In complex analysis, the open mapping theorem states that if
U
{\displaystyle U}
is a domain of the complex plane
C
{\displaystyle \mathbb {C} }
and
f
:
U
→
C
{\displaystyle f:U\to \mathbb {C} }
is a non-constant holomorphic function, then
f
{\displaystyle f}
is an open map (i.e. it sends open subsets of
U
{\displaystyle U}
to open subsets of
C
{\displaystyle \mathbb {C} }
, and we have invariance of domain.).
The open mapping theorem points to the sharp difference between holomorphy and real-differentiability. On the real line, for example, the differentiable function
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
is not an open map, as the image of the open interval
(
−
1
,
1
)
{\displaystyle (-1,1)}
is the half-open interval
[
0
,
1
)
{\displaystyle [0,1)}
.
The theorem for example implies that a non-constant holomorphic function cannot map an open disk onto a portion of any line embedded in the complex plane. Images of holomorphic functions can be of real dimension zero (if constant) or two (if non-constant) but never of dimension 1.
Proof
Black dots represent zeros of
g
(
z
)
{\displaystyle g(z)}
. Black annuli represent poles. The boundary of the open set
U
{\displaystyle U}
is given by the dashed line. Note that all poles are exterior to the open set. The smaller red disk is
B
{\displaystyle B}
, centered at
z
0
{\displaystyle z_{0}}
.
Assume
f
:
U
→
C
{\displaystyle f:U\to \mathbb {C} }
is a non-constant holomorphic function and
U
{\displaystyle U}
is a domain of the complex plane. We have to show that every point in
f
(
U
)
{\displaystyle f(U)}
is an interior point of
f
(
U
)
{\displaystyle f(U)}
, i.e. that every point in
f
(
U
)
{\displaystyle f(U)}
has a neighborhood (open disk) which is also in
f
(
U
)
{\displaystyle f(U)}
.
Consider an arbitrary
w
0
{\displaystyle w_{0}}
in
f
(
U
)
{\displaystyle f(U)}
. Then there exists a point
z
0
{\displaystyle z_{0}}
in
U
{\displaystyle U}
such that
w
0
=
f
(
z
0
)
{\displaystyle w_{0}=f(z_{0})}
. Since
U
{\displaystyle U}
is open, we can find
d
>
0
{\displaystyle d>0}
such that the closed disk
B
{\displaystyle B}
around
z
0
{\displaystyle z_{0}}
with radius
d
{\displaystyle d}
is fully contained in
U
{\displaystyle U}
. Consider the function
g
(
z
)
=
f
(
z
)
−
w
0
{\displaystyle g(z)=f(z)-w_{0}}
. Note that
z
0
{\displaystyle z_{0}}
is a root of the function.
We know that
g
(
z
)
{\displaystyle g(z)}
is non-constant and holomorphic. The roots of
g
{\displaystyle g}
are isolated by the identity theorem, and by further decreasing the radius of the disk
B
{\displaystyle B}
, we can assure that
g
(
z
)
{\displaystyle g(z)}
has only a single root in
B
{\displaystyle B}
(although this single root may have multiplicity greater than 1).
The boundary of
B
{\displaystyle B}
is a circle and hence a compact set, on which
|
g
(
z
)
|
{\displaystyle |g(z)|}
is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum
e
{\displaystyle e}
, that is,
e
{\displaystyle e}
is the minimum of
|
g
(
z
)
|
{\displaystyle |g(z)|}
for
z
{\displaystyle z}
on the boundary of
B
{\displaystyle B}
and
e
>
0
{\displaystyle e>0}
.
Denote by
D
{\displaystyle D}
the open disk around
w
0
{\displaystyle w_{0}}
with radius
e
{\displaystyle e}
. By Rouché's theorem, the function
g
(
z
)
=
f
(
z
)
−
w
0
{\displaystyle g(z)=f(z)-w_{0}}
will have the same number of roots (counted with multiplicity) in
B
{\displaystyle B}
as
h
(
z
)
:=
f
(
z
)
−
w
1
{\displaystyle h(z):=f(z)-w_{1}}
for any
w
1
{\displaystyle w_{1}}
in
D
{\displaystyle D}
. This is because
h
(
z
)
=
g
(
z
)
+
(
w
0
−
w
1
)
{\displaystyle h(z)=g(z)+(w_{0}-w_{1})}
, and for
z
{\displaystyle z}
on the boundary of
B
{\displaystyle B}
,
|
g
(
z
)
|
≥
e
>
|
w
0
−
w
1
|
{\displaystyle |g(z)|\geq e>|w_{0}-w_{1}|}
. Thus, for every
w
1
{\displaystyle w_{1}}
in
D
{\displaystyle D}
, there exists at least one
z
1
{\displaystyle z_{1}}
in
B
{\displaystyle B}
such that
f
(
z
1
)
=
w
1
{\displaystyle f(z_{1})=w_{1}}
. This means that the disk
D
{\displaystyle D}
is contained in
f
(
B
)
{\displaystyle f(B)}
.
The image of the ball
B
{\displaystyle B}
,
f
(
B
)
{\displaystyle f(B)}
is a subset of the image of
U
{\displaystyle U}
,
f
(
U
)
{\displaystyle f(U)}
. Thus
w
0
{\displaystyle w_{0}}
is an interior point of
f
(
U
)
{\displaystyle f(U)}
. Since
w
0
{\displaystyle w_{0}}
was arbitrary in
f
(
U
)
{\displaystyle f(U)}
we know that
f
(
U
)
{\displaystyle f(U)}
is open. Since
U
{\displaystyle U}
was arbitrary, the function
f
{\displaystyle f}
is open.
. Black annuli represent poles. The boundary of the open set
U
{\displaystyle U}
, centered at
z
0
{\displaystyle z_{0}}
.
