Nonlinear Equations - UFRJ

40 [CH. 3: TOPOLOGY AND ZERO COUNTING
dimensional.
[
∂
D(s) = det ∂s t(s)
D t H(t(s), x(x))
]
∂
∂s x(s)∗ ≠ 0
D x H(t(s), x(s))
and in particular this determinant has the same sign at the boundaries
of X i .
Again, because ỹ is a regular value of f, the tangent vector of X i
at s 0 is of the form
[
Thus,
v
−vD x H(t, x) −1 (g(x) − f(x))
([ ] [ ])
v 0 1 −w
∗
D(s 0 ) = det
0 Df(x) w I
with w = Df(x) −1 (g(x) − f(x)) and x = x(s 0 ). The reader shall
check that the rightmost term has always strictly positive determinant
1 + ‖w‖ 2 . Therefore, det D(s 0 ) has the same sign of det Df(x).
When s = s 1 , we have exactly the same situation with v < 0.
Thus,
sign det Df(x(s 0 )) + sign det Df(x(s 1 )) = 0
The second case t(s 0 ) = t(s 1 ) = b is identical with signs of v
reversed. In the third case, we assume that t(s 0 ) = a and t(s 1 ) = b,
and hence v > 0 in both extremities. There we have
sign det Df(x(s 0 )) − sign det Df(x(s 1 )) = 0
The fourth case is trivial.
We conclude that
⎛
deg(g, y) − deg(f, y) = ∑ i
⎝
∑
]
sign det DH(b, x)−
(b,x)∈∂X i
⎞
−
∑
sign det DH(a, x) ⎠ = 0.
(a,x)∈∂X i