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.