COMPLETING THE FUNCTIONAL CALCULUS
9
and a 8 0 such that f [ B(z0 ; 8) ] c B(o
0
; e) e Q . Hence D u B(z0 ; 8} is a
connected subset of H that includes D . Since D is a component of H ,
B(z0 ; 8) e D , a contradiction to the fact that z
0
e 3D . Hence f(3D) c dQ .
Claim 4. If K is a compact subset of Q , then
f_1(K)
n H is compact.
If {zk } c f
X(K)
n H , then, by passing to a subsequence if necessary, it may
be assumed that z
k
- z
0
in cl H . Hence f(zk) - f(z0) . Thus f(z0) e K c Q
and so z
0
e
f_1(K)
n o(A) c H . So z
0
e
f_1(K)
n H and this set is compact.
Claim 5. If D is a component of H and co0 and co1 e Q, , then the
equations f(z) = co0 and f(z) = co1 have the same number of solutions in D ,
counting multiplicities.
By Claim 4 these equations have only a finite number of solutions in D .
Indeed, f_1(cOj) n H is a compact set and f is a non-constant analytic function.
Let Pj = the number of solutions of f(z) = C0j in D (counting multiplicities).
Let y be a path in Q from co0 to co1 . Again Claim 4 implies that
f_1(y)
n D is
compact. Thus there is a smooth Jordan system r in D such that
f_1(y)
n D c
the inside of T . Hence
r
1
Pi =
27U
f -
COi
r
= n( f o r ; coi)
But the winding number n(f o r ; Q is constant on components of C \ f o r
and co0 and cOj are both contained in y which is contained in the same
component of C \fo T . Hence p
0
= p
:
.
Claim 6 If D is a component of H , then there is a natural number p
such that f is a strictly p-valent map of D onto Q .
This is a direct consequence of combining claims 5 and 1.
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