23.1. SARD’S LEMMA AND APPROXIMATION 757

Lemma 23.1.5 Let f ∈ C(Ω;Rp

)and let {yi}∞

i=1 be points not in f(∂Ω) and let δ > 0.Then there exists g ∈C∞

(Ω;Rp

)such that ∥g− f∥

∞,Ω < δ and yi is a regular value for gfor each i. That is, if g(x) = yi, then Dg(x)−1 exists. Also, if δ < dist(f(∂Ω) ,y) for somey a regular value of g ∈C∞

(Ω;Rp

), then g−1 (y) is a finite set of points in Ω. Also, if y is

a regular value of g ∈C∞ (Rp,Rp) , then g−1 (y) is countable.

Proof: Pick g̃ ∈ C∞(Ω;Rp

),∥g̃− f∥

∞,Ω < δ . g≡ g̃− e where e is from the aboveLemma 23.1.4 and η so small that ∥g− f∥

∞,Ω < δ . Then if g(x) = yi, you get g̃(x) = yi+ea regular value of g̃ and so det(Dg(x)) = det(Dg̃(x)) ̸= 0 so this shows the first part.

It remains to verify the last claims. Since ∥g− f∥Ω,∞ < δ , if x ∈ ∂Ω, then

∥g(x)−y∥ ≥ ∥f(x)−y∥−∥f(x)−g(x)∥ ≥ dist(f(∂Ω) ,y)−δ > δ −δ = 0

and so y /∈ g(∂Ω), so if g(x) = y, then x∈Ω. If there are infinitely many points in g−1 (y) ,then there would be a subsequence converging to a point x ∈Ω. Thus g(x) = y so x /∈ ∂Ω.However, this would violate the inverse function theorem because g would fail to be oneto one on an open ball containing x and contained in Ω. Therefore, there are only finitelymany points in g−1 (y) and at each point, the determinant of the derivative of g is nonzero.For y a regular value, g−1 (y) is countable since otherwise, there would be a limit pointx ∈ g−1 (y) and g would fail to be one to one near x contradicting the inverse functiontheorem.

Now with this, here is a definition of the degree.

Definition 23.1.6 Let Ω be a bounded open set in Rp and let f : Ω→ Rp be continuous.Let y /∈ f(∂Ω) . Then the degree is defined as follows: Let g be infinitely differentiable,∥f−g∥

∞,Ω < dist(f(∂Ω) ,y) , and y is a regular value of g.Then

d (f,Ω,y)≡∑{

sgn(det(Dg(x))) : x ∈ g−1 (y)}

From Lemma 23.1.5 the definition at least makes sense because the sum is finite andsuch a g exists. The problem is whether the definition is well defined in the sense that weget the same answer if a different g is used. Suppose it is shown that the definition is welldefined. If y /∈ f(Ω) , then you could pick g such that ∥g− f∥

Ω< dist

(y, f(Ω))

. However,this requires that g

(Ω)

does not contain y because if x ∈Ω, then

∥g(x)−y∥ = ∥(y− f(x))− (g(x)− f(x))∥ ≥ ∥f(x)−y∥−∥g(x)− f(x)∥> dist

(y, f(Ω))−dist

(y, f(Ω))

= 0

Therefore, y is a regular value of g because every point in g−1 (y) is such that the deter-minant of the derivative at this point is non zero since there are no such points. Thus iff−1 (y) = /0,d (f,Ω,y) = 0.

If f(z) = y, then there is g having y a regular value and g(z) = y by the above lemma.

Lemma 23.1.7 Suppose g, ĝ both satisfy the above definition,

dist(f(∂Ω) ,y)> δ > ∥f−g∥∞,Ω ,dist(f(∂Ω) ,y)> δ > ∥f− ĝ∥

∞,Ω

23.1. SARD’S LEMMA AND APPROXIMATION 757Lemma 23.1.5 Let f € C (Q;R?) and let {y;}7, be points not in £(Q) and let & > 0.Then there exists g € C* (Q;R?) such that |\g— f\|..5 < 6 and y; is a regular value for gfor each i. That is, if g(x) = yj, then Dg(x)~' exists. Also, if 5 < dist (f(Q) ,y) for somey a regular value of g € C” (Q; R’) , then g! (y) is a finite set of points in Q. Also, ify isa regular value of g € C® (IR”, IR”) , then g~! (y) is countable.Proof: Pick § € C* (Q;R?) ,|/&—fl|..5 < 5. g=%—e where e is from the aboveLemma 23.1.4 and 7 so small that ||g —f\|,, 9 < 6. Then if g(x) =y;, you get (x) =y; +ea regular value of & and so det (Dg (x)) = det (D&(x)) 4 0 so this shows the first part.It remains to verify the last claims. Since ||g—fl|5 .. < 6, if x € 0Q, thenIlg (x) — yl] > |If(x) — yl] — |lf(x) — g(x)|] > dist (f(9Q) ,y)-6 > 6-5 =0and so y ¢ g(0Q), so if g(x) =y, then x € Q. If there are infinitely many points in g~! (y) ,then there would be a subsequence converging to a point x € Q. Thus g(x) = y sox ¢ aQ.However, this would violate the inverse function theorem because g would fail to be oneto one on an open ball containing x and contained in Q. Therefore, there are only finitelymany points in g ! (y) and at each point, the determinant of the derivative of g is nonzero.For y a regular value, g~! (y) is countable since otherwise, there would be a limit pointx € g !(y) and g would fail to be one to one near x contradicting the inverse functiontheorem. JNow with this, here is a definition of the degree.Definition 23.1.6 Let Q be a bounded open set in R? and let f : Q— R? be continuous.Let y ¢ (OQ). Then the degree is defined as follows: Let g be infinitely differentiable,I|f—g]|..6 < dist (f(0Q) ,y) , and y is a regular value of g.Thend(f,Q,y) =) {sgn (det (Dg(x))) :x eg! (y)}From Lemma 23.1.5 the definition at least makes sense because the sum is finite andsuch a g exists. The problem is whether the definition is well defined in the sense that weget the same answer if a different g is used. Suppose it is shown that the definition is welldefined. If y ¢ f(Q) , then you could pick g such that ||g — f\|5 < dist (y, f (Q)) . However,this requires that g (Q) does not contain y because if x € ©, thenIIig(x)—yll = \(y—f(x)) — (g(x) —£(x))|| 2 If) —yll — lle) -f@) |> dist (y,f(Q)) — dist (y,f(Q)) =0Therefore, y is a regular value of g because every point in g~! (y) is such that the deter-minant of the derivative at this point is non zero since there are no such points. Thus iff(y) =0,d(f,Q,y) =0.If f(z) = y, then there is g having y a regular value and g(z) = y by the above lemma.Lemma 23.1.7 Suppose g,& both satisfy the above definition,dist (f(AQ),y) > 6 > ||f—gl|...g dist (f(AQ) ,y) > 6 > ||f-gll..0