1999, polished up 2007.
Theorem Let Y be a connected open subset of the complex plane,
and let B be the closed ball of radius 1 centered at the origin.
Suppose B is not entirely contained in Y. Then for any point x in
Y−B,
the homomorphism from π1(Y−B,x) to
π1(Y,x) is injective.
Proof
If this homomorphism isn't injective, there's a loop in
Y−B that's homotopic to a constant
in Y, but not in Y−B. The homotopy h can be considered as
a map of the closed unit disc E2 to Y.
The loop has a distance of closest approach d to B.
Define B' to be the ball of radius R = 1 + d/2. Then the loop
is in π1(Y−B').
Suppose first that Y contains the boundary of B'. Then
ε > 0 can be found such that
{reiθ : R−ε < r < R+ε}
⊂ Y.
Apply the
Seifert-Van Kampen theorem to the open sets
U = Y ∩ {reiθ: r
> R−ε}
and
V = Y ∩ {reiθ: r < R+ε}.
If U were
not connected it would be the union of two disjoint open sets U1
and U2 with
{reiθ : R−ε < r < R+ε}
⊂ U1, and U2 would be disjoint from
U1 ∪ V; so Y would not be connected. So U
is connected. Similarly V is connected, and because U and V are open
sets in
the plane, they are path connected.
The maps
i1: π1 (U ∩ V) →
π1(U) and
i2: π1 (U ∩ V) →
π1(V) are injections, since
π1 (U ∩ V) is generated by a circular
path that has winding number 1 around all points
reiθ with r ≤ R - ε.
Since int(B') ⊄ Y and U ∩ V ⊂ Y,
Y doesn't contain all such points. The
Seifert-Van Kampen theorem says that
π1(U ∪ V)
= the free product π1(U) * π1(V),
mod the relations {i1(k) = i2(k), ∀
k
in U ∩ V}.
Also π1(U) = π1(U) *
π1(U ∩ V)/ {i1(k) = k},
applying Seifert-Van Kampen to U and U ∩ V.
Since i2 is injective,
π1(U) *
π1(U ∩ V)/ {i1(k) = k}
is isomorphic to the image of π1(U) in
π1(U) *
π1(V)/ {i1(k) = i2(k)}.
So the map
π1(U) → π1(U ∪ V) is
injective.
π1(Y−B') → π1(U) is
injective, so that proves the loop can't be homotopic to a constant if
bdry(B') ⊂ Y.
So
assume that bdry(B') ⊄ Y.
Consider the connected component U containing the boundary of
E2 in h−1 (Y−B').
E2−U is a closed set.
Let {Vi} be
the connected components of int(E2−U).
The boundary of a Vi isn't empty since E2 is
connected.
Lemma: The boundary of a Vi is connected.
Proof:
If a boundary point of a Vi has a neighborhood with no
intersection
with U, the neighborhood is contained in
int(E2−U). So the neighborhood and
Vi
would be in the same connected component of
int(E2−U). So
bdry(Vi) = cl(U) ∩ cl(Vi).
Suppose cl(U) ∩ cl(Vi) isn't
connected. Then it's a union
of two disjoint sets, which are closed in
cl(U) ∩ cl(Vi), and so also closed in
E2. Let
b1 and b2 be points from the two closed sets.
Here is an illustration of the
proof of the lemma.
By the
Urysohn lemma, a function f can be defined on E2 with
f = 1 on the
closed subset of bdry(Vi) containing b1,
and
f = −1 on the
closed subset of bdry(Vi) containing
b2.
U and the Vi's are path connected because they are connected and
open. Since f = −1 on part of
bdry(Vi) = cl(U) ∩ cl(Vi) and
f = 1 on the rest of bdry(Vi),
f = 0 on some point in U, and f = 0
on some point in Vi.
But f isn't 0 on any point of
bdry(Vi). So f −1(0) is a
union of two disjoint closed sets,
one contained in Vi and one contained in
int(E2 − Vi).
So we can use the Urysohn lemma again to define
g: cl(E2 − Vi) →
[0,1] with
g = 1 on the subset of
f −1(0) in
int(E2 − Vi), and
g = 0 on
bdry(Vi). Also define
g: cl(Vi) → [−1,0]
with g = −1 on the subset of
f −1(0) in Vi, and
g = 0
on bdry(Vi).
Since the two definitions agree on a closed set, g is continuous.
Now, define F: E2 → C by F(x) = f(x) + ig(x).
F is continuous,
and uniformly continuous because it's a function from a compact
metric space into a metric space. F is never 0 on E2, so
F'(x) = F(x)/|F(x)| is a (uniformly) continuous map from
E2 → bdry(E2).
Find a point u1 in U within a small distance d of b1,
and a point u2 in U within a distance d of b2.
Similarly, find a point v1 in Vi within a distance
d of b1,
and a point c2 in Vi within a distance d of
b2.
Since F'(b1) = f(b1) = 1 and
F'(b2) = f(b2) = −1, if d
is small enough
f(u1), f(v1) > 0 and
f(u2), f(v2) < 0, and also
F' on the straight line from u1 to v1 doesn't
wrap around the origin, and F' on the straight line
from u2 to v2 doesn't wrap around the origin.
Define a path P1 in U from u1 to u2,
and a path P2 in Ci from v2 to
v1.
Then there is a loop L from u1 to u2 via P1,
from u2 to v2
by a straight line, v2 to v1 via P2, from
v1 to u1 by a straight line.
F'(P1) is a path from F'(u1)
in the top right quarter-circle through i to
F'(u2) in the
top left quarter-circle, with the
imaginary part never negative.
F' on the straight line from u2 to v2 is a
path from F'(u2) in the top left
quarter-circle to F'(v2) in the
bottom left quarter-circle,
which never has positive real part.
Similarly F'(P2) is a path from
F(v2) in the bottom left quarter-circle
to
F'(v1) in the bottom right
quarter-circle,
with the imaginary part never positive; and F' on the line from
v1 back to u1 is a path from
F'(v1) in the bottom right
quarter-circle
to F'(u1) in the top right quarter-circle, with
non negative real part.
So F' is a map of E2 onto bdry(E2), which
isn't possible. So bdry(Vi) is connected. ∴
Let {Bj} be the connected components of
bdry(U) in
E2.
The Bj's are contained in h−1
(Y ∩ bdry(B')).
Since bdry(Vi) is connected, it is contained in some
Bj. The Bj's are closed, being the
components of a closed set.
Some part of the boundary of B' isn't in Y.
So, Y ∩ bdry(B') is a union of disjoint open
intervals of bdry(B').
Now define a closed set Aab for each connected component
(a,b) of Y ∩ bdry(B') :
Aab = cl (∪ Bj such that
Bj ⊃ bdry(Vi) for some
Vi and h(Bj) is contained in
(a,b))
(a,b) is a closed set in the topology of Y, since
(a,b) = [a,b] ∩ Y.
And h−1(Y ∩ bdry(B')) is closed in
E2,
so the h−1((a,b))'s are closed sets which don't
share connected components of
h−1(Y ∩ bdry(B')).
Since h−1((a,b)) is closed,
h maps Aab into (a,b).
Lemma: Define
Cab = Aab ∪ Vi such that
bdry(Vi) ⊆ some Bj in
Aab. Then Cab is closed.
Proof: Let p be a limit point of Cab. If every
neighborhood of p intersects some Bj in Aab, then
p is in Aab. So suppose a neighborhood N of p doesn't
intersect Aab. Then N intersects a Vi whose
boundary is in Aab. If p is not in Cab, p is in
int(E2 − Vi), so a neighborhood
N1 ⊂ N of p is contained in
int(E2 − Vi). If
N = (N ∩ Vi) ∪
(N ∩ int(E2 − Vi)), it
would be a union of disjoint open sets, so not connected. So N contains
a point of bdry(Vi), so it intersects
Aab.
So Cab is closed.
By the
Tietze extension theorem, h' can be defined sending
Cab → R, with h' agreeing with h on
Aab. Since Aab is compact, h
actually sends Aab into a closed interval [c,d] contained in
(a,b). There is a continuous function
j: R → (a,b) with j(x) = x, ∀ x in [c,d].
Then jh' : Cab → (a,b) and
jh'(x) = h(x), ∀ x ∈ Aab.
Lemma: Define h' on E2 agreeing with h on cl(U),
and h' on the Vi's from the Tietze extension theorem
applied to each Cab, so that h'
doesn't send any of E2 into int(B').
Then h' is continuous on E2.
Proof:
h' is continuous on U and on the Vi's. Suppose x is in
bdry(U), so that h'(x) = h(x) ∈ (a,b) ⊂
Y ∩ bdry(B').
Because h' = h on bdry(U), a
δ-ball around x can be found small enough so that
h'(bdry(U) ∩ δ-ball) is contained in
(a,b).
Suppose a Vi intersects this δ-ball. If
bdry(Vi) ∩ δ-ball = ∅ then
δ-ball = (δ-ball ∩ Vi) ∪ (δ-ball
∩ int(E2 − Vi)), so it would be
disconnected. So bdry(Vi) intersects the
δ-ball. So
h'(bdry(Vi)) ⊂ (a,b),
so bdry(Vi) is contained in Aab and
Vi is contained in Cab. Since
h' = h on cl(U), and h' is defined by the
Tietze extension theorem
on the closed set Cab, and
cl(U) ∪ Cab ⊃ δ-ball, by the
glueing lemma h' is continuous at x.
h' is a homotopy of the loop to a constant and h' doesn't go into
int(B'). But then the loop is homotopic to a constant
in π1(Y−B); remember, B is the ball of
radius 1. ∴!
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