Adjunction spaces and k-spaces
by Dale W Behrens
A thesis submitted to the Graduate Faculty in partial fulfillment of the requirements for the degree of
DOCTOR OF PHILOSOPHY . Mathematics
Montana State University
© Copyright by Dale W Behrens (1973)
Abstract:
Let X ⊂f Y denote the adjunction space of X and Y via f:A → Y where A is a closed subset of X and
let p:X + Y → X ⊂f Y be the identification from the disjoint union of X and Y to the adjunction space.
Let kX be the k-space associated with Hausdorff X. The question of when k distributes through the
adjunction is settled for the following cases. An example is given for which k(X ⊂f Y) ≠ kX ⊂kf kY
where X ⊂f Y = X/A (Theorem 2.2). It is proven that if p is compact covering (Theorem 2.4), for
example when X is paracompact (Corollary 2,8), then k(X ⊂ Y) = kX ⊂kf kY. Let βX be the
Stone-Cech compactification of completely regular X. Using the fact that β distributes over the
adjunction, (β (X ⊂f Y) ⊂ βX ⊂βf βY), whenever X is normal and whenever Y and X ⊂f Y are
completely regular, it is shown that k will distribute through the adjunction by enlarging the space X,
that is, there exists a space X1 where X ⊂ X1 ⊂ βX such that k(X ⊂f Y) ⊂ kX1 ⊂kg kY where g
extends f from A to A1 ⊂ X1. ADJUNCTION SPACES AND k-SPACES
by
DALE W BEHRENS
A t h e s i s s u b m itte d t o th e G ra d u ate F a c u lty i n p a r t i a l
f u l f i l l m e n t o f th e re q u ire m e n ts f o r . t h e d eg ree
of
DOCTOR OF PHILOSOPHY ' . :
M a th e m a tic s
A p p ro v e d : /
H ead , M a jo r D e p a r tm e n t
(flUxeAl 'ff!
C h a irm a n , Exam lrfing Com m ittee
G r a d u a te Dean
MONTANA STATE UNIVERSITY
Bozeman, M ontana
■ ■ ' ' D ecem ber, 1973
ill
ACKNOWLEDGMENT
I w o u ld l i k e t o g i v e s p e c i a l t h a n k s t o my t h e s i s
a d v i s o r D r 0 R i c h a r d M. G i l l e t t e f o r h i s many v a l u a b l e
s u g g e s t i o n s and. t o t h e m a t h e m a ti c s d e p a r t m e n t a t
W e s te r n W a s h in g to n S t a t e C o l l e g e w h ic h e n c o u r a g e d me t o
c o n t i n u e g r a d u a t e w ork i n m a t h e m a t i c s .
Iv
TABLE OF CONTENTS
Page
ABSTRACT. .
v
INTRODUCTION.. .
I
C h a p te r
2 . . . . ......... ............................................................................ * . . . .
3 . . . . ........................................................... .............................. ..
BIBLIOGRAPHY................................................ ....................... > --------. . . .
IO
18
28
V
ABSTRACT
Let X
Y d e n o t e t h e a d j u n c t i o n s p a c e o f X and Y v i a
f : A — > Y w h e re A i s a c l o s e d s u b s e t o f X an d l e t
• •
p :X + Y — > X U p Y b e t h e i d e n t i f i c a t i o n fro m t h e d i s j o i n t
u n io n o f X and Y t o th e a d ju n c ti o n sp a c e .
L e t kX b e t h e
k - s p a c e a s s o c i a t e d w i t h H a u s d o r f f X. The q u e s t i o n o f when
k d i s t r i b u t e s th ro u g h th e a d ju n c tio n i s s e t t l e d f o r th e
fo llo w in g c a s e s .
An exam ple i s g i v e n f o r w h ic h
k(X U f Y) ^ kX U k f kY w here X U f Y = X/A (T heorem 2 . 2 ) .
It
i s p r o v e n t h a t i f p i s co m p a ct c o v e r i n g (T heorem 2 . 4 ) , f o r
exam ple when X i s p a r a c o m p a c t ( C o r o l l a r y 2 . 8 ) 5 t h e n
k(X U f Y) = kX U k f kY. L e t PX b e t h e S to n e - C e c h
c o m p a c t i f i c a t i o n o f c o m p l e t e l y r e g u l a r X. U s in g t h e f a c t
t h a t p d i s t r i b u t e s over, t h e a d j u n c t i o n ,
( P (X Uf Y) s PX U p f P Y ) , w h e n e v e r X i s n o rm a l an d w h e n e v e r Y
and. X Uf Y a r e c o m p l e t e l y r e g u l a r , i t i s shown t h a t k w i l l
d i s t r i b u t e t h r o u g h t h e a d j u n c t i o n b y e n l a r g i n g t h e s p a c e X,
t h a t i s , t h e r e e x i s t s a s p a c e X1 w h e re X C X1 C f3X su c h t h a t
k(X Uf Y) s kXj. Uk kY w h e re g e x t e n d s f fro m A t o A1 C X1 ..
\
'
■.
V
V
INTRODUCTION
H i s t o r i c a l l y , J 0H0Cs W h ite h e a d [1 3 ] s y s t e m a t i z e d t h e
c o n c e p t o f a d j u n c t i o n s p a c e s an d now t h e c o n c e p t i s u s e d
e x t e n s i v e l y i n A l g e b r a i c T o p o lo g y ,
The a d j u n c t i o n s p a c e
i s a n i d e n t i f i c a t i o n s p a c e and h e n c e l n h e r i t e s t h e p r o b le m s
o f a l l i d e n t i f i c a t i o n s p a c e s i n t h a t many p r o p e r t i e s o f t h e
o r i g in a l spaces a re n o t p re s e rv e d ,
lis ts
Hu [ 5 J f o r exam ple
s e v e r a l p r o p e r t i e s t h a t h o l d an d p r o v i d e s some
c o u n te re x a m p le s,
We a r e i n t e r e s t e d i n H a u s d o r f f s p a c e s
a n d u n f o r t u n a t e l y i d e n t i f i c a t i o n s do n o t p r e s e r v e t h i s
p r o p e r t y , s o we w i l l assum e t h r o u g h o u t t h i s p a p e r t h a t a l l
spaces a re H au sd o rff, in c lu d in g th e a d ju n c tio n sp a c e ,
.=
W h ite h e a d [1 3 ] d o e s h a v e a th e o r e m s t a t i n g s u f f i c i e n t
c o n d i t i o n s f o r a n a d j u n c t i o n s p a c e t o be H a u s d o r f f w h ic h
r e q u i r e s X . t o b e n o r m a l an d Y t o b e U ry so h n ( d i s t i n c t
p o i n t s h a v e n e i g h b o r h o o d s whose c l o s u r e s a r e d i s j o i n t ) .
For
t h e c o n v e n i e n c e o f t h e r e a d e r , we w i l l i n c l u d e n o t a t i o n an d
some known f a c t s o f a d j u n c t i o n s p a c e s .
L e t A b e a c l o s e d s u b s e t o f X, l e t ■f : A :—
Y be a
c o n t i n u o u s m ap, a n d l e t X + Y d e n o t e t h e d i s j o i n t u n i o n o f
X a n d Y,
The a d j u n c t i o n s p a c e o f X and Y v i a f , X U ^Y ,
i s f o rm e d b y i d e n t i f y i n g a e A w i t h f ( a ) e Y,
p:X +’■Y — > X Uf Y b e t h e i d e n t i f i c a t i o n .
Let
Our t e r m i n o l o g y
2
w i l l f o l l o w t h a t o f Rc Brown [ 1 ] .
1
The c o m m u ta tiv e d ia g r a m
X ---------— > X U f. Y
if
fr '
A ------- j ------- * Y -
w h e re i a n d T a r e e m b e d d in g s f o l l o w s fro m t h e d ia g r a m
X + Y
A s u b s e t U o f X U f Y i s o p e n i f and o n l y i f f -1 (U) and
I -1 (U) a r e o p e n .
T h e r e f o r e , f (X - A) i s o p e n i n X U f Y,
f| (X - A) i s a homeom orphlsm i n t o , and T(Y) i s c lo s e d o
The a d j u n c t i o n s p a c e i s a p u s h o u t i n the. s e n s e t h a t i f
X ------- —------^ Z
it
•
a —
v |i. ;
f — ^
y
com m utes, t h e n t h e r e i s a u n i q u e g m apping X U f Y t o Z s u c h
t h a t g o i = i 1 a n d g o f = f 1..
The p u s h o u t p r o p e r t y i s an
3
im p o rta n t f e a tu r e o f a d ju n c tio n s p a c e s .
I f Y i s a s i n g l e p o i n t , t h e n X U f Y = X/A w h ere X/A i s
th e space X w ith A i d e n t i f i e d to a p o in t.
I n A l g e b r a i c T o p o lo g y , one i s i n t e r e s t e d i n w o r k in g
w ith a c e r t a i n c a te g o r y o f sp a c e s i n w h ic h 's e v e r a l s ta n d a r d
o p e ra tio n s a re c lo se d .
K e l l e y [7 ] V
The c o n c e p t o f k - s p a c e s a p p e a r s i n
S t e e n r o d [1 2 ] a n d many o t h e r s h a v e d e m o n s t r a t e d
t h a t th e c a te g o ry o f k -sp a c e s i s c o n v e n ie n t and i t in c lu d e s
a l l l o c a l l y c o m p a c t an d a l l f i r s t c o u n t a b l e s p a c e s .
are c e r ta in
T h ere
d ra w b a c k s t o t h e c a t e g o r y o f k - s p a c e s "in t h a t
sub s p a c e s o f ' a k - s p a c e n e e d n o t b e a k - s p a c e , p r o d u c t s o f
k - s p a c e s n e e d n o t b e k - s p a c e s , and f u n c t i o n s p a c e s of.
k - s p a c e s w i t h c o m p a c t- o p e n t o p o l o g y n e e d n o t b e k - s p a c e s ,
b u t s u i t a b l e m o d if ic a tio n s , o f th e s e to p o lo g ie s a llo w s th e
c a t e g o r y o f k - s p a c e s and c o n t i n u o u s maps t o b e c o n v e n i e n t .
F o r t h e c o n v e n i e n c e o f t h e r e a d e r , we w i l l I n c l u d e n o t a t i o n
and some known f a c t s o f k - s p a c e s .
A s u b s e t A o f X i s c o m p a c t l y c l o s e d p r o v i d e d i t m e e ts
e a c h c o m p a ct s u b s e t o f X i n a c l o s e d s e t , and X i s a
■
..
.
.
. ..
.
.
-
■
.
;
.
•
.
k - s p a c e i f X i s ,H a u s d o r f f and a l l c o m p a c tly c l o s e d s e t s a r e
C lo se d 0
T h e re i s a u n i q u e k - space- kX a s s o c i a t e d w i t h e a c h
H a u s d o r f f s p a c e X.
The s p a c e s X an d kX h a v e t h e same
u n d e r l y i n g s e t , a n d t h e c l o s e d , s e t s o f kX a re . t h e c o m p a c tly
4
c lo s e d , s u b s e t s o f X.
c o n t i n u o u s 5: kX a n d
The i d e n t i t y mapfrom . kX t o
X have th e
i s a k - s p a c e , t h e n kX = X.
X is
same c o m p a c t s e t s an d
if X
I f f : X — > Y s t h e n t h e same
f u n c t i o n k f : kX — > kY i s c o n t i n u o u s and t h e d ia g r a m
e
X ------- - — > Y
I
.A
kX - - - ^
co m m u tes.
kY
One v ie w s k a s a f u n c t o r fro m t h e c a t e g o r y o f
H a u s d o r f f s p a c e s and c o n t i n u o u s maps' i n t o t h e c a t e g o r y o f
k - s p a c e s and c o n t i n u o u s m aps.
I n r e l a t i n g a d ju n c ti o n sp a c e s and k - s p a c e s , th e
q u e s t i o n a r i s e s a s t o when k d i s t r i b u t e s t h r o u g h t h e
a d ju n c tio n .
An ex am p le i s g i v e n f o r w h ic h
..
k(X U f Y) 7^ kX U k f kY w h e re X U f Y = X/A (T heorem 2 . 2 ) .
It
i s p r o v e n t h a t i f p i s c o m p act c o v e r i n g (Theorem 2 . 4 ) 5 f o r
exam ple when X i s p a r a c o m p a c t ( C o r o l l a r y 2 . 8 ) , t h e n
k(X U f Y) = kX U k f kYo
L e t pX be t h e S to n q - C e c h
c o m p a c t i f i c a t i o n o f c o m p l e t e l y r e g u l a r X.
U s in g t h e f a c t
t h a t p d i s t r i b u t e s o v e r th e . a d j u n c t i o n
(P (X U f Y) s pX L ^ f pY) w h e n e v e r X i s n o rm a l and w h e n e v e r Y
a n d X U f Y a r e c o m p l e t e l y r e g u l a r , i t i s shown t h a t k w i l l
5
d i s t r i b u t e , t h r o u g h t h e a d j u n c t i o n b y e n l a r g i n g t h e s p a c e X,
t h a t i s ,, t h e r e e x i s t s a s p a c e X1 w h e re X C X1 C PX su c h t h a t
k(X U f Y) s XX1
kY w here g e x t e n d s f fro m A t o A1 C. X1 .
F o r n o t a t i o n 5 Cl^A w i l l d e n o t e t h e c l o s u r e o f a s u b s e t
A o f Xo
B d ry A w i l l d e n o t e t h e b o u n d a r y o f A a n d I n t A w i l l
d e n o te t h e i n t e r i o r o f Ao
I f a map i s d e n o t e d b y a l e t t e r
u n d e r a b a r , we w i l l a lw a y s assum e t h e p u s h o u t d ia g r a m
X
A
A
T ■
f .
X Uf Y
A
T
^
Y
We e m p h a siz e ' a g a i n t h a t a l l s p a c e s c o n s i d e r e d a r e H a u s d o r f f
an d a l l maps a r e c o n t i n u o u s .
A m apping f :X —
Y i s c o m p a c t c o v e r i n g i f e a c h com pact
s u b s e t o f Y i s t h e im age o f some co m p a ct s u b s e t o f X, 'f i s
a c o m p act m ap p in g i f t h e p r e im a g e o f e a c h co m p a ct s u b s e t
o f Y i s c o m p a c to
CHAPTER I
I n C h a p t e r I 3 we r e l a t e t h e c o m p a c tn e s s o f X3 A3 f 3 Y3 .
X Uf Y3 a n d s u b s e t s o f X Uf Y3 an d e s t a b l i s h a u s e f u l .
n e c e s s a r y an d s u f f i c i e n t c o n d i t i o n f o r t h e i d e n t i f i c a t i o n
p:X + Y —
X Uf Y t o be co m p a ct c o v e r i n g .
U sin g t h e
n e c e s s a r y c o n d i t i o n . , we t h e n show a n y c o m p act s u b s e t o f an
a d ju n c ti o n sp a ce i s i t s e l f an a d ju n c tio n sp a c e .
I t i s c l e a r t h a t i f X a n d Y a r e c o m p a c t3 t h e n so i s
X U f Y0
I f X U f Y i s c O m pact3 t h e n X h e e d h o t b e c o m p a c t.
H ow ever3 Y i s c o m p a c t b e c a u s e i : Y ——> X U f Y i s a n e m bedding
an d T(Y) i s a c l o s e d s u b s e t o f X U f Y.
In fa c t, T is a
c l o s e d c o m p a c t map.
P ro p o sitio n I o l 0
I f A i s a co m p a ct s u b s e t o f X3 t h e n
p:X + Y — > X. Uf Y i s a c l o s e d Compact map.
P ro o f.
I f A i s c o m p a c t, t h e n p i s c l o s e d b e c a u s e f i s
c l o s e d [ 2 , p l 2 8 } 3 a n d p i s c o m p a c t b e c a u s e p - 1 ( z ) i s com pact
f o r each z e
X Uf Y [4 ].
C o r o l l a r y 1 . 2 .'
an d Y a r e c o m p a c t.
I f X Uf Y a n d A a r e c o m p a c t, t h e n X
7
P r o*p o s i t i o n 1 . 3 .
S u p p o se I n t A = $ o r A i s n o t o p e n .
T hen n e c e s s a r y a n d s u f f i c i e n t c o n d i t i o n s f o r
Y t o he
co m p a ct i s t h a t Y b e c o m p act a n d c l o s e d s u b s e t s o f X
m i s s i n g A b e c o m p a c t.
P ro o f.
L e t .{ U a } b e a n o p e n c o v e r o f X.-Uf Y.
S in c e.
T(Y) i s C om pact3 a f i n i t e num ber o f Ua l S c o v e r T ( Y ) , so t h e
c o v e r c a n b e r e d u c e d t o (U a ) U ( U^)
w h e re t h e Ua ' s
n
_
■
n _
c o v e r f (X - A ).
S i n c e U U. ) f ( A ) s A C U. = U f - 1 ( U , ) a n d
i= l 1
i= l
■. 1 '
sin c e I n t
A
= 0o r A i s n o t o p e n , U Pl (X- A) ^ 0 . Now
X - U i s c l o s e d a n d m i s s e s A5 so X - U i s c o m p a c t.
T h e r e f o r e , t h e r e i s a f i n i t e num ber o f U ' s , s a y U .,
m_
^m5 s u c h t h a t U f ~ x ( U .) O X - U . ■ Thus
J= I
J
j = I,
.o .
(Uj ) U
u U V i= !
covers X Uf Y.
P ro p o s itio n 1 .4 .
L e t K b e a c o m p act s u b s e t o f X U f Y.
I f Yv i (K) D A i s c o m p a cts t h e n Tv i (K) i s c o m p a c t.
P ro o f.
L e t X0 = T "1 ( K ) , Y0 = T ' 1 ( K ) , a n d
A0 = f ”1 (K) H A a n d a p p l y C o r o l l a r y I „2 u s i n g t h e f a c t t h a t
K = X0 U Y0 .where g = f ] A0 [ 8 ] .
c
8
Lemma 1 . 5 «
I f F i s a c lo s e d , s u b s e t o f X
Cl^G - G i s a s u b s e t o f A w h e re G = f ’"1 (F
P roof.
f(X - A ) .
L e t x e Clx G — G.
Y3 t h e n '
f (A )) .
A3 th e n 7 (x ) i s in
If x
A l s o , f ( x ) e T ( C l x G) C Cl (F - T (A )) C F .
T h e r e f o r e 3 x e G s i n c e T] (X - A) i s a homeo m orphism ; a
c o n tra d ic tio n .
Thus 3 x e A.
P ro p o s itio n 1 .6 .
The i d e n t i f i c a t i o n
p:X + Y — > X U f Y i s c o m p a ct c o v e r i n g i f and" o n l y i f
ClxT -1 (K - T ( A ) ) i s c o m p a c t f o r a l l c o m p act s u b s e t s K o f
X Uf Y.
.
■
P r o o f . L e t K b e a c o m p a ct s u b s e t o f X U p Y a n d l e t
.
■/
'■'I
■
'
G = T"1 (K - T ( A ) ) .
I f p i s c o m p a ct c o v e r i n g , t h e r e e x i s t s
a c o m p a c t s u b s e t C o f X + Y s u c h t h a t p(C ) •= K.
Now CPi X
a n d C P Y a r e c o m p a ct a n d T (C P X) U i ( C P Y). = K.
G=
S in ce
( C P X) - A3 Clx G i s c o m p a c t.
S u p p o s e K i s a co m p a ct s u b s e t o f X Uf Y a n d Clx G i s
c o m p a c t.
Now Clx G U T- 1 (K) i s c o m p a ct and b y Lemma 1 . 5 ,
C lx G - G C A3 so p (C lx G U I " 1 (K )) = K.
9
C o ro lla ry 1 . 7 »
I f K i s a c o m p a ct s u b s e t o f X Uf Y
a n d p i s c o m p a c t c o v e r i n g , t h e n K i s an a d j u n c t i o n s p a c e
fo rm e d fro m two c o m p a c t s e t s „
P r o o f ..
I f K O f (A) = 0 , t h e r e i s n o t h i n g t o show.
I f K O f (A) £ 0 , l e t X0 = f - 1 (K) , A0 = X0 O- Aj' Y0 = T "1 (K) 5
an d g = f I A0 .
Then K = X0
Y0 w h e re Y0 i s c o m p a c t.
X1 = CI x Gk Where Gk = f - 1 (K - f (A ) ) .
s i n c e p i s c o m p a ct c o v e r i n g .
X1 ’ - Gk i s c o n t a i n e d i n A.
Let
Then X1 i s com pact
We a l s o know t h a t
I f X1 = Gk j t h e n K = X1 + Y0 .
S u p p o se X1 ^ Gk , t h e n l e t A1 = X1 O A0 and h = f | A1 .
Then X1 U h Y0 = K .
The h y p o t h e s i s o f C o r o l l a r y . 1 . 7 c a n n o t b e w eak en ed b y '
rem o v in g t h e c o n d i t i o n t h a t p i s c o m p act c o v e r i n g .
e x a m p le 3 l e t X b e t h e exam ple u s e d i n t h e p r o o f o f
T heorem 2 . 2 a n d c o n s i d e r X/A.
For an
CHAPTER 2
I n C h a p t e r 2 , we g i v e a n e x a m p le .when k d o e s n o t
d i s t r i b u t e th ro u g h th e a d ju n c tio n .
We p r o v e t h a t i f t h e
■
i d e n t i f i c a t i o n p:X + Y — > X U f Y i s c o m p act c o v e r i n g , f o r
exam ple when X i s p a r a c o m p a c t , t h e n k w i l l d i s t r i b u t e .
C o n s i d e r t h e d ia g r a m
kX
A
kA
B e c a u s e f and k f a r e t h e same p o i n t s e t map, t h e u n iq u e
c o n t i n u o u s map g w h ic h i s g i v e n b y t h e p u s h o u t p r o p e r t y o f
kX U ^ f kY i s t h e i d e n t i t y .
T h e r e f o r e , kX U ^ f kY i s
H a u s d o r f f p r o v id e d . X Uf Y i s H a u s d o r f f and kX Ufef kY i s a
k - s p a c e b e c a u s e i t i s t h e i d e n t i f i c a t i o n im age o f t h e
k - s p a c e kX + kY.
I t now f o l l o w s t h a t kg i s t h e i d e n t i t y .
The p r o b le m i s t o show when k(X Uf Y) = kX Ufef kY.
P ro p o sitio n 2 .1 .
I f kX Ufef kY an d X Uf Y h a v e t h e
same co m p a ct s e t s , t h e n k(X Uf Y) = kX Ufef kY.
' P roof.
The s p a c e kX Ufef kY i s . a H a u s d o r f f k - s p a c e
w i t h t h e same co m p a ct s e t s a s X Uf Y.
11 '
Theorem 2 . 2 .
I n g e n e r a l , k(X U f Y) ^ kX U fcf kY.
The p r o o f I s b a s e d upon t h e f o l l o w i n g lemma w h ere
/
X U f Y = X/A.
Lemma 2 . 3 .
T h e r e i s a s p a c e X an d a c l o s e d s u b s e t A
su c h t h a t kX/kA i s n o t c o m p a ct a n d X/A i s c o m p a ct i f and
o n l y i f t h e r e . i s a s p a c e Y and a c l o s e d s u b s e t B w i t h em pty
i n t e r i o r su c h t h a t ( a ) e a c h c l o s e d s u b s e t o f Y m i s s i n g B i s
c o m p a c t a n d (b ) Y - B i s c o m p a c t l y c l o s e d b u t n o t c l o s e d .
P r o o f .. . S u p p o se X/A i s c o m p a c t an d kX/kA i s n o t
c o m p a c t.
L e t (kX - kA)°° b e t h e one p o i n t c o m p a c t l f i c a t i o n
o f kX - kA.
F o l l o w i n g M ic h a e l . [9 ] •> (kX - kA)°° S kX/kA i f
a n d o n l y i f c l o s e d s u b s e t s o f kX m i s s i n g kA a r e c o m p a c t.
S i n c e kX/kA =/ (kX - kA)00 , t h e r e i s a s u b s e t M o f X su c h
t h a t M m i s s e s A a n d i s c l o s e d i n kX and i s h o t c o m p a c t.
Y = M U B b e a sub s p a c e o f X w h e re B = B dry A.
Y - B = M i s c o m p a c t l y c l o s e d . i n Y,'
i n X. Then
t h e n Clx M w o u ld b e com pact
M i P C l xM = M .would b e c l o s e d i n X.
Thus ClyM P B / 0 .
s u p p o s e F i s a c l o s e d s u b s e t o f Y m i s s i n g B.
Clx F P A = 0 .
C le a rly ,
To show Y - B i s n o t
c l o s e d i n Y, we o b s e r v e t h a t M i s n o t c l o s e d
Cl^M Pi B / , 4> f o r i f i t w e r e ,
Let
Then .
T h e r e f o r e , Clx F i s co m p a ct an d we h a v e
an d
Now
12
Clx F n M c l o s e d , b u t Clx F H M = Clx F D Y = F i s c l o s e d i n X
Thus F i s c o m p a c t i n X a n d h e n c e i n Y s i n c e F m i s s e s A.
F o r th e c o n v e rs e , s in c e c lo s e d s u b s e ts o f Y m is sin g B
a r e c o m p a c t, Y/B ^
(Y - B)00 and h e n c e Y/B i s c o m p a c t.
For
n o t a t i o n , c o n s i d e r t h e d ia g ra m '
Y --------P" " ■-» Y/B
h
'
A
■kY ------- -— ^ kY/kB.
S u p p o se kY/kB i s c o m p a c t.
we h a v e Y - B c l o s e d i n kY.
S in c e Y - B i s c o m p actly c lo s e d ,
H ence q(Y - B) i s c o m p a ct i n
kY/kB, w h ic h i m p l i e s p(Y - B) i s Compact i n Y/B.
Y - B is
Thus
c l o s e d i n Y, a c o n t r a d i c t i o n . .
P r o o f o f Theorem 2 . 2 .
L e t X = N U A b e a sub s p a c e o f
SN, t h e S t o n e - C e c h c o m p a c t i f i c a t i o n o f t h e p o s i t i v e
i n t e g e r s , w h e re A i s a f o l l o w s .
in fin ite } .
F ix
A =■ { Xg| E e S } .
Any i n f i n i t e
x^ e C ^ E
L e t S = { E C N| E i s
- E fo r: e a c h E e S
and l e t
The c a r d i n a l i t y o f A and X i s c [ 3 , p 9 7 ] .
c l o s e d s u b s e t o f SN h a s c a r d i n a l i t y
2 ° [ 2 , p 2 4 4 ] , so i f K w e re a n i n f i n i t e c o m p a c t s u b s e t o f X,
t h e n K, a l s o c o m p a c t i n SN, w o u ld h a v e c a r d i h a l i t y 2 ° .
T h e r e f o r e , c o m p a c t s u b s e t s o f X m u st b e f i n i t e an d we
I
13
c o n c l u d e t h a t X i s n o t a k - s p a c e b e c a u s e kX i s d i s c r e t e
and X i s h o t d i s c r e t e .
M o re o v e r , X - A
is . c o m p a ctly c lo s e d
b u t n o t c l o s e d , an d b y t h e d e f i n i t i o n o f A, c l o s e d s u b s e t s ■
o f X m is sin g .A a re f i n i t e .
By Lemma 2 . 3 , X/A i s com pact
a n d kX/kA i s n o t c o m p a c t.
Our ex am p le p r o v i d e s a c o m p l e t e l y r e g u l a r non k - s p a c e ■
X, a c o m p a c t m e t r i c s p a c e Y = X/A, a n d a c o n t i n u o u s , c l o s e d ,
o n t o map f : X — > Y w h ic h i s n o t c o m p a c t c o v e r i n g ( f o l l o w s
fro m Theorem 2 . 4 w h ic h i s b e l o w ) *
M o re o v e r, B dry f - 1 ( f (A ) )
w h ic h i s A i s n o t c o m p a c t.
Any exam ple w h e re k(X
Y) ^ kX
kY c o n t a i n s
a n ex am p le w h e re Z/B ^ k Z /k B , Z a s u b s p a c e o f X, a n d ■
I n t B = 0 , f o r s u p p o s e k(X U j, Y) ^ kX
kY*'
T hen t h e r e
i s a c o m p a c t s u b s e t K o f X Uf Y s u c h t h a t g _1 (K) i s n o t
c o m p a c t i n kX Ufcf, kY ( r e f e r t o t h e d ia g r a m a t t h e
b e g in n in g o f C h a p te r 2 ) .
N ote t h a t K H f ( A ) , ^ 0,. . L e t
Xo = J - 1 (K) , Ao = Xo H A , Y0 = i "1 ( K ) , a n d . h = ''f | A0 .
Then
K = X0 U h lY0 i s co m p a ct and g "1 (K) = kX0 U kh Y0 i s ' h o t
c o m p a c t.
N ote t h a t Y0 i s c o m p a c t.
We how. h a v e a s p a c e X, a c l o s e d s u b s e t .A. o f X y ,and a
c o m p a c t . s p a c e Y s u c h t h a t X Uf Y i s c o m p a c t, kX Ukf Y i s
n o t c o m p a c t.
N ote t h a t X - A c a n n o t b e c l o s e d f o r i f i t
14
W ere5 t h e n t h e i d e n t i f i c a t i o n X + Y —
■compact c o v e r i n g .
S u p p o se X - A
i s c o m p a c t l y c l o s e d . ' Then
t h e c o n d i t i o n s o f Lemma 2 . 3 a r e m e t .
c o m p a ctly c lo s e d .
X U f Y w o u ld be
S u p p o se X - A i s n o t
Then kA i s n o t o p e n i n k X . . By
P r o p o s i t i o n 1 . 3 and b e c a u s e kX
Y i s n o t c o m p a c t, t h e r e
i s a s u b s e t . M o f X su c h t h a t M m i s s e s A an d i s c l o s e d i n kX
a n d i s n o t c o m p a c t.
L e t Z = M U B dry A b e a s u b s p a c e o f X.
Then Z a n d B d ry A s a t i s f y t h e c o n d i t i o n s o f Lemma 2 .3 »
Theorem 2 . 4 .
I f p:X + Y -—=> X U f Y i s c o m p act
c o v e r i n g , t h e n k(X U f Y) = kX U ^ f kY.
P roof.
C o n s i d e r t h e d ia g r a m
X. +
A
:
Y --------£ — - 7>X U f Y
A
kX + .k Y ---- ----- kX U k f kY. ■'
We show X U f Y a n d kX U ^ f kY h a v e t h e same, c o m p a c t s e t s .
L e t K be.- a c o m p a c t s u b s e t o f kX U ^ f kY..
i n X U p Y b ecau se th e i d e n t i t y i s
Now K- i s com pact
c o n tin u o u s.
C o n v e rsely ,
s u p p o s e K i s c o m p a c t i n X U f Y. , S i n c e p i s c o m p a c t
c o v e rin g ,, t h e r e i s a com pact su b s e t. C of X + Y such t h a t
p(C ) = K . ' Thus C i s co m p a ct i n kX + kY w h ic h i m p l i e s q(C)
i s c o m p a c t.
15
.
T h e r e f o r e , q(C) = p (C ) = K b e c a u s e p and. q a r e
t h e same p o i n t s e t map.
C o ro lla ry 2 ,5 .
k(X U f
y)
I f A i s a c o m p a ct s u b s e t o f X, t h e n
= kX U kf kY.
P ro o f.
I f A i s c o m p a c t, t h e n p i s c o m p a ct c o v e r i n g
by P ro p o s itio n 1 .1 .
C o ro lla ry 2 . 6 .
I f f : A — > Y i s a' co m p a ct m a p p in g ,
t h e n k(X U f Y) = kX U kf kY.
P r o o f . . L e t K b e a co m p a ct s u b s e t o f X U f Y.
Then
T "1 (K) i s c o m p a c t, a n d f -1 (K) i s co m p a ct b y ■,
P ro p o s itio n 1 .4 .
'
Hence p i s c o m p a ct c o v e r i n g .
M ic h a e l [ 9 , 1 0] p r o v e s t h a t i f X i s p a r a c o m p a c t and
f : X — ^ Y i s c l o s e d an d o n t o , t h e n f i s c o m p a ct c o v e r i n g , .
a n d i f X a n d Y a r e p a r a c o m p a c t , t h e n p:X + Y — > X U f Y i s
co m p a ct c o v e r i n g .
I t i s e v i d e n t fro m the. p r o o f , g i v e n i n
[1 0 ] t h a t X p a r a c o m p a c t i s enough f o r t h e l a t t e r .
c o n v e n i e n c e , we i n c l u d e a p r o o f .
For
16
P ro p o sitio n 2 .7 .
C o n s i d e r t h e d ia g r a m
X ----- !-----------:— 5> X + Y ------ E -------> X L L Y
P a \/
-> (X + Y ) /( A + Y ) .
X/A
T hen
(i)
X/A = (X + Y ) /( A + Y ) 3
(ii)
g is
(iii)
i f p a i s co m p a ct c o v e r i n g , t h e n s o i s p .
P ro o f.,
c l o s e d w h e re pog = q 5
P a rts
( I ) and ( i i )
a re e v id e n t.
For ( i i i ) ,
le t p
h e co m p a ct c o v e r i n g , a n d l e t K be a co m p a ct s u b s e t
fi­
e f X U f Y. Then h '"1 og(K) i s a c o m p a c t s u b s e t o f X/A. ByP r o p o s i t i oh 1 . 6 , s i n c e C l^ p a "1 ( h -1 0g (K) - p ( A ) ) ='
Clx? " 1 (K - f ( A ) ) i s c o m p a c t, p i s co m p a ct c o v e r i n g .
Co r o l l a r y 2 . 8 .
I f X i s p a ra co m p a ct, th e n
k(X U f Y) = kX U k f kY.
P ro o f.
c o v erin g [9 ].
(
.
I f X i s p a r a c o m p a c t ^ t h e n p a i s c o m p act
-
The c o n v e r s e o f P r o p o s i t i o n 2 . 7 ( i i i )
does n o t h o ld .
F o r e x a m p le , l e t A b e a c l o s e d s u b s e t o f X Such t h a t
P a :X — 5» X/A i s n o t c o m p a c t c o v e r i n g (N U A. i n t h e p r o o f o f
■Theorem 2 . 2 ) .
L e t f :A — 5> A b e the- i d e n t i t y .
Then
p:X + A — ^.X U j, A = X i s c o m p a ct c o v e r i n g . ( i n f a c t , c l o s e d
a n d c o m p a c t) and p & i s n o t c o m p a c t c o v e r i n g .
H ow ever, t h e r e i s a p a r t i a l c o n v e r s e a s f o l l o w s .
R e f e r t o t h e d ia g r a m o f P r o p o s i t i o n 2 . 7 .
P ro p o sitio n 2 . 9 .
I f q i s co m p a ct c o v e r i n g , t h e n so
13 v
P ro o f.
L e t K b e a compact, s u b s e t o f X/A.
S i n c e 1q i s
co m p a ct c o v e r i n g , t h e r e i s a compact, s u b s e t C o f X + Y s u c h
t h a t q (C) = Ki
By Lemma 1 . 5 a n d s i n c e
(C A X ) - A =
Pa "1 (K - P ( A ) ) , Clx P ^- 1 (K - p ( A ) ) C C i s c o m p a c t.
C o ro lla ry 2 .1 0 .
t h e n so i s p a «
C o ro lla ry 2 .1 1 .
I f p an d g a r e c o m p a c t c o v e r i n g ,
.
I f p i s com pact c o v e rin g and Y o r
X U f Y i s c o m p a c t, t h e n p a i s co m p a ct c o v e r i n g .
P roof.
an d c o m p a c t.
I f Y o r X U f Y i s c o m p a c t, t h e n g i s c l o s e d
CHAPTER 3
L e t X b e c o m p l e t e l y r e g u l a r an d l e t 0 X . d e n o t e t h e
S t o n e - C e c h c o m p a c t i f i c a t i o n o f X.
F o r c o n v e n i e n c e , we
w i l l u s e C l A t o d e n o t e t h e S t o n e - C e c h compact i f i c a t i o n
c l o s u r e o f a s u b s e t A a n d Cl^A w i l l d e n o t e t h e X - c l o s u r e
o f A.
I f X .and Y a r e c o m p l e t e l y r e g u l a r , t h e n one e a s i l y
shows t h a t pX + £Y = p (X + Y) a n d t h a t Cl F ' - F C PX - X
w h e n e v e r F i s a c l o s e d s u b s e t o f X.
P ro p o s itio n 3 .1 .
L e t X an d kX b e c o m p l e t e l y r e g u l a r .
Then i n g e n e r a l p (k X ) ^ k ( p x ) .
P roof.
B e c a u s e PX i s c o m p a c t, we h a v e k ( p x ) = p x .
L e t X = N U A a s i n t h e exam ple i n t h e p r o o f o f Theorem 2 . 2
S i n c e PX = PN,
|kX|
|pX|
= 2°.
= c and t h e r e f o r e ,
H ow ever, kX i s d i s c r e t e and
pC
| P (k X ) | = 2
[3, p l3 0 ] .
I t s h o u l d b e n o t e d t h a t e v e n th o u g h X-may be
c o m p l e t e l y r e g u l a r , kX n e e d n o t b e [ 1 1 ] .
W i l l a r d [ l 4 ] showed t h a t f o r a m e t r i c s p a c e X,
P (X/A) y p X /C l A.
We g e n e r a l i z e t h i s r e s u l t t o t h e
g e n e r a l a d j u n c t i o n s p a c e w i t h s u i t a b l e r e s t r i c t i o n s and
o b t a i n t h e r e s u l t t h a t p (X/A) ss p x / C l A w h e n e v e r .X and X/A
a re c o m p le te ly r e g u l a r .
The g e n e r a l i z a t i o n i s u s e d t o
,
■ e s t a b l i s h Theorem 3 . 5 .
"
'
'/
Theorem 3 . 4 and Theorem 3 . 5 g i v e
r
19
c h a r a c t e r i z a t i o n s o f fc(X U j* Y) e v e n th o u g h k may o r may n o t
d i s t r i b u t e th ro u g h th e a d ju n c tio n .
L e t X5 Y5 a n d X U f Y be c o m p l e t e l y r e g u l a r .
Assume
t h a t F e x t e n d s f so t h a t t h e d ia g r a m
Cl A -------------------- > p Y
/N
f
A -----------f -------- > Y •
com m utes.
F = P f.
F o r . e x a m p le 5 when X i s n o r m a l 5 PA = Cl A and
C o n s i d e r t h e d i a g r a m ..
'
> X U fTY
w here t h e u n i q u e c o n t i n u o u s map g i s g i v e n b y t h e p u s h o u t
p r o p e r t y o f X U f Y.
The map g i s 1 - 1 b e c a u s e i n g e n e r a l ,
i f one h a s t h e com m uting d ia g r a m .
20
—
w h e re ■f 11 (X - A) a n d I '
The im age o f X
U 4, Y
a r e I'^ I , t h e n g i s 1 - 1 .
Y u n d e r g i s dense b e c a u se i f U i s
o p e n i n PX U 51PY9 t h e n X H F "1 (U) ^ 0 o r Y H T - 1 (U) ^ '0
s i n c e X a n d Y a r e d e n s e i n pX a n d PY r e s p e c t i v e l y , h e n c e
U H g ( X -Uf l Y). ^ 0 .
The map P g i s c o n t i n u o u s a n d i f t h e d ia g r a m
p x ----A
Cl A -
Pf
P (X U , Y
■
AN0 pi
F — >
PY '
commutes ( f o r e x a m p le 9 i f X i s n o r m a l ) , t h e n h i s t h e .
u n i q u e c o n t i n u o u s map g i v e n b y t h e p u s h o u t . p r o p e r t y o f
. . P X U p PYe
We now show t h a t Pgoh| g(X U f Y) i s t h e i d e n t i t y on
PX U p P Y . r e s t r i c t e d t o g(X U f . Y) an d t h a t ho|3g| (X U f Y) is.
t h e i d e n t i t y on p (X U f Y) r e s t r i c t e d t o X U f Y an d c o n c l u d e
21
t h a t p g o h a n d h 0p g a r e i d e n t i t y maps "because g(X U
X Uf Y a re d en se.
Y) and
N ote t h a t p g d h i s t h e i d e n t i t y on
g(X U f Y) i f a n d o n l y i f p g o h o g = g w h ic h i n . t u r n ' i s
e q u i v a l e n t t o p g o h o g o f = F an d ggohogoT = I .
Now
p g o h o g o f = P g o P f = p ( g o f ) = PF = F and
pgohogoi = PgoPi = p (g o i)
= P I = To
A l s o j h opg. i s t h e
i d e n t i t y on X Uf Y i f a n d o n l y i f h o P g o j = j w h ic h i s
e q u i v a l e n t t o h o P g o j o f = j of. a n d h o P g o j o i - J 0T 6
Now
hoPgojof" = . h o g o f = J of and h o P g o j o i = h o g o i = j o i .
C o n s e q u e n t l y 5 we h a v e t h a t h i s a homeom orphism and
we h a v e e s t a b l i s h e d t h e f o l l o w i n g th e o r e m .
Theorem 3 . 2 .
I f X3 Y a n d X Uf Y a r e c o m p l e t e l y
r e g u l a r , F-:Cl A' —
Y e x te n d s f ,
a n d (* ) com m utes, t h e n
P(X Uf Y) & pX 1 4 pY.
C o ro lla ry 3 . 3 »
I f X an d X/A a r e c o m p l e t e l y r e g u l a r ,
t h e n p X /C l A s p ( X /A ) .
P roof.
L e t Y b e a s i n g l e t o n s p a c e and. a p p l y
Theorem 3 . 2 .
U- '
r'
- '
-I. ,
I
22
L e t X5 Y 3 and. X U ^1 Y be c o m p l e t e l y
Theorem 3 . 4 .
r e g u l a r , and l e t p :X + Y —
X Uf Y be c lo s e d .
Then
k(X Uf Y) s kXo U^g kY w h ere g : A0 — » Y e x t e n d s f ,
A0 C Gl -A,, and X0 = X U A0 .
P roof.
'
C o n s i d e r t h e d ia g r a m
PX + PY
Pp
P ( X V f Y)
A
A "
g p ^ ( X U f Y)
X Uf Y
w here p 0 = p p | P p -1 (X U p Y ).
We h a v e
A C U Cl L -1 Cy) C C l U f - i (y ) =. Cl A.
y e f (A)
y e f (A)
Let
A0 = U Cl f ”1 (y) and l e t X0 = X U A0 b e a s u b s p a c e o f
y e f (A)
PX.
N ote t h a t Cl A PiX 0 = A0 b e c a u s e i n g e n e r a l , i f F i s
a c l o s e d s u b s e t of. X, t h e n Cl F - F C P-X - X.
A0 i s
a c l o s e d s u b s e t , o f X0 .
T h erefo re,
W ith p c l o s e d , we h a v e
:
P p - 3V(Z) = Cl p ”1 ( z ) = Cl T - 1 .(z ) U T”1 ( z ) f o r e a c h •
z e X Uf Y [6 ]:.
Thus P p ”i (X Uf Y) = X0 + Y b e c a u s e i f
w e X - A , t h e n P p "1 ( p ( w ) ) = Cl p "1 (p('w) j. = wj. i f w e Y,
th en
Pp " 1
( p ( w ) ) = Cl f "1 (p(w) )-U T"1 ( p ( w ) ) ' =
Cl f "1 ( p ( w ) ) U 1W; i f w E A0 , t h e n , w e Cl f "1 (y) =
. 23
Cl f 1 ( T ( y ) ) f o r some y e f (A) , h e n c e w e P p "1 ( T ( y ) ) .
For
t h e r e v e r s e I n c l u s i o n , i f w e p p "1 (X U f Y ), t h e n
w e P p " 1 (z ) = Cl f - 1 ( z ) U T "1 ( z ) f o r some z e X U f Y, so
w e X0 + Y.
Now l e t g = T "10P0 1A0 : A0 — > Y.
T hen g| A = f and
g(Ao) = f (A) s i n c e P0 (C l f " 1 ( y ) ) . = T ( y ) f o r e a c h y e f (A)
( f 1 (y) C P p "1 ( T ( y ) ) i m p l i e s t h a t Cl f "1 (-y) C P p "1 ( i ( y ) ) ) .
L e t q b e t h e i d e n t i f i c a t i o n m ap p in g X0 +. Y — > X0 U
Y.
O
We h a v e p0 an i d e n t i f i c a t i o n b e c a u s e i t i s c l o s e d [ 3 , p l 4 7 ] ,
and a l s o p 0 i s c o m p a ct ' [ 3 , p l 4 7 ] .
Hence t h e r e i s a .
homeomorphism h su c h t h a t t h e d ia g ra m
' X0 + Y --------- 9-------> X0 U n. Y
P o \
X Uf
commutes [-2 ,p l2 3 ] .
T h e r e f o r e , k(X U f Y) ^ k(X0 U ^ Y) =
kX° U fcg kY b y Theorem 2 . 4 .
j|;
■
From : [ 3 , p l 4 7 ] , P p "1 (X U f Y) i s t h e l a r g e s t s u b s p a c e o f
PX + PY t o w h ic h p h a s a c o n t i n u o u s e x t e n s i o n i n t o X U p Y
and t h i s e x t e n s i o n i s t h e o n l y c l o s e d and com pact
e x te n sio n .
I t w o b ld .b e d e s i r a b l e t o h a v e A0 = A w henever
p i s c o m p act c o v e r i n g b u t t h i s i s not.- a lw a y s t r u e .
For
24
exam ple., i f p i s a c o m p act m ap, t h e n Ag = A -w here as
i f X = R e a l s , A = I n t e g e r s , and p:X
w hich p r o p e r l y c o n t a i n s A.
^ X/A, t h e n A0 = Cl A
For a c h a ra c te riz a tio n
a g r e e i n g w i t h c o m p a ct c o v e r i n g ,, we h a v e t h e f o l l o w i n g
th e o r e m . .
Theorem 3 . 5 »
re g u la r.-
L e t X "be n o r m a l , Y- and X U f Y c o m p l e t e l y
Then k(X U f Y) g ^X1 U kh kY w here A1 C Cl A,
X1 = X U A1 , and. AzA1 —
P ro o f.
Y. e x t e n d s f .
F o r e a c h com pact s u b s e t K o f X U f Y, l e t
Gk = f -1 (K - f ( A ) )..
C o n s i d e r t h e d ia g ra m
P X + P Y — -------^ ------- -—-5s P (X U f .Y)
A
P p "1 (X U f Y)
A .
X1 + Y
A
X + Y -----:---------- 1 ------ > '
X Uf Y
w here p0 == P p l P p "1 (X UL Y ) , X1 . = A U ( U Cl- Gk ) , t h e u n i o n
i
K
b e i n g o v e r a l l co m p a ct s u b s e t s K o f X U f Y, and
q = Pol (X1 + Y ) .
s u b s e t o f X1 .
L e t A1 = X1 Pi Cl A. ' Then A1 i s a c lo s e d ,
B e c a u s e X i s n o r m a l , P(X U f -Y) ^ p X U ^ f PY.
25
Thus Pp
( z ) . € X - A i f z € f (X - A) and. P p - 1 ( z ) C Cl A + Y
i f z e T(Y) .
Hence P p ”1 (X U f Y) C (X U Cl A) + Y.
If K is .
a c o m p act s u b s e t o f X U f Ya t h e n K i s a l s o co m p a ct i n
PX U ^ f PY.
T h erefo re,
Cl P f -"1 (K - P f (C l A)) - P f -1 (K' - P f (C l A)) = Cl Gk - GK' i s
c o n t a i n e d i n Cl A b y Lemma 1 .5 .
T h erefo re,
Xi = A1 U ■(X - A ) .
We h a v e t h a t q i s an i d e n t i f i c a t i o n b e c a u s e i f
U C X Uf Y is
su c h t h a t Q^1 (U) i s o p e n , t h e n
q- 1 (U) H (X + Y) = p ”1 (U) i s o p e n i n X + Y.
open..
H ence U i s
L e t h = i ”1 °q| Ax : Ai — > Y^ t h e n h| A - f .
Let
qi :Xi + Y — > X1 U ^ Y b e t h e . a d j u n c t i o n i d e n t i f i c a t i o n .
C o n s i d e r t h e d ia g r a m s
X1 U
X U, Y
Y-
'
26
X1 +-Y------------ ^
> X1 U h Y
w h e re g i s t h e u n i q u e c o n t i n u o u s map g i v e n .b y th e. p u s h o u t
p r o p e r t y o f ' X U f Y.
M s o 5 g = Q1 oq- 1 .-
T h erefo re5 g is
o p e n [ 2 5p l :2 3 ] . ’ I t i s c l e a r t h a t g i s 1 - 1 a n d s i n c e
X1 - Ai = X - A5 g i s o n t o .
To show t h a t q i s com pact c O v e r i n g 5 s u p p o s e K i s a
com pact s u b s e t o f X U f Y5 t h e n h -1 (K - Ii(A1 ) ) = G^..
T h e r e f o r e 5 Gl^
h - 1 (K - H(A1 )) = Cl G^ C X1 i s c o m p a c t.
So'
b y P r o p o s i t i o n 2 . IO 5 q i s c o m p act c o v e r i n g .
A p p ly in g Theorem 2 . 4 5 k(X U f Y) s XX1 U ^ h kY.
||
■.!
I n Theorem 3 . 5 5 X n o rm a l c a n b e r e p l a c e d b y t h e
c o n d i t i o n s t h a t X i s c o m p l e t e l y r e g u l a r and d ia g r a m (*)
co m m u tes. ‘
I f p i s c o m p act c o v e r i n g , t h e n b y P r o p o s i t i o n 2 . 9 ,
X1 = X.
We a l s o h a v e t h a t t h e l a r g e s t s p a c e t o w hich p h a s
a c o n t i n u o u s e x t e n s i o n i s no l a r g e r t h a n (X U C l A) + Y.
I n g e n e r a l , A0 o f Theorem 3 . 4 and A1- o f Theorem 3• 5
are n o t r e la te d .
I f f h a s co m p a ct p o i n t i n v e r s e s
(fo r
exam ple When f i s 1 - 1 ) , t h e n Ap ,= A5 and i f p i s n o t com pact
■27
c o v e r i n g ( r e f e r t o t h e exam ple f o l l o w i n g C o r o l l a r y 1 , 7 ) ,
'
•
'
t h e n t h e r e e x i s t s a c o m p a c t s u b s e t K o f X Uf Y -s u c h t h a t
C I ^ f " 1 (K
A.
f (A ) ) i s n o t c O m pact5 so A1 p r o p e r l y c o n t a i n s
Ori t h e o t h e r h a n d , f o r X = R e a l s , A = I n t e g e r s , we h a v e
t h a t t h e i d e n t i f i c a t i o n X — > X/A i s c o m p act c o v e r i n g , so
A1 = A b u t ' A0 = Cl A w h ic h p r o p e r l y c o n t a i n s A.
BIBLIOGRAPHY
1.
Brown, R . , E le m e n ts o f G e n e r a l T o p o lo g y . . M c G ra w -H ill,
1968.
2.
D ugundji , J . ,
3.
G il i m a n , L. and J e r i s o n , M ., R in g s o f C o n tin u o u s
T o p o lo g y . A llyn- and B a co n , 1 9 6 6 .
F u n c tio n s.
4.
Van N o s t r a n d , i 9 6 0 .
H a l f a r , E . , "Com pact M a p p in g s ," Proc.- Amer. M ath. S o c .
8 (1957) , 828- 8 3 0 .
5.
Hu, S . T . , Homotopy T h e o r y .
Academ ic P r e s s , 19.59-
6.
I s i w a t a , T . , "M appings an d S p a c e s , " P a c i f i c M ath. J , 20
( 1967) , 455- 4 8 0 .
7.
K e lle y , J , ,
G e n e r a l T o p o lo g y .
Van N o s t r a n d C o . , I n c . ,
19558.
M cC a n d le ss , B . , " A d j u n c t i o n S p a c e s a n d - t h e H e r e d i t a r y
P r o p e r t y , " P r o c . Amer. M ath. S o c . l 8 ( 1 9 6 7 )» 6 9 5 - 7 0 0 .
9.
M i c h a e l , E . , "A N ote on C l o s e d Maps and Compact S e t s , "
I s r a e l J . M ath. 2 ( 1 9 6 4 ) , 1 7 3 -1 7 6 .
1 0 . M i c h a e l , E . , "K0 - S p a c e s , " J , M ath, and Mech. 15 ( 1 9 6 6 ) ,
9 8 3 -IOO2 .
1 1 . N o b le , N . , "Two E xam ples on P r e im a g e s o f M e t r i c S p a c e s , "
P r o c , Am er. M ath. S o c . 36 ( 1 9 7 2 ) , 5 8 6 - 5 9 0 .
1 2 . S t e e n r o d , N . , "A C o n v e n ie n t C a te g o r y o f S p a c e s , " M ich.
M ath. J . I 4 ( 1 9 6 7 ) , 1 3 3 - 1 5 3 .
1 3 . W h it e h e a d , J . H . C . , ” C o m b i n a t o r i a l H om dtopy," B u l l . Am er.
M ath. S o c . 55 ( 1 9 4 9 ) , 213-245.
14. W illa r d , S . ,
;
" M e t r i c S p a c e s A l l o f Whose D e c o m p o s itio n s
Are M e t r i c , " P r o c . Amer. M ath. S o c . 21 ( 1 9 6 9 ) ,
. 126- 1 2 8 .
arch D378NB395ST,TE W m R sm
RL
Illll
D378
B395
Behrens, Dale W.
cop.2
Adjunction spaces and
k-spaces
0 >37 #
S 39fT
(L~
2
< —
I
.