Based sphere maps are classified by degree

Every $S^j$, j≥0, has a finite cross-polytope boundary triangulation transported radially to the unit sphere. Its vertex e1 can be moved to any b by the orthogonal formula $x\mapsto x-2⟨x,e_1-b⟩(e_1-b)\mathrm{/}\mathrm{\parallel }{e}_1-b{\mathrm{\parallel }}^2$ if b≠e1, and by the identity otherwise. Thus F7 proves well-pointedness at every chosen point, including both points of S0, without a general-CW HEP assertion.

On $S^0=\{p,q\}$, F5 gives $H_0=\mathbb{Z}[p]\oplus \mathbb{Z}[q]$; its point clause follows from the one-generator chain complex with alternating zero and identity differentials, and the finite sum clause from the two summand chain complexes. F4 makes reduced $H_0$ the kernel of (a,b)↦a+b, generated by [p]−[q]. Swapping p and q acts as −1 on this kernel. Apply F8 at n=0 with source basepoint p and target basepoint q, allowed by step 1.1. Its cone-cover connecting isomorphism is independent of that auxiliary basepoint: the cover and its intersection projection use only suspension height. Naturality therefore makes the suspension swap act as −1 on $H_1$. The suspension swap is a coordinate reflection of S1.

For a self-map g of $S^j$, j≥1, regard it as based from a to g(a), both well-pointed by step 1.1. F8 gives $s(\mathrm{\Sigma }g{)}_{\ast }=g_{\ast }s$ for the same cone-cover isomorphism s on source and target. Since top homology is cyclic, conjugation by this isomorphism preserves its integer multiplier. Hence the two-apex suspension preserves degree. Under $[x,t]\mapsto (\sqrt{1-t^2}x,t)$ for −1≤t≤1 it carries a coordinate reflection to the same coordinate reflection in the next sphere. Starting with step 2.1 proves reflection degree −1 in every dimension. Coordinate permutations conjugate one reflection to another and their degrees cancel with those of their inverses by F3. Identity degree is 1 by F2; constant degree is 0 because it factors through the point, whose positive homology is zero by F5.

F1 supplies a finite affine representative with matrices $A_j$. Let $k$ be the number of positive determinants minus the number of negative determinants. In the cube interior choose the same finite number of disjoint closed coordinate cubes, with centers $c_j$ and a common half-width $\epsilon >0$. Define a second map to be infinity outside them and, inside the jth cube, $Q_{\epsilon }(B_j(x-c_j))$, where $B_j=I$ when $\det A_j>0$ and $B_j=J$ otherwise, with J the first-coordinate reflection. These are continuous based bubbles by F1's displayed formula; since $\mathrm{\parallel }Jv{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$, their supports are exactly the chosen cubes. This is an affine normal form with common radius $R=\epsilon$ and matrices $I$ or $J$. F11 therefore identifies both the original representative and this standard-bubble map with the same signed product of the identity generator, so they are based homotopic and have the same degree by F3. Zero in the target chart has precisely these centers as its preimages. In centered coordinates, near the jth center the standard bubble is $B_{j}v\mathrm{/}(1-\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\epsilon )$. On $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$, replacing the denominator by $1-t\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\epsilon$, for $0\le t\le 1$, is a homotopy of punctured pairs between $B_{j}v$ and that germ. Translation of the source center and positive coordinate scalings preserve the local orientation. The local/global orientation comparison in F6's proof now identifies its local multiplier with the degree of the whole-sphere identity or reflection, namely +1 or -1 by step 3.1. F6 gives total degree k; the empty family gives the constant map and degree zero. Source and target use the same oriented quotient identification in F9, so the positive generator is the identity.

F11 says the normalized representative of the original class is $k$ times the identity generator, and step 4.1 with F3 says its degree is $k$. Thus equal degrees give equal classes. Conversely based homotopic maps have equal degree by F3. For any integer $m$, concatenate $m$ identity representatives if $m\ge 0$ and $-m$ reversed representatives if $m<0$; F10 supplies these finite products, including the empty product. The same finite-fibre calculation gives degree $m$. Products add the exponents of a single generator, so degree is an isomorphism; F9 transfers the assertion to the oriented pinch operation.