MORSE THEORY AND THE GAUSS-BONNET FORMULA Alina ...

Proof. Suppose G −1 (v) was infinite. Since ν 1 M is compact, we can extract
a convergent sequence {ai} ⊂ G −1 (v) converging to some element a ∈ ν 1 M. By
continuity we must have G(a) = v as well, so a is a regular point of G. Therefore, for
any neighborhood V1 of a there exists some element ai that also maps to v. However,
if a is a regular point of G then there exists a neighborhood V2 of a such that G|V2 is
a homeomorphism; this contradicts the existence of the sequence ai.
As v is regular and has finitely many preimages, we can conclude that v has a
neighborhood U such that G−1 � �
(U) = U1 ... Uk is a finite disjoint union of open
sets Ui, each of which is homeomorphic to U. �
Corollary 4.7. For any function f : ν 1 M → R we have
�
ν 1 M
�
f(u)|det dG|dVν1M(u) =
�
SN−1 u∈G−1 (v)
f(u)dV S N−1(v)
Technically, the corollary should only allow us to compute integrals over S0. How-
ever, S0 has full measure in S N so we can replace integrals over S0 with integrals over
S N .
Proof of theorem 4.1
Define Ck(v) = {(v, p) ∈ S0 × M : p is a critical point of index k for hv} for every
n�
v ∈ S0. Then (−1) k #Ck(v) = χ(M) by the Morse inequalities.
k=0
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