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Reduced degree into the 0-sphere is homotopy invariant and multiplicative

Statement

Let (P,s) be a balanced oriented finite 0-manifold, so that every map P→S0 has a reduced degree, and let f:P→S0 be a map (The reduced degree of a map into the 0-sphere).

(i) If F:P×[0,1]→S0 is continuous and Ft(x):=F(x,t), then Ft=F0 for every t∈[0,1]; in particular deg⁡Ft=deg⁡F0.

(ii) If g:S0→S0 is any map and deg⁡g denotes its reduced degree as a self-map of S0, then deg⁡(g∘f)=deg⁡(g)deg⁡(f).

Facts & Assumptions

Given: A balanced oriented finite 0-manifold (P,s) with ∑x∈Ps(x)=0 and a map f:P→S0.

[F1]

S0={−1,+1} is oriented by the boundary orientation of [−1,1]: the point +1 has sign +1 and −1 has sign −1. A reduced degree is defined only for a balanced source: ∑x∈Ps(x)=0, and then deg⁡(h)=12∑x∈Ps(x)h(x)∈Z for a map h:P→S0. For P=S0 itself this reads deg⁡(h)=h(+1)−h(−1)2∈{−1,0,+1} (The reduced degree of a map into the 0-sphere).

[F3]

In S0={−1,+1}⊆R the two singletons are open for the subspace topology: {+1}=S0∩B2(+1,1) and {−1}=S0∩B2(−1,1) (Euclidean spheres and closed balls as subspaces of Rn, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). A continuous map from a connected space into S0 is therefore constant: the preimages of the two disjoint open singletons would otherwise separate the domain.

Proof

Proof technique: direct; split the composition law by the three possible reduced degrees of the self-map g.

1.1F2F3algebra

The product P×[0,1] is the disjoint union of the connected subsets {x}×[0,1], one for each x∈P; a continuous map from a connected space into S0 is constant, so F is constant on each {x}×[0,1], hence Ft(x)=F(x,t)=F(x,0)=F0(x) for all x∈P and t∈[0,1]; equal maps have equal reduced degree, which proves (i).

1.2F1algebra

Suppose g is not constant. A map S0→S0 is determined by the pair (g(+1),g(−1)), so a nonconstant g sends +1 and −1 to different values; hence g is either the identity, with σ:=+1 and g(y)=y for both y, or the antipodal map g(y)=−y, with σ:=−1, so that g(y)=σy for all y∈S0; then g∘f=σf and, by linearity of the defining signed count, deg⁡(g∘f)=12∑xs(x)σf(x)=σdeg⁡(f), while σ=deg⁡(g) by [F1], the identity and the antipodal map having reduced degrees +1 and −1; thus deg⁡(g∘f)=deg⁡(g)deg⁡(f).

2.1F1step 1.2algebra∎

If g is constant with value c∈S0, then g∘f is the constant map c and deg⁡(g∘f)=12c∑x∈Ps(x)=0 by balance, while deg⁡(g)=12(c−c)=0 by [F1]; hence again deg⁡(g∘f)=0=deg⁡(g)deg⁡(f), which completes (ii).

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