Serge Ballif
MATH 502 Homework 7
March 21, 2008
(1)
(The inverse function theorem) Let
U
and
V
be open subsets of
C
and let
f
:
U
→
V
be a holomorphic bijection.
Show that
f
is a homeomorphism (that is, show that the inverse function
f

1
is continuous; use the open mapping
theorem). Show further that
f
is a biholomorphic equivalence (that is, show that
f

1
is differentiable).
Since
f
is a bijection, the inverse function
f

1
:
V
→
U
exists and is also a
bijection. For simplicity of notation, rename
f

1
=
g
.
Claim 1.
g
is continuous.
Proof.
Since
f
is a bijection, each subset of
U
can be written in the form
g
(
W
) for
some unique subset
W
⊂
V
. Note that
f
is a nonconstant holomorphic map on
an open set. Thus,
f
is an open map (by the open mapping theorem). Therefore,
if
g
(
W
) is an open subset of
U
,
g

1
(
g
(
W
)) =
f
(
g
(
W
)) =
W
is an open subset of
V
. Therefore,
g
is a continuous function.
Claim 2.
g
is diﬀerentiable.
Proof.
At each
w
∈
U
,
f
has some power series representation
f
(
z
) =
f
(
w
) +
c
1
(
z

w
) +
c
2
(
z

w
)
2
+
···
because
f
is holomorphic on
U
. Some algebraic
manipulation yields the equation
f
(
z
)

f
(
w
)
z

w
=
h
(
z
)
(
*
)
for some power series
h
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 Summer '08
 JOHNRIDENER
 Math, Topology, Sets, Continuous function, Metric space, Topological space, biholomorphic equivalence

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