G.1
Normality of quotient spaces
For a quotient space, the separation axioms--even the
are difficult to verify.
ausdorff property--
We give here three situations in which the quotient
space is not only Hausdorff, but normal.
Theorem
G.1.
is normal, then
Proof.
p: X-pY
be a closed quotient map.
X
containing
A
such that
p
A
*(A),
p
The proof is easy.
(U)
is a subset of
is contained in
The set
C = X-
N
so that the set
of
If
x
no point of
C,
A.
p
so that it lies in
be disjoint closed sets of
p
(B)
N2
of
and
X
containing them.
B,
sch that
N1
and
Definition.
X;
and let
and
Y
N2
p
Let
f
Now the map
X
and
Zf
Y
p
U,
(p(C))
f
(C)
is an
p
U. of
The set
then
(U)
Y
p(C)
is
is an open set
p
(x)
contains
is contained in
Y.
NOW let
is continuous,
p
A
(A)
N,
and
U2
N1
beIopen sets of
and
U1
p' (U2 )
and
Y
X
and
and
Choose disjoint open sets
N1
and
containing
lies in
N2'
U2 0
be disjoint spaces; let
be a continuous function.
A
be a closed subset
We. define the
to be the quotient space obtained from the union of
(f(a)).
ply
Let
a
of
p: XU Y
A
with the point
Zf
C
and
If
f
is closed in
is closed in
XY,
C
(C).
Y
C
Y,
Zf
then
is closed in
A
and hence closed in
s
that
p(C)
f(a) and with all
into
is a closed set of
The set
X
be the quotient map.
is a continuous injection of
that it is also a closed map.
set
p
are disjoint, so are
Let
equals the union of
X.
lies in
by identifying each point
the points of
the
thus
Since
U1
(U1 )
f: A-SY
adjunction space
N
Since one-point sets are closed in
are disjoint closed sets of
Because
of
N;
Y.
Y.
ad
p(C)
is a closed map, one-point sets are closed in
B
A
U = Y-
is a point of
Now we verify normality of
and
X
N.
is closed.
A,
that contains
Y,
then there is an open set
closed and disjoint from
Y
If
is normal.
First we show that if
open set of
containing
Y
Let
X.
is closed in
Y,
We show
p- (p(C))
so
Therefore,
Zf,
by definition
of the quotient topology.
It now follows that
is a homeomorphism of
Y
p(Y)
is a closed subspace of
with
p(Y).
Zf,
and that
ply
G.2
X
Theorem G.2.
Proof.
If X and
Y
are notial, then so is
Zf.
A direct proof, using the definition of normality, is a bit
elaborate.
(See [D], p. 145.)
An easier proof uses the Tietze theorem,
as we now show.
First, we note that
belongs to
p(Y), then
Zf is T 1.
z3}
Let
z be a point of
Zf.
If z
is closed because one-point sets are closed in
Y, and p)Y : Y-->Zf is a closed map. Otherwise, p-l (z) is a one-point set
in X, and therefore closed; it.follows from the definition of a quotient map
that
z) is closed.
Now let
and
B
CX = p
by setting
and
C be disjoint closed sets of
(C)nX.
Similarly, let
h = g AX o
-
A,
C
By = p
and h = 0
on
(B) Y
BX ,
and
7f.
and
O
on By
Using normality of
k : X.
Y, choose a continuous function
and
X
1 on
Cy.
into
Then
g
[0,1];
and
g : Y -[,]
Then define
h
X
ad h=continuous1
on C
and h is unambiguously
is continuous, by the pasting lemma.
and the Tietze theorem, extend
[-O1].
XUY
.
4
defined when two of the sets overlap,
function from
XC
C
byfunsetting h = gof on A [0,1]h
andto = Oinduces
on
Because each of these three sets is closed in
function
on
/ 2
Using normalilty of
that equals
BX = p(B) X
Cy = p- (C) Y.
h = 1
_
Y -&
Let
k
h
to a continuous
together define a continuous
it induces a continuous function
G.3
F: Zf-;[O,1]
on the quotient space that equals
F
([O,½))
and
F
((½,1])
0
on
B
and
1
on
C.
TLe sets
are then disjoint open sets about
B
and
C,
I
respectively.
Oe application of adjunction spaces occurs in point-set topology, when
one is studying absolute retracts.
(See Exercise 8, p.224.)
Another application
occurs in algebraic topology, when one constructs a CW complex; we will discuss
this application shortly.
Definition.
of
X
Let
whose union is
sitbspaces
X, if a set
for each
[.
is open in
X
be a space and let
X.
A
for each
be a family of subspaces
The topology of
X
is closed in
whenever
X
(Or, equivalently,if a set
X
Xj
U
is said to be coherent with the
A
XC
is open in
X
is closed in
XX
Un XO
whenever
.)
There is a strong connection between coherent topologies and quotient
spaces.
It is described as follows:
To begin, let us give
topology and consider the product space
of
XX J
X
that is the union of the subspaces
J.
J
the discrete
Tlen we consider the subspace
z J.
X4X [41, for all
This
space is called the topological sum (or sometimes the disjoint union) of the
spaces
X.
It is denoted
JX
.
If we project
Xx J
onto X,1
we obtain
a continuous map
p:
which maps each space
p
X
X
XpZh by the obvious homeomorphism onto
is a quotient map if and only if the topology of
subspaces
X
.
subspaces
X
, then a map
of the functions
It follows that if
fXK
Theorem G.3.
subspaces
Let
Xi , for
these subspaces.
i
X
f : X--Y
X
X
. The map
is coherent with the
has the topology coherent with the
is continuous if and only if each
is continuous.
X
Z.
If each
be a space that is the union of countably many cosed
Suppose the topology of
Xi is normal, then so is
1
X
X.
is coherent with
G.4
Proof.
i,
so
p
is a point of
is closed in
p}
Let
for
If
A
arnd
n>O,
B
X.
is closed in
~Pn X
X
Xi
for each
is a T1 space.
be closed disjoint sets in
X.
IDefine
YO
=
AU B, and
define
AVBUXl1 ...UX
f : - YO - >[C0,1] by letting it equal 0
Define a continuous function
1
thlen
Therefore
Yn
and
X,
on
B.
on
A
In general, suppose one is given a continuous function
].
fn : Yn-[- ,
The space Xn+l is normal and Yn Xn+l iS closed in
Xn+
If gn denotes the restriction of fn to the subspace YnCIXn+l,
1.
we use the Tietze theorem to extend
g: X+
1
-
[0,1]
functions
Because
f
and
n
g
Yn
to a continuous function
g
ar:d
Xn+ 1
are closed subsets of
Yn+l'
the
combine to define a continuous function-
fn+l: Yn+l-
[ °'
1
C)
l]
that is an extension of
f . Te; functions f
in turn combine to define
n
n
f: X-->[0,1] that equals 0 orn A ar;d 1 on B. Because X
a function
has the topology coherent with the subspaces
Given an element Sk of
x
all elements
in
Sa
f
is continuous.
of
S
S
let
,
x <
such that
be the subspace consisting of
X
.
lTen
is a closed interval
X.
so it is compact.
The space
S x
S
We show that
SL, each of which
X
iS the union of the spaces
is compact Hausdorff and thus normal.
S - SD- , which is not
normal, has the topology coherent with these subspaces.
Let
U
be a subset of
subspace, for each 4.
such that
S xS
U
(X x s
is open in this
T-,en the intersection
U n (SX Sl )
is open in
Sax Sfor each
,
and hence open in
the union of the sets
Un (
it is open in
EJ
The preceding theorem does not extend to uncountable coherent
Example 1.
unions.
the map
Xn ,
S
Sag,
SQ ),
as desired.
S
S
Since
U
is
G.5
It is an interesting question to ask under what conditions coherent
topologies exist.
One has the following two theorems:
Theorem G.4.
spaces
X
,
Let
for
X
be a set that is the union of the topological
o(C-J.
If there is a topological space
as its underlying set, such that each
is a topological space
the topology of
X
X
Proof.
in
X
is a subspace of
such that each
X,
D
to be closed in
It is immediate that
.
XT ,
then there
X.
X
and
The topology of
XT
.
We define a set
for each
having X
is a subspace of
is coherent with the subspaces
is finer than that of
XC
X~
XT
0
Xc
and
if
Dn X
is closed
are closed.
X
The
required properties about unions and intersections follow from the equations
4.,
E
is closed in
so that
E
of
XC .
Isn't this obvious?
in
XC ,
then
A
is closed in
for some set
X
A
that of
XTI
some
closed in
he set
Theorem G.5.
X; ,
for
(J..
is closed in both
A
XO
is closed in
Thus the topology of
X.
Xc
First, note that if
Xi by definition.
Xo4 is a subspace of
XT.
X4
XC
XT ,
Thus
.
A
is closed
Conversely, suppose
we have
aBcause the topology of
is also closed in
is a subspace
XC
B = A
Xi
is finer than
B = A AX4
for
XC) as desired. 0
Let
X
be a set that is the union of the topological spaces
If for each pair of indides
X, aad
from each of them, then
Each
Not quite.
BEcause
closed in
X.
E
We must prove that each
is closed in
X.
then
XT
.
What else is there to prove?
A
,
)
),
(ru X
XT ,
is closed in
is finer than the topology of
B
X)U *...V (Dn X
is an arbitrary collection of closed sets.
Note that if
for each
ID eb
=
(R LAX Djox,
where o
(D 1
DrI) nAXL =
U.
(D1V
X
Xa is a closed set in
d,,
the set
XAn XA
, and inherits the same topolcxy
X
has a topology coherent with the subspaces
X
in this topology.
X.
G.6
set
to be closed in
D
is closed in
A
B
Convarsely, let.
AnXi
then
X,
be a
BnX
To do this, we must show that
is a subspace of
X
a closed subspace of
;
B
Then
X
XA
B
the set
X,
i. closed in
B
nX
X1 by definition of
is closed in
X ;
closed set of
X
is closed in
Xn X
X
Since
s.
for each
is closed in
X.
XC .
is closed in
B
we show
is closed in
First,
XC.
XA is a closed subspace of
We show that each space
It is
i.
for each
X
X.
immediate that this is a topology on
if
is closed in
DhXd
if
XC
by declaring a
XC
Once again, we define a topological space
Proof.
because the latter
because
X
Xd
is
a
This theorem does not hold if the word "closed" is omitted from the
that is the union of three
X
There is an example of a set
hypothesis.
spaces such that the intersection of any two of the spaces is a subspace
of each of them; but there is no topology on X
(ee
of the spaces are subspaces!
Example 2.
R
subspace of
R
[Mu], p. 213.)
in the product topology.
consisting of all points
inn.
for
Xi = 0
Consider
at all of which all three
Fn
1
such that
x = (x],x 2 ,...)
...
each of which is a closed subspace of the next.
Their union is
by Theorem G.5 has a topology coherent with the subspaces
Theorem G.3 implies that IRl
inherits as a subspace of IR6
other
,
n.
topologies as well, ones that it
in its various topologies.
The subset
inherits its usual topology from each of these topologies on
subspaces
R
is normal in this topology.
Now 1RO also has several
Theorem
denote the
Then one has the sequence of subspaces
,
R C 2C
which
Let
G.4 also applies to show that
R
Re.
R
Hence
has a topology coherent with the
IR ; this theorem also implies that the coherent topology is finer
than each of these topologies on
R.
Since the one derived from the box
topology is the finest of these, one has the following:
Challenge question:
subspaces
IR
Rn
Is the topology on
tlhe same as the topology that
in the box topology?
R
that is coherent with the
IRO inherits as a subspace of
G.7
Here is a quick outline of how these notions are used in
Final remark.
algebraic topology. (See §38 of [Mu] for more details.)
is the subset of
n
The unit ball in
Rn
consisting of all points
whose euclidean distance from the origin is less than or equal to one; the
unit sphere consists of those points for which this distance equals one.
spaces are denoted
h: Bn-->C,
then
Bn
and
Si
, respectively.
If there is a homeomorphism
Bd C
is called an n-cell, and we denote by
C
These
the subspace
h(Sn-).
There is a class of spaces that is very important in algebraic topology
called CW complexes; they were invented by J.H.C.Whitehead.
In algebraic
topology, one defines for a given space a number of groups, such as the homology
H (X), the cohomology groups
groups
Hn(X), and the homotopy groups
InX).
Defining is one thing, but computing (or even getting useful information)
The structure of CW complex gives one a tool for dealing with these
is another.
Ci complexes are quite versatile--many useful spaces, such as the
groups.
Grassman manifolds we shall mention shortly, have the
complex.
structure of a CW
Or! the other hand, if one has some prescribed groups and wants
to find a space for which the homology groups, say, are isomorphic to these
groups, one can construct a CW complex that will do the job.
A CW complex is constructed as follows:
f_ :Bd C -X0.
and a family of continuous maps
sum
C
, and uses the maps
So far, so good.
and a continuous map
rD, and
X1
X0 o
C
f :
Bd C
-
X0
by means of this map.
it is called a 1-dimensional CW complex.
X1;
Now one takes a collection
g:
,
Ole forms the topological
to define a continuous map
f,
One forms the adjunction space obtained from
This space is denoted
C
Then one takes a collection of disjoirt i-cells
X .
space, which we call
One begins with a discrete
2 Bd D
X1
-
bI, means of this map.
D
of disjoint 2-cells
, and forms an adjunction space from
This space is denoted
and is a
X
2-dimensional CW complex.
One has eventually an n-dimensional CW complex
It is clear how to continue.
X n,
for each
n.
Are we finished?
of the adjunction space Xn ,
n
X-1
X.
construction
Recall that in the
the projection map defines a homeomorphism of
with a closed subspace of
closed subspace of
No.
X.
We normally identify
n -
X
with this
G.8
With this convention, we now have a sequence of spaces
X1 C
XO
C X '-...
...
each of which is a closed subspace of the next.
Their union is given the
topology coherent with these subspaces; it is called an (infinite-dimensional)
CW complex.
In order to work with the space we obtain, it is essential that it be
a Hausdorff space.
(Basically, so we know that compact sets are closed.)
The theorems we have proved in this section do much more than that; they show
that every
CVT complex is normal.
Final final remark.
We have talked a lot about how quotient spaces are
used in algebraic topology.
Let us close by giving an example of how they are
used in differential geometry.
A very important space in differential geometry is the space
n
k-dimensional vector subspaces of
R .
of k-planes in n-space.
The space
through the origin in
.
R
Gnk
of
It is called the Grassman manifold
G
is thus the space of all lines
n,1
It is fairly intuitive what one means by saying that
two k-planes are "close" tc one another, but how does one topologize this space
rigorously?
One topologizes it as a quotient space.
To be specific, let
V
be the set of all k by n matrices, where k n,
n,k
whose rows are orthonormal vectors. Such a matrix satisfies the equation AA t = Ik
There is an obvious topology on this set, for it can be considered as a
subspace of
R
.
as the equation
In this topology, it is compact, for it is closed and bounded,
AA
= Ik
shows.
Let
p: Vk
Gn
be
k
sends each matrix to the k-dimensional vector subspace of
spanned by its rows.
We topologize
n
k
by requiring
the map that
JR
p
that is
to, be a quotient
map.
Because
p
is continuous, it is immediate that
next question is this:
map
p
Is: it Hausdorff?
is compact.
The answer is "yes,"
The
because the
is in fact a closed map, so Theorem G.1 applies.
To show
p
two matrices
is closed, we examine first what the relationship is between
A' and
B
whose rows span the same vector subspace of
This occurs precisely when each row of
the rows of
A = CB,
Gn,k
B, and conversely.
where
C
A
Rn
equals a linear combination of
This statement can be expressed by the matrix equation
is a nonsingular k by k matrix. It follows that
C
satisfies the
G.9
equation
CC
= Ik,
which means that
C
belongs
computation with matrices verifies this fact:
AA t = BB
with the equations
.
ABt = C
= Ik ,
ard
to
Vk,k'
A quick
The equation
A.= CB, along
implies that
t
Ik = C(BAt).
t
BA The first equation ives us, by transposing, the equation
t
C
substituting this result into the second equation gives us the equation
I k = CCt.
Now we show that
p
then the set' p-l(p(S))
C
belongs to
image of
Vk
Vk kX S
is compact
ard
S
If
S
is a closed set in
is the set of all matrices of the form
and
k
is a closed map.
A
is an element of
S.
Thus
p
CA,
(p(S))
under the map given by matrix multiplication.
V
where
is the
Now
Vk1
is compact (being closed in
Vn). Their cartesian
n,kI
product is compact, so the image under matrix multiplication (which is
continuous) is also compct and therefore a closed
bset of
By definition of the quotient topology, it follows that
in
G
V
n,k
p(S) is closed
as desired.
k
There is of course a great deal more to say about Grassman manifolds.
The space
G
is in fact a manifold (as the terminology implies); it is a
manifold of dimension
It we replace
of
n
into
n+l;
k(n - k).
IRn
throughout by
iR ,
then there is the obvious inclusion
it gives rise to an incllsionmap of Vn,
into
Vn+l
.
k
This in turn induces a continuous injective map on the quotient spaces
i: G
-G+,
k
Since all the spaces involved are compact Hausdorff, we can thus consider
G
to be a closed subspace of
n,k
the spaces
Gk,k
one has the space of all
the coherent topology.
k planes
G
n+lk
If now we take thfe union of
G+l,k ...
in
Arid Theorem G.3
IR3 .
Gn,k
.
)
As you would expect, we give it
implies that this space is normals