$ is used.

Each of $W_N,W_L$ is closed by compactness and the Hausdorff property. Thus every point of either has an ambient neighbourhood missing the other. On that neighbourhood $W=W_N\cup W_L$ is exactly the corresponding neat embedded submanifold. Therefore $W$ is a compact neat embedded submanifold, with boundary $(N_0\bigsqcup L_0)\times \{0\}\bigsqcup (N_1\bigsqcup L_1)\times \{1\}$. The normal quotient restricts on each open-and-closed piece to its original normal quotient.

Set $\epsilon =\min ({\epsilon }_N,{\epsilon }_L)\in (0,1\mathrm{/}2)$. The product ends of the two pieces give product ends of $W$ with this width. Their framings paste smoothly on its disjoint open-and-closed pieces and restrict to the combined endpoint framings throughout those collars. Hence $(W,\epsilon ,{\mathrm{\Psi }}_N\bigsqcup {\mathrm{\Psi }}_L)$ is the asserted framed cobordism. For disjoint finite sets, summing one per point, or the orientation sign per point, splits into the sums over the two sets; reducing cardinalities modulo two gives parity additivity. This includes either set being empty and rank-zero cobordisms.

Work in $B$, put $z_i=u(x_i)$, $d=\mathrm{\mid }{z}_1-z_0\mathrm{\mid }>0$, and choose a constant orthonormal basis $(e_1,\dots ,e_m)$ with $e_1=(z_1-z_0)\mathrm{/}d$, completing it by finite elimination and normalization. The segment between $z_0,z_1$ lies in $B$. Let $\sigma$ be [F2], fix $0<h<1\mathrm{/}4$, and set $a(s)=d\sigma (3s-1)$, $t(s)=hs(1-s)$, and $\gamma (s)=(z_0+a(s)e_1,t(s))$. The standard step function is strictly increasing on $(0,1)$: differentiating $\sigma (r)=\beta (r)\mathrm{/}(\beta (r)+\beta (1-r))$ gives a positive numerator ${\beta }^{\mathrm{\prime }}(r)\beta (1-r)+\beta (r){\beta }^{\mathrm{\prime }}(1-r)$ there, since $\beta (r)=e^{-1\mathrm{/}r}$ for $r>0$ has ${\beta }^{\mathrm{\prime }}(r)>0$. Thus $a$ is strictly increasing between its two constant endpoint legs. One has $t(0)=t(1)=0$ and $0<t(s)\le h\mathrm{/}4<1$ inside; $t^{\mathrm{\prime }}(s)=h(1-2s)$ vanishes only at $s=1\mathrm{/}2$, where $a^{\mathrm{\prime }}(s)>0$. Hence $\gamma$ is an injective immersion: the middle is separated by its horizontal coordinate, and the two distinct vertical legs have strictly monotone heights. Compactness and Hausdorffness give continuity of the inverse on the image. The image is a compact embedded arc with literal vertical product collars of any sufficiently small width $\epsilon <t(1\mathrm{/}3)$, and no top endpoint.

Write $q(s)=\sqrt{a^{\mathrm{\prime }}(s{)}^2+t^{\mathrm{\prime }}(s{)}^2}>0$. The function $q$ is smooth, since the positive square root is $C^1$ by the scalar inverse function theorem and repeated differentiation of its derivative gives smoothness. Use normal vectors $w_1=(-t^{\mathrm{\prime }}(s)e_1,a^{\mathrm{\prime }}(s))\mathrm{/}q(s)$ and $w_j=(e_j,0)$ for $2\le j\le m$. Their quotient classes form a basis: the tangent $(a^{\mathrm{\prime }}{e}_1,t^{\mathrm{\prime }})$ and $w_1$ have determinant $q(s)>0$ in the $(e_1,t)$-plane, and the remaining vectors span the complementary spatial directions. On the first leg $a^{\mathrm{\prime }}=0$, $t^{\mathrm{\prime }}>0$, so $w_1=(-e_1,0)$; on the second $a^{\mathrm{\prime }}=0$, $t^{\mathrm{\prime }}<0$, so $w_1=(e_1,0)$. Thus the framing is constant on both product collars. Use the inverse of this basis map as the normal trivialization. This gives a framed null-cobordism of the model pair at the actual points; reversing the first normal vector throughout reverses both endpoint signs if their order needs to be switched.

The prescribed frames have the respective signs of one of those two model choices. By [F2] join each prescribed inverse framing to the corresponding model frame at the same fixed point, making the paths constant near their ends. The two stationary cylinders $\{x_i\}\times I$ have disjoint images; on each, the inverse frame path trivializes the normal quotient $T_{x_i}M$. Their union therefore is an embedded framed cobordism from the prescribed pair to the model pair, with product ends and constant collar framings. This construction needs no claim that unrelated cobordisms can be made disjoint.

Glue the stationary-cylinder cobordism to the staple by [F3]. The result is supported in $U\times I$, has the prescribed pair as its bottom end and empty top end, and has the specified normal framing on the bottom collar. Thus the pair is framed null-cobordant. All paths and integrals are finite constructions; the countable-choice hypothesis is inherited from [F1] and [F3].

(Orientations of $W$ and of its normal bundle.) Orient the normal bundle $\nu (W\subseteq M\times I)$ by the framing $\mathrm{\Psi }$, and orient the $1$-manifold $W$ by the rule that a positive normal frame followed by a positive tangent frame of $W$ is a positive frame of $T(M\times I)$; this orientation exists and is unique because $W$ is connected componentwise and the rank of $\nu (W)$ is $m$. With this choice the orientation of $W$ is determined by the framing and the product orientation, and no orientation is imposed on the individual points of the $N_i$.

On an end collar a normal frame given by ${\mathrm{\Psi }}^{-1}$ is a frame $b$ of $T_{x}M$ of sign $\epsilon (x)$. In the product orientation, $(b,\epsilon (x){\mathrm{\partial }}_t)$ is positive, so the rule in step 1.1 makes $\epsilon (x){\mathrm{\partial }}_t$ the positive tangent of $W$ there. At the bottom, the outward tangent is $-{\mathrm{\partial }}_t$, so the boundary point sign is $-\epsilon (x)$; at the top it is $+\epsilon (x)$. Consequently the signed boundary count is $-{\mathrm{\Sigma }}_0+{\mathrm{\Sigma }}_1$. This computes the signs directly on $W$, without suppressing the dimension-dependent signs of the ambient faces.

By [F2] the signed count of the boundary of a compact oriented $1$-manifold is zero, so $-{\mathrm{\Sigma }}_0+{\mathrm{\Sigma }}_1=0$ and ${\mathrm{\Sigma }}_0={\mathrm{\Sigma }}_1$: the signed count is a framed cobordism invariant. Consequently a closed framed $0$-manifold consisting of exactly two points of the same framing sign and no other points has signed count $\pm 2\mathrm{\ne }0$, while the empty framed $0$-manifold has signed count $0$, so it is not framed null-cobordant; a pair of opposite signs in a common chart ball is framed null-cobordant by [F3].

The count descends to classes by [F1]. For any $k\in \mathbb{Z}$, a chart ball contains $\mathrm{\mid }k\mathrm{\mid }$ distinct points carrying frames of sign $\mathrm{sgn}(k)$; the empty set represents $0$. Their signed count is $k$. This proves surjectivity without choosing a preferred framing at every point of $M$.

Suppose $m\ge 2$. A Euclidean ball minus finitely many points is path-connected: for two allowed endpoints choose an intermediate point off the finitely many lines through an endpoint and a removed point; the two straight segments lie in the convex ball and avoid the removed points. Such an intermediate point exists because a finite union of lines has empty interior in dimension at least two (in a small ball choose a line direction distinct from the finitely many directions, then exclude its finitely many intersections). Consequently a frame path can be replaced by one avoiding any prescribed finite set $P\subset M$ disjoint from its endpoint base points: subdivide the original path into finitely many tangent trivializations, move its subdivision frames slightly off $P$ inside the chart overlaps, and join the new endpoints inside each punctured chart, keeping the frame coordinate in its original determinant component by [F6]. Small moves in the overlaps preserve that component. The finitely many local paths glue; smoothing with fixed endpoints as in [F2] on the open manifold $M\setminus P$ gives a smooth frame path avoiding $P$. Its graph cobordism is disjoint from every stationary cylinder $\{p\}\times I$, $p\in P$, and their union is therefore embedded by [F4].

For $m=1$, identify $M$ with an oriented circle using [F5]. If both signs occur in a finite configuration, some cyclically adjacent pair has opposite signs. The arc between them with a small extension at either end is a chart interval containing no other occupied point. Apply [F3] inside that interval and adjoin the stationary cylinders of the other points, whose images are disjoint from its support. Repeat until only $\mathrm{\mid }k\mathrm{\mid }$ points of one sign remain. To compare two remaining configurations of $r=\mathrm{\mid }k\mathrm{\mid }>0$ points, choose cyclically increasing real lifts $a_1<\cdots <a_r<a_1+1$ and $b_1<\cdots <b_r<b_1+1$ in a period-one circle coordinate, matching the cyclic orders. The paths $a_i(t)=(1-t)a_i+t{b}_i$ remain distinct modulo one: successive gaps, including the final cyclic gap, are convex combinations of positive gaps. The oriented circle coordinate supplies a smooth nonzero tangent frame; choose the constant model frame of the required sign along each trajectory. At the fixed endpoints, adjust the prescribed frames to these models by [F6] on disjoint stationary cylinders. Flatten the parameter at the ends and use the quotient graph framing from [F2]. The resulting disjoint graphs and endpoint cylinders give a cobordism of the two configurations. For $r=0$ both reduced configurations are empty.

For $m\ge 2$, fix a chart ball $U$ and distinct target points there avoiding $N$. Apply step 1.2 successively to move each framed point of $N$ to a target point of the same sign, taking $P$ to be the other currently occupied points. Thus every move extends to a cobordism of the entire configuration. Arrange each positive-negative pair at two points in its own small ball in $U$, disjoint from all the other points. By [F3] cancel these pairs one at a time, adjoining only stationary cylinders outside the supporting ball. The remaining configuration has $\mathrm{\mid }k\mathrm{\mid }$ points all of sign $\mathrm{sgn}(k)$, where $k=\mathrm{\Phi }(N,\phi )$. Two such configurations with the same $k$ can both be moved to the same distinct target points (chosen to avoid both initial finite sets), with the same chosen frames there, again using step 1.2. Thus their classes agree.

Steps 2.1 and 1.3 show that configurations with the same signed count are cobordant; [F1] gives the converse. Together with step 1.1 this proves the bijection and the null-cobordism criterion. Every two classes admit disjoint representatives by placing the required finite sets in separate small balls. Define their sum by the class of that union: its count is the sum of the two counts, so the bijection proves independence of the disjoint representatives. Associativity, commutativity and the empty unit follow from integer addition, giving the asserted monoid isomorphic to $(\mathbb{Z},+)$. This uses no disjointness inference for arbitrary cobordisms.

A framed cobordism from $N_0$ to $N_1$ is a compact $1$-manifold with boundary $N_0\bigsqcup N_1$. By [F1], $\mathrm{\mid }{N}_0\mathrm{\mid }+\mathrm{\mid }{N}_1\mathrm{\mid }$ is even, so parity is invariant under framed cobordism. This argument does not assume orientability of the ambient manifold.

Nonorientability excludes the empty manifold. It also excludes $m=1$: by [F1], a nonempty closed connected $1$-manifold is a single circle, and its period coordinate supplies a global positive tangent ray, making it orientable. Thus $m\ge 2$. In a Euclidean ball of that dimension with finitely many points removed, two allowed points can be joined by two straight segments through an intermediate point avoiding the finitely many lines through either endpoint and a removed point. A finite union of lines has empty interior: choose a direction different from all their directions and remove its finitely many intersections inside a small ball. Hence the required intermediate point exists.

Given any frame path from [F2] and a finite set $P$ of forbidden base points disjoint from its endpoints, subdivide it into finitely many tangent trivializations. Move subdivision frames slightly off $P$ inside the chart overlaps and preserve their local determinant component. Within each chart, join their base points in the punctured ball by step 1.2, and join their frame coordinates by [F5]; the original path ensures that the local signs of the two endpoints agree. The resulting paths glue and can be smoothed with endpoints fixed on $B(M\setminus P)$ by the smoothing argument in [F2]. The graph cobordism therefore avoids all stationary cylinders at $P$. Adjoining those cylinders gives an embedded cobordism of the full finite configuration, with the normal framings defined separately on the disjoint pieces.

Choose distinct target points in a chart ball, avoiding the initial configuration, with opposite chart framings at each chosen pair. Move the initial points successively to those targets by step 2.1, taking the forbidden set to be all other currently occupied points. The global frame bundle is connected by [F2], so no initial sign restricts the chosen terminal frame. Arrange the pairs in separate small balls and cancel each using [F3], adjoining only stationary cylinders outside that ball. An even configuration reduces to the empty one; an odd configuration reduces to a single framed point. Any two singleton configurations are cobordant by [F2], whereas a singleton is not null-cobordant by step 1.1.

Empty and singleton configurations realize the two parities, and step 3.1 proves that these are precisely the two classes. Thus parity is a bijection to $\mathbb{Z}\mathrm{/}2$ and null-cobordism is equivalent to even cardinality. Every two classes have disjoint representatives by using distinct points. The class of their union depends only on the sum of their parities, by the bijection, so addition on classes is well defined and additive. No orientation of $M$ and no disjoint union of intersecting cobordisms is used.

For $x\in f^{-1}(y)$ the normal quotient $\nu (x)=T_{x}M\mathrm{/}{T}_x\{x\}=T_{x}M$ is identified with $T_{x}M$, and the differential $d{f}_x:T_{x}M\to T_y{S}^m$ is an isomorphism because $y$ is a regular value of an equidimensional map; the induced framing is the composite $\beta \circ d{f}_x$, by [F1], where $\beta :T_y{S}^m\to {\mathbb{R}}^m$ sends the positive basis $b$ to the standard basis.

Since $b$ is a positive basis, [F2] gives $\epsilon (x)=\mathrm{sgn}(d{f}_x)$ for every $x$ of the fibre, and the fibre is finite; summing and applying the regular value formula of [F3] to the proper map $f$ gives ${\sum }_{x\in f^{-1}(y)}\epsilon (x)={\sum }_{x\in f^{-1}(y)}\mathrm{sgn}(d{f}_x)=\mathrm{deg}(f)$.

Hence the signed count of the framed regular preimage is exactly the compact-support degree of the original map, with no new definition of degree and no use of an orientation of $S^m$ beyond the fixed positive basis.

(The bump profile.) By [F1] in dimension one fix a smooth $\chi :\mathbb{R}\to [0,1]$ with $\chi =1$ on $[-1\mathrm{/}2,1\mathrm{/}2]$ and $\chi =0$ outside $(-3\mathrm{/}4,3\mathrm{/}4)$, and define $W(s):=1-\chi (s{)}^2(1-s)$ for $s\ge 0$. Then $W$ is smooth, $W(s)=s$ for $0\le s\le 1\mathrm{/}2$, $W(s)=1$ for $s\ge 3\mathrm{/}4$, and $0\le W(s)\le 1$ with $W(s)>0$ for $s>0$; moreover $1-W(s)=\chi (s{)}^2(1-s)$ is the square of the smooth function $\delta (s):=\chi (s)\sqrt{1-s}$, read as $0$ where $\chi$ vanishes.

(The model pinch.) Put $G(s):=W(s)\mathrm{/}s$ for $s>0$ and $G(0):=1$; since $W(s)=s$ for $s\le 1\mathrm{/}2$, the function $G$ is smooth and positive on $[0,\mathrm{\infty })$. Define $\mathrm{\Phi }(s):=\chi (s)\sqrt{W(s)(1-s)\mathrm{/}s}$ for $0<s\le 3\mathrm{/}4$ and $\mathrm{\Phi }(0):=1$ and $\mathrm{\Phi }(s):=0$ for $s\ge 3\mathrm{/}4$; the radicand $Q(s)=W(s)(1-s)\mathrm{/}s$ is smooth and positive on $[0,3\mathrm{/}4)$ and equals $1-s$ near $0$, so $\sqrt{Q}$ is smooth and positive there by [F2], and since $\chi$ is flat at $3\mathrm{/}4$ and vanishes beyond it, $\mathrm{\Phi }=\chi \sqrt{Q}$ is smooth on $[0,\mathrm{\infty })$. Set $F(x):=(2\mathrm{\Phi }(\parallel x{\parallel }^2)x,2W(\parallel x{\parallel }^2)-1)$ for $x\in {\mathbb{R}}^m$. With $s=\parallel x{\parallel }^2$ one has $\mathrm{\Phi }(s{)}^{2}s=\chi (s{)}^{2}W(s)(1-s)=W(s)(1-W(s))$, hence $\parallel F(x){\parallel }^2=4\mathrm{\Phi }(s{)}^{2}s+(2W(s)-1{)}^2=4W(1-W)+4W^2-4W+1=1$, so $F$ maps into $S^m$. Each component is smooth, and since $F$ is locally constant with value $N=(0,\dots ,0,1)$ for $s\ge 3\mathrm{/}4$, its expressions in the two stereographic charts of $S^m$ ([F3]) are smooth; thus $F:{\mathbb{R}}^m\to S^m$ is smooth.