306 CHAPTER 13. DEGREE THEORY

Lemma 13.1.8 Let g : Ω→ Rp be C2 where Ω is an open subset of Rp. Then

p

∑j=1

cof(Dg)i j, j = 0,

where here (Dg)i j ≡ gi, j ≡ ∂gi∂x j

. Also, cof(Dg)i j =∂ det(Dg)

∂gi, j.

Next is an integral representation of ∑{

sgn(det(Dg (x))) : x ∈ g−1 (y)}

but first is alittle lemma about disjoint sets.

Lemma 13.1.9 Let K be a compact set and C a closed set in Rp such that K ∩C = /0.Then

dist(K,C)≡ inf{∥k−c∥ : k ∈ K,c ∈C}> 0.

Proof: Let d ≡ inf{∥k−c∥ : k ∈ K,c ∈C}. Let {ki} ,{ci} be such that

d +1i> ∥ki−ci∥ .

Since K is compact, there is a subsequence still denoted by {ki} such that ki → k ∈ K.Then also

∥ci−cm∥ ≤ ∥ci−ki∥+∥ki−km∥+∥cm−km∥

If d = 0, then as m, i→ ∞ it follows ∥ci−cm∥ → 0 and so {ci} is a Cauchy sequencewhich must converge to some c ∈ C. But then ∥c−k∥ = limi→∞ ∥ci−ki∥ = 0 and soc= k ∈C∩K, a contradiction to these sets being disjoint. ■

In particular the distance between a point and a closed set is always positive if the pointis not in the closed set. Of course this is obvious even without the above lemma.

Definition 13.1.10 Let g ∈ C∞(Ω;Rp

)where Ω is a bounded open set. Also let

φ ε be a mollifier.

φ ε ∈C∞c (B(0,ε)) , φ ε ≥ 0,

∫φ ε dx = 1.

The idea is that ε will converge to 0 to get suitable approximations.

First, here is a technical lemma which will be used to identify the degree with an inte-gral.

Lemma 13.1.11 Let y /∈ g (∂Ω) for g ∈C∞(Ω;Rp

). Also suppose y is a regular value

of g. Then for all positive ε small enough,∫Ω

φ ε (g (x)−y)detDg (x)dx = ∑{

sgn(detDg (x)) : x ∈ g−1 (y)}

Proof: First note that the sum is finite from Lemma 13.1.5. It only remains to verifythe equation. If y /∈ g (Ω) , then for ε < dist

(g(Ω),y), φ ε (g (x)−y) = 0 for all x ∈ Ω

so both sides equal 0.I need to show the left side of this equation is constant for ε small enough and equals the

right side. By what was just shown, there are finitely many points, {xi}mi=1 = g−1 (y). By

the inverse function theorem, there exist disjoint open sets Ui with xi ∈Ui, such that g is one

306 CHAPTER 13. DEGREE THEORYLemma 13.1.8 Let g :Q— R?” be C? where Q is an open subset of R?. ThenPy? cof (Dg); ; =0,j=lwhere here (Dg); =2,j;= ge. Also, cof (Dg);; = aeNext is an integral representation of Y {sgn (det (Dg (a))) : « € g~! (y)} but first is alittle lemma about disjoint sets.Lemma 13.1.9 Let K be a compact set and C a closed set in R? such that KNC = 0.Thendist (K,C) = inf {||k—cl|:k €K,cEC}>0.Proof: Let d = inf {||k —c|| :& € K,c € C}. Let {k;}, {c;} be such that1d+— > \|ki — el -Since K is compact, there is a subsequence still denoted by {k;} such that k; > k € K.Then also\|¢i — Em|| < lle: — kil] + |i — Km || + [len — KmIf d = 0, then as m,i > © it follows ||¢; —m|| > 0 and so {e;} is a Cauchy sequencewhich must converge to some c € C. But then ||c—&|| = lim;-_,.. ||c; — &;|| = 0 and soc=k€CNK, acontradiction to these sets being disjoint. HiIn particular the distance between a point and a closed set is always positive if the pointis not in the closed set. Of course this is obvious even without the above lemma.Definition 13.1.10 Le gEc (Q; R?) where Q is a bounded open set. Also let¢, be a molilifier.be ECP(B(0.€)), 20, f Ode =I.The idea is that € will converge to 0 to get suitable approximations.First, here is a technical lemma which will be used to identify the degree with an inte-gral.Lemma 13.1.11 Let y ¢ g (0Q) for g € C” (Q;R?). Also suppose y is a regular valueof g. Then for all positive € small enough,[/%(9(@) v) det Dg (@) dx = ¥ {sgn detDg (x)):2 9°" (y)}Proof: First note that the sum is finite from Lemma 13.1.5. It only remains to verifythe equation. If y ¢ g(Q), then for € < dist (g (Q) ,y), (g(x) —y) =0 for all e EQso both sides equal 0.Ineed to show the left side of this equation is constant for € small enough and equals theright side. By what was just shown, there are finitely many points, {x;}”"., =g~!(y). Bythe inverse function theorem, there exist disjoint open sets U; with x; € U;, such that g is one