Kenneth Kuttler

13.4. BORSUK’S THEOREM 315

Lemma 13.4.3 Let g ∈C∞(

Ω;Rp)

be an odd map. Then for every ε > 0, there existsh ∈C∞

(Ω;Rp

)such that h is also an odd map, ∥h−g∥

∞< ε , and 0 is a regular value of

h,0 /∈ g (∂Ω) . Here Ω is a symmetric bounded open set. In addition, d (g,Ω,0) is an oddinteger.

Proof: In this argument η > 0 will be a small positive number. Let h0 (x) = g (x)+ηxwhere η is sufficiently small but nonzero that detDh0 (0) ̸= 0. See Lemma 9.14.1. Notethat h0 is odd and 0 is a value of h0 thanks to h0 (0) = 0. This has taken care of 0.However, it is not known whether 0 is a regular value of h0 because there may be otherx where h0 (x) = 0. By Lemma 13.4.2, there are vectors y j with

∥∥yk∥∥ ≤ η and 0 is a

regular value of h(x)≡ h0 (x)−∑pj=1y

jx3j . Then

∥h−g∥∞, Ω

≤ maxx∈Ω

{∥ηx∥+

∑k=1

∥∥∥yk∥∥∥∥x∥}

≤ η ((p+1)diam(Ω))< ε < dist(g (∂Ω) ,0)

provided η was chosen sufficiently small to begin with.So what is d (h,Ω,0)? Since 0 is a regular value and h is odd,

h−1 (0) = {x1, · · · ,xr,−x1, · · · ,−xr,0} .

So consider Dh(x) and Dh(−x).

Dh(−x)u+o(u) = h(−x+u)−h(−x) =−h(x+(−u))+h(x)

=−(Dh(x)(−u))+o(−u) = Dh(x)(u)+o(u)

Hence Dh(x) = Dh(−x) and so the determinants of these two are the same. It followsfrom the definition that d (g,Ω,0) = d (h,Ω,0)

∑i=1

sgn(det(Dh(xi)))+r

∑i=1

sgn(det(Dh(−xi)+ sgn(det(Dh(0)))))

= 2m±1 some integer m ■

Theorem 13.4.4 (Borsuk) Let f ∈C(

Ω;Rp)

be odd and let Ω be symmetric with0 /∈ f (∂Ω). Then d (f ,Ω,0) equals an odd integer.

Proof: Let ψn be a mollifier which is symmetric, ψ (−x) = ψ (x). Also recall that fis the restriction to Ω of a continuous function, still denoted as f which is defined on all ofRp. Let g be the odd part of this function. That is,

g (x)≡ 12(f (x)−f (−x)) = f (x) on Ω

Thus d (f ,Ω,0) = d (g,Ω,0). Then

gn (−x)≡ g ∗ψn (−x) =∫

Ω

g (−x−y)ψn (y)dy

=−∫

Ω

g (x+y)ψn (y)dy =−∫

Ω

g (x−(−y))ψn (−y)dy =−gn (x)

Thus gn is odd and is infinitely differentiable. Let n be large enough that

∥gn−g∥∞,Ω < δ < dist(f (∂Ω) ,0) = dist(g (∂Ω) ,0)

Then by definition of the degree, d (f,Ω,0) = d (g,Ω,0) = d (gn,Ω,0) and by Lemma13.4.3 this is an odd integer. ■

13.4. BORSUK’S THEOREM 315Lemma 13.4.3 Let g © C” ( Q;R? ) be an odd map. Then for every € > 0, there existshec ( Q; R?) such that h is also an odd map, ||h — g||,, < €, and 0 is a regular value ofh,0 ¢ g (OQ). Here Q is a symmetric bounded open set. In addition, d(g,Q,0) is an oddinteger.Proof: In this argument 7 > 0 will be a small positive number. Let ho (x) = g (x) +N xwhere 1) is sufficiently small but nonzero that det Dho (0) 4 0. See Lemma 9.14.1. Notethat ho is odd and O is a value of ho thanks to hg(O) = 0. This has taken care of 0.However, it is not known whether O is a regular value of ho because there may be otherx where ho (x) = 0. By Lemma 13.4.2, there are vectors y/ with Ily* || <7 and Oisaregular value of h(a) = ho (x) —YF_, y/x;. ThenP|h—gl..a < nay {nel +5:” k=1< n((p+1)diam(Q)) < € < dist(g (AQ) ,0)provided 7 was chosen sufficiently small to begin with.So what is d (h,Q,0)? Since 0 is a regular value and h is odd,ho! (O) = {a1,---,@,,—a1,---,—a,, 0}.So consider Dh (x) and Dh (—2).Dh(—a)u+o(u) =h(-a2+u)—h(-a) = —h(a+(—u))+h(ax)=~ (Dh () (—u)) +0(—u) = Dh(a@) (u) +0(w)Hence Dh (a) = Dh(—2) and so the determinants of these two are the same. It followsfrom the definition that d(g,Q,0) =d(h,Q,0)sgn (det (Dh (a;))) + y sgn (det (Dh (—a;) +sgn (det (Dh (0)))))= 2m+1 some integer m HfTheorem 13.4.4 (Borsuk) Let f EC ( Q; R’) be odd and let Q be symmetric with0 ¢ f (OQ). Then d(f,Q,0) equals an odd integer.Proof: Let y,, be a mollifier which is symmetric, y(—a) = y(a). Also recall that fis the restriction to Q of a continuous function, still denoted as f which is defined on all ofR?. Let g be the odd part of this function. That is,g(x) =5(f («)—-f (-#)) =f (x) on QThus d(f,Q,0) =d(g,Q,0). ThenIiveIn (—2) =g* Y, (—@) = [9-e-w) Wn (y) dy= - [ g(e+y) Vn (y)dy = - [ g(e- (—y)) Wn (—y) dy = —Gn (@)Thus g, is odd and is infinitely differentiable. Let n be large enough thatIn — Flog < 6 < dist (f (AQ) ,0) = dist (g (AQ) ,0)Then by definition of the degree, d(f,Q,0) = d(g,Q,0) = d(g,,Q,0) and by Lemma13.4.3 this is an odd integer.