Proof It follows from Proposition 1.3 that the identity theorem
tion 1.2) holds for meromorphic functions on X too. Hence an /
f ^ 0, has only isolated zeros. Therefore 1/f is meromorphic on X
Proposition 1.5 (Local form of holomorphic maps). Let X, Y be
surfaces and f : X — F a holomorphic map that is not constant, w
and b :— f(a).
There are then a natural number k 1, a chart ip : U —
around a with p(a) = 0, and a chart t/; : U' — V ofY with ip(
f(U) C U', such that the map F := ip o / o
: V —
in terms of the coordinate z inV.
Proof. Let p\ : XJ\ -» Vi, ip : U' — V' be charts of X resp. y w
(pi(a) = 0, 6 £ J/', ^(&) = 0 and /(C/i) C
Consider the function
By the identity theorem /i is not constant. Since /i(0) = 0, there is
k 1, such that
for a holomorphic function g : Vi — » C with g(0) / 0. Choose £
that for all z £ C with \z\ £ we have z E V\ and |g(^) — g(0)|
the ^-neighborhood of 0 there is a holomorphic function h with h(z)
(i.e., a branch of yfg(z) is defined).
There are open neighborhoods V2, V of 0 such that the map
a:V2 - • V
Z H-• z / i ( z )
is holomorphic. Let U := (/^(V^), p = a ° ¥i|i/ : U — V. By c
of the map F := ip o / o
we have F(^) — 2^.
As in Remark 1.3 one can show that the number k is independe
choice of charts p\ : U\ — Vi and xji :
It follows from this proposition that there is a neighborhood U
that each y £ f(U) with y 7^ 6 has exactly k preimages under / .
Definition. We call the number k in Proposition 1.5 the order or
ity of / at the point a. Symbolically, k — ordaf.