9.8 JordanBrouwer Separation Theorems
Denition 9.8.1 Suppose A X.We say that A separates X if X rA is disconnected (i.e,
not path connected),or equivalently if
~
H
(X rA) 6= 0.
Terminology:If B is homeomorphic to D
k
then B is called a kcell.
Theorem 9.8.2 Let B S
n
be a kcell.Then S
n
rB is acyclic.(i.e.
~
H(S
n
rB) = 0 8 q.)
In particular,B does not separate S
n
.
Remark 9.8.3 B' and S
n
rfg = R
n
,but in general A'B does not imply that XrA'
X rB.
Proof:By induction on k.
k = 0 is trivial since then B = and S
n
rfg = R
n
.
Suppose that the theorem is true for (k 1)cells.
Let h:I
k
!B be a homeomorphism.
Write B = B
1
[ B
2
where B
1
:= h(I
k1
[0;1=2]) and B
2
:= h(I
k1
[1=2;1]).
Let C = B
1
\B
2
;a (k 1)cell.
Let i:(S
n
rB)!S
n
rB
1
,j:(S
n
rB)!(S
n
rB
2
).
Suppose 0 6= 2
~
H
p
(S
n
rB).
Lemma 9.8.4 Either i
() 6= 0 or j
() 6= 0.
Proof:S
n
rB
1
and S
n
rB
2
are open so they have a MayerVietoris sequence.
(S
n
rB
1
)\(S
n
rB
2
) = S
n
rB (S
n
rB
1
) [ (S
n
rB
2
) = S
n
r(B
1
\B
2
) = S
n
rC.
~
H
p+1
(S
n
rC)
 ~
H
p
(S
n
rB) >
(i
;j
)
 ~
H
p
(S
n
rB
1
)
~
H
p
(S
n
rB
2
)
0
(by hypothesis)
wwww
so either i
() 6= 0 or j
() 6= 0.
Proof of Theorem (cont.):By the lemma,continuing to subdivide we obtain a nested
decreasing sequence of closed intervals I
n
s.t.if we let j
m
:(S
n
rB)

(S
n
rQ
m
),where
Q
m
:= h(I
k1
I
m
),then j
m
6= 0.
By the Cantor Intersection Theorem,\
m
I
m
= a single point feg.
147
H
p
(S
n
rB)

:::

H
p
(S
n
rQ
m
)

:::

H
p
S
n
rh(I
k1
feg)
0
(induction)
www
where we have used that E:= h(I
k1
feg) is a (k 1)cell.Since S
n
rQ
m
is open and nested
and S
n
rE = [
1m=1
(S
n
rQ
m
),H
(S
n
rE) = lim
!
H
(S
n
rQ
m
).
Therefore 7!0 in H
(S
n
rE) implies that j
m
() = 0 in H
(S
n
rQ
m
) for some m,which
is a contradiction.Hence 69 nonzero 2 H
p
(S
n
rB).
Theorem 9.8.5 Suppose h:S
k

S
n
.Then
~
H
i
S
n
rh(S
k
)
=
(
Z i = n k 1;
0 otherwise.
Proof:By induction on k.
If k = 0,
~
H
p
S
n
rh(S
0
)
=
~
H
p
S
n
rf2 pointsg
=
~
H
p
R
n
rfpointg
=
~
H
p
(S
n1
).
p
Suppose that the theorem is true for k 1.
Let E
k
+
,E
k
be the upper and lower hemispheres of S
k
.Notice that by compactness,h is a
homeomorphism onto its image,so h(E
k
+
) and h(E
k
) are kcells.
Also S
n
rh(E
k
+
),S
n
rh(E
k
) are open so MayerVietoris applies.
S
n
rh(E
k
+
)
[
S
n
rh(E
k
)
=
S
n
rh(E
k
+
\E
k
)
=
S
n
rh(S
k1
)
S
n
rh(E
k
+
)
\
S
n
rh(E
k
)
=
S
n
rh(E
k
+
[ E
k
)
=
S
n
rh(S
k
)
0k
~
H
p
S
n
rh(E
k
+
)
~
H
p
S
n
rh(E
k
)
!
~
H
p
S
n
rh(S
k1
)
=
 ~
H
p1
S
n
rh(S
k
)
!
~
H
p1
S
n
rh(E
k
+
)
~
H
p1
S
n
rh(E
k
)
k0
Theorem 9.8.6 (Jordan Curve Theorem) Suppose n > 0.Let C be a subset of S
n
which is
homeomorphic to S
n1
.Then S
n
rC has precisely two path components and C is their common
boundary.(Furthermore,the components are open in S
n
:)
148
Proof:By the preceding theorem,
~
H
0
(S
n
r C)
= Z,so S
n
r C has two path components.
Denote these components W
1
and W
2
.
C is closed in S
n
so S
n
rC is open.Hence by local path connectedness of S
n
,its components
W
1
and W
2
are open.Thus
W
1
W
c
2
.
If x 2 @W
1
=
W
1
rW
1
,then x =2 W
2
(since x 2
W
1
= W
c
2
) and x =2 W
1
.So x 2 (W
1
[W
2
)
c
=
C.Hence @W
1
C.
Conversely let x 2 C.
Let U be an open neighbourhood of x.Show U\
W
1
6=;.Since U arbitrary,it will follow
that x is an accumulation point of
W
1
so that x 2
W
1
.But x 2 C so x =2 W
1
,resulting in
x 2
W
1
rW
1
= @W
1
.
To show U\
W
1
6=;:
U\C is homeomorphic to an open subset of S
n1
(since C
= S
n1
by hypothesis) so it
contains the closure of an (n 1)sphere.Let C
1
be this closure.Under the homeomorphism
C
= S
n1
,C
1
= N
r
[x] for some r and x.Thus C
1
C is an (n 1)cell.Let C
2
=
C rC
1
.
Then C
2
is also an (n 1)cell (up to homeomorphism it is the closure of the complement of
N
r
[x] in S
n1
) and C
1
[C
2
= C which is closed.By Theorem 9.8.2,C
2
does not separate S
n
so
9 path in S
n
rC
2
joining p 2 W
1
to q 2 W
2
.(Im)\(
W
1
rW
1
) =
1
(
W
1
)r
1
(W
1
)
.If
this is empty then
1
(
W
1
) =
1
(W
1
).However the equality of these open and closed subsets
of I means that either (W
1
) =;or
1
(W
1
) = I.We know
1
(W
1
) 6=;since 0 2
1
(W
1
)
(since p = (0) 2 W
1
).And 1 =2
1
(W
1
) since q =2 W
1
.Therefore (Im)\(
W
1
rW
1
) 6=;.
Thus 9y 2 (Im)\(
W
1
r W
1
) @W
1
C = C
1
[ C
2
.Since Im S
n
r C
2
,y =2 C
2
so
y 2 C
1
U.Hence y 2 U\
W
1
.
p
So @W
1
= C.Similarly @W
2
= C,as desired.
Corollary 9.8.7 (Jordan Curve Theorem  standard version):Supppose n > 1.Let C be
asubspace of R
n
which is homeomorphic to S
n1
.Then R
n
rC has precisely two components
(one bounded,one unbounded  known as the\inside of C"and\outside of C"respectively)
and C is their common boundary.
Proof:Include R
n
into S
n
,writing R
n
= S
n
= fPg.Then S
n
r C is the union of two
components W
1
,W
2
whose common boundary is C.One of the components,say W
1
contains P
so W
1
rfPg,W
2
are the components of R
n
rC and their common boundary is C.
Theorem 9.8.8 (Invariance of Domain):Let V be open in R
n
and let f:V!R
n
be contin
uous and injective.Then f(V ) is open in R
n
and f:V!f(V ) is a homeomorphism.
Remark 9.8.9 Compare the inverse function theorem which asserts this under the stronger
hypothesis that f is continuously dierentiable with nonsingular Jacobian,but also asserts
dierentiability of the inverse map.
149
Proof:
Include R
n
into S
n
.Let U be an open subset of V.Let y 2 f(U).We show that f(U)
contains an open neighbourhood of y.
Write y = f(x),Find s.t.N
[x] U.Set A:= N
[x] r N
(x).So A is homeomorphic
to S
n1
.Since f
N
[x]
U is a homeomorphism (an injective map from a compact set to
a Hausdor space),f(A) is homeomorphic to S
n1
.Therefore f(A) separates S
n
into two
components W
1
and W
2
which are open in S
n
.
N
(x) is connected and disjoint from A,so f
N
(x)
is connected and disjoint from f(A).
Thus f
N
(x)
is contained entirely within either W
1
or W
2
.Say f
N
(x)
W
1
.
S
n
rf(A) rf
N
(x)
= S
n
rf
A[ N
(x)
= S
n
rf
N
[x]
(which the later argument will show is equal to S
n
rW
c
2
= W
2
).Since f
N
[x]
is an ncell,it
does not disconnect S
n
,i.e.S
n
rf
N
[x]
is connected.Because f
N
[x]
W
1
W
c
2
which
is equivalent to W
2
S
n
rf
N
[x]
,we get W
2
= S
n
rf
N
[x]
(as remarked earlier),since
W
2
is a path component of S
n
.Hence f
N
[x]
= W
c
2
=
W
1
.Thus f
N
(x)
= W
1
.(i.e.If
z 2 W
1
rf
N
(x)
then z 2 S
n
rf(A) rf
N
(x)
= S
n
rf
N
[x]
= S
n
rW
c
= W
2
,which
contradicts W
1
\W
2
=;.)
Therefore we have shown that 9 an open set W
1
s.t.y 2 W
1
f(U) and thus f(U) is
open.Applying the above argument with U:= V gives that f(V ) is open.It also shows that
f:V!f(V ) is an open map,so it is a homeomorphism.
150
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