By Boris G. Konopelchenko
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35) where d(suppφλ ) = supx,y∈suppφλ |x − y|. Since our partition of unity is subordinate to the cover {Bρ(x) (x)}x∈X\K we can find a x˜ ∈ X\K such that suppφλ ⊂ Bρ(˜x) (˜ x) and hence d(suppφλ ) ≤ ρ(˜ x) ≤ dist(K, suppφλ ). 36) as expected. By our choice of δ we have |F (xλ ) − F (x0 )| ≤ ε for all λ with φλ (x) = 0. Hence |F (x) − F (x0 )| ≤ ε whenever |x − x0 | ≤ δ and we are done. ✷ Note that the same proof works if X is only a metric space. Finally, let me remark that the Brouwer fixed point theorem is equivalent to the fact that there is no continuous retraction R : B1 (0) → ∂B1 (0) (with R(x) = x for x ∈ ∂B1 (0)) from the unit ball to the unit sphere in Rn .
By our choice of δ we have |F (xλ ) − F (x0 )| ≤ ε for all λ with φλ (x) = 0. Hence |F (x) − F (x0 )| ≤ ε whenever |x − x0 | ≤ δ and we are done. ✷ Note that the same proof works if X is only a metric space. Finally, let me remark that the Brouwer fixed point theorem is equivalent to the fact that there is no continuous retraction R : B1 (0) → ∂B1 (0) (with R(x) = x for x ∈ ∂B1 (0)) from the unit ball to the unit sphere in Rn . In fact, if R would be such a retraction, −R would have a fixed point x0 ∈ ∂B1 (0) by Brouwer’s theorem.
15 we can assume f ∈ C(U , Rn ) and since n is odd we have deg(−1l, U, 0) = −1. Now if deg(f, U, 0) = −1, then H(t, x) = (1−t)f (x)−tx t0 must have a zero (t0 , x0 ) ∈ (0, 1) × ∂U and hence f (x0 ) = 1−t x0 . Otherwise, if 0 deg(f, U, 0) = −1 we can apply the same argument to H(t, x) = (1−t)f (x)+tx. ✷ In particular this result implies that a continuous tangent vector field on the unit sphere f : S n−1 → Rn (with f (x)x = 0 for all x ∈ S n ) must vanish somewhere if n is odd. Or, for n = 3, you cannot smoothly comb a hedgehog without leaving a bald spot or making a parting.