784 CHAPTER 23. DEGREE THEORY
where h(x) ≡ g(x+xi)− y. There is no restriction on the size of r. Consider one of theterms in the sum. d (h,B(0,r) ,y) . Note Dh(0) = Dg(xi) .
h(x) = Dh(0)x+o(x) (23.7.16)
Now let
L(z1, ...,zp)≡
∇h1 (z1)...
∇hp (zp)
By the mean value theorem applied to the components of h, if x ̸= 0,∥x∥ ≤ r, there existz1, ...,zp ∈ B(0,r)×·· ·×B(0,r)
∥h(x)−h(0)∥∥x∥
=∥h(x)−0∥∥x∥
=
∥∥L(z1, ...,zp)x∥∥
∥x∥=
∥∥∥∥L(z1, ...,zn)
(x∥x∥
)∥∥∥∥ (23.7.17)
Since Dh(0)−1 exists, we can choose r small enough that for all (z1, ...,zp) ∈ B(0,r)×·· ·×B(0,r)
det(L(z1, ...,zp)) ̸= 0
Hence, for such sufficiently small r, there is δ > 0 such that for S the unit sphere,
{x :∥x∥= 1} ,
inf{∥∥L(z1, ...,zp)x
∥∥ : (z1, ...,zp,x) ∈ B(0,r)×·· ·×B(0,r)×S}= δ > 0
Then it follows from 23.7.17 that if ∥x∥ ≤ r, ∥h(x)∥∥x∥ ≥ δ so ∥h(x)−0∥ ≥ δ ∥x∥. Letting∥x∥= r so x ∈ h(∂B(0,r)) ,
dist(h(∂B(0,r)) ,0)≥ δ r
Now pick ε < δ . Making r still smaller if necessary, it follows from 23.7.16 that if∥x∥ ≤ r,
∥h(x)−Dh(0)x∥ ≤ ε ∥x∥< δ r ≤ dist(h(∂B(0,r)) ,0)
Thus,∥h−Dh(0)(·)∥
∞,B(0,r) < dist(h(∂B(0,r)) ,0)
and so d (h,B(0,r) ,0) = d (Dh(0)(·) ,B(0,r) ,0).Say A = Dh(0) an invertible matrix. Then if U is any open set containing 0, it follows
from the properties of the degree that d (A,U,0) = d (A,B(0,r) ,0) whenever r is smallenough. By the product formula and 3. above,
1 = d (I,B(0,r) ,0) = d(A−1A,B(0,r) ,0
)= d (A,B(0,r) ,AB(0,r))d
(A−1,AB(0,r) ,0
)+d(
A,B(0,r) ,AB(0,r)C)
d(
A−1,AB(0,r)C,0)