ON THE lWJEBRAIC INV .ARI.ANCE

OF THE

DIMENSION OF A MANIFOLD.

by

Jacques Fortin

A thesis submitted to the Faculty of Gradua.te studies and Research in partial fUlfilment of the requirements for the

degree of Doctor of Philosophy

Department of Ma.thema.tics, McGill University, M::>ntrea.l.

April 4,196 3

TABLE OF CON'l'ENTS

Part I Pre11minaries •••••••••••••••••PP• 1-9

Part II •••••••••••••••••••••••••••••••••••PP• 10-13 THEOREM: Q(In) ~ Q(I), for n f: 1 •••••• page 12

The main theorem ••••••••••••••••••••••••Page 13

Part III ••••••••••••••••••••••••••••••••••PP• 14-24

Discussion of the general case ••••••••••• pages l4ss.

Main result •••••••••••••••••••••••••••••Page 24

Appendix ••••••••••••••••••••••••••••••••••PP• 25-29

Bib1iography ••••••••••••••••••••••••••••••Page ;o

PREFACE

The main abject of' this thesis is to sol.ve the f'ollow­

ing probl.em, wb.ich vas raised, a f'ew years ago, by Fine, Gillm.e.n and

Lambek. If' X and Y are triangulable manifolds of' dimensions m and n

re~ectively, and if' Q(X) is isomorphic to Q(Y), then mFn. ( Q(X)

is the max:i.mal ring of' quotients ot the ring of' all continuous real

valued f'unctions def'ined on X. )

The contents of' the thesis is divided into tb.ree parts.

In the :first, ire prove some general propositions, Y.b.ich will be needed

later.

In the second part, we prove that, if' X and Y are tri­

angulabl.e manifolds such that the dimension of' X is 1 and Q(X) is iso­

morphic to Q(Y), then the dimension ot Y is also 1.

Finally, assuming the validity of' a plausibl.e hypothe­

sis, we can solve the probl.em in the general case, where m and n are

arbitrary positive integers.

I wish to e~ress ~ thanks to Prof'essors J. Lambek

and B.A. Rattray of' McGil.l University, and to Prof. F. Rothberger

of' Laval University, f'or their encouragement and suggestions.

MOntreal, April. 196;.

PART I

THE RING C(X)

Let X be a topological space. As usual, C(X)

'Will denote the ring of ali continuous :t'unctions f'rom X into the

real line al under point'Wise multiplication and addition. It is

obvious that C(X} is a co:mmu.tative ring Wi.th zero element .Q and

unity element.! ( _ indicating constant :t'unctions}.

THE RING Q(C(X))

The maxl.maJ. ring of' quotients Q(C(X)) of' C(X)

may be constructed :from lJ Hom.(D,C(X)) (D ranging over ali dense

ideals in C(X) ) by identif'ying cp1

E.Hom(D1,c(x)) and -q>2 €

Rom.(D2,c{x)}, Yhenever Cl>]. and cp2 agree on J\n D2• (An ideal I of'

a co:mmu.tative ring R is dense if' { Vr € R) [ri=O • r=O] ).

FONCTIONS ON DENSE OPEN SETS AND THE RING Q(X)

If' V and v' are dense open sets of' a topologi-

cal spa.ce x vith vt: v, then the ma.pping

f' + f/V'

is a. ring monomorphism of the ring C(V} into the ring c(v' ).

.f2

We shall consider the direct limit Q(X) = lim c(v), +V

V ranging over all dense open subsets of x. Thus Q(X) may be constructed fromUc(v) by identity­

V 1ng f € C(V ) and f € C(V ) vhenever f and f agree on V n V (no-

l 1 2 2 1 2 12

tice that the intersection of two open dense subsets is open and

dense).

We Vill denote by (f)fl. the equivalence class contai­

ning f and, if fE a(v), ve Vill vrite v= dom f. If f and f agrée 1 2

on dom f n dom f , i.e, if f and f are equivalent, then ve vill 1 2 1 2

vrite as usual f- f • 1 2

PROPOSITION 1 ( rrJ:;) If X is a coçletely regular topologicaJ. space, then

Q(C(X)) is isomcrphic to Q(X).

(A proof is given in Appendi~

Q(X) can be turned into a partia.l.J..;y ordered ring in

the folloVing vay:

f(x)> g(x) on some dense open set V ëontained in dom fn dom g.

It is obvious that this partial order is vell de:f'ined and tha.t:

(:r)* ~Cs>*• { V(h)*) [{t)* + (h)* ~(s)* + (h)*l

(t)* ~(g)* .. ( \f(h)*2:fO )*) ({f)*• (h)* ~(g)*• (h)«j •

·/3

PROPOSITION 2

If (f)* and (g)* are elements of Q(X), then (f}~ (g)*

iff there exists an element (h)* 1n Q(X) such that

(f}* - (g}* = (h)*2.

Proof

If (f}* ~(g)*, then there exists an open dense set in

dom fn dom g, such that f{x} .. g{x)> 0 for ali x 1n V. It follows

that, for every x 1n V, there exists a non-negative real numbera (x),

such that f'(x) - g(x) = (a(x))2. Put h(x) = c(x} for ali x 1n v.

It is clear that h is a. continuous real va.lued :function defined on V

and that f-g -h2• It follows that (f)*- (g) *= (h)*2.

PROPOSITION ;;

For every space X, Q(X} contains the field R 1 of real

numbers.

Pro of

Let' s define cp : R 11-+Q(X) by cp(a} = {a)*, (~ -being the

constant :function on X taking the value a). It is obvious that cp is a

monomorphism of R l1nto Q(X).

PROPOSITION 4

Let X and Y be topological spaces and cp an isomorphism

between Q(X} and Q(Y). T'nenCf/R # is the identity mapping on R 1.

.J4

Proof

According to Prop. 21 every isomorphism between Q(X)

and Q(Y) preserves arder. Since q> ( (!) *) = (!,) * and since the equa­

tion (m)*•(x}* = (~)*, with m, n integers, has a unique so~ution in

Q(X}, it follows that q> is the identity mapping on the rationals.

Now let a be any non-rational positive real nu.mber. We will prove

that q> ((~)*} = (.S:}*. For every natural nu.mber n 1 there existe a ra­

tional nu.mber a such that n

o< a-9n < 1/n,

hence (,2)* < ~)* - {an)*< (1/n}~

Therefore - ..

(.2)*< cp({a}*) - (an)*< (~/n),

and so -C"Y.!:.f< cp((~)*) - {~)*< (~/nf. Consequently, if fe:q>((~)*), there existe, for each natural nu.mber n,

a dense open set Vn in Y, such that

-1/n< f(y) - O< lfn, for ail Y in Vn•

Now Vn is dense in V= dom f, hence the ~ast inequ.ality holds for a1l

y in v. It follows that f(y) =a for a1l y in v, which içlies (f)*=(a)*.

We shall use the following notation:

{~) The range of a :f'u.nction f will be written R(f), i.e.,

R{f) = {f(x) 1 xE dom r} (2) If (rf e Q(X) and (e) * is an ideçotent of Q(X) 1 then we will de­

note by R (ef) the set R(f/dom fn e-1 (1) ). *

·/5

If' V is a dense subset of X and if f is a continuous

mapping of X into Y, then f(X) = "f{'V}.

Pro of

As f(X) = "f{'VJ u f(X-V), it suffices to show that

f(X-V) s"f{'VJ. Now suppose f{X-V) tf.: "f{'V), then there is an x in X-V

such that f{x) f "f{'VJ. It follows that xe f-1 (Y-f(V)) and that

t-1{Y-f"(VJ) n V = rp. This contradicts the denseness of V in X.

PROPOSITION5

* * * * If (f)·:·, {g) , (e1 ) , (e2 ) are elements of Q(X) such

that ' {t)* = (g)*, {e )* = (e )* and {e )*, (e

2)* are idempotents, then

l 2 l R*{e1t) = R*(e2f).

Proof

Let dom f = v, dom g = U, eï1{l) = E1 and e21 (l)=E2

•

Clearly, the set (Vn u) n (E1 n E.2) is dense in both E1n V and

E2n u. Therefore

R*(e1f) = R(f/VnE1 ) = R(f/(Vf\U)n (E1n E2 ))

= R(gf(V l' U) n (E1

n E2

))

= R{gfUnE2) = ~(e2g).

PROPOSITION 6

If (t)*, (e)*e Q(X), with (e)~ = {e)*, then

.J6

e'2=e' & ee'- e' & j(f-~)e' l< ,~on some dense open set] .. } •

Proof

Let us denote R*(ef) by A. Th.en

R*(ef') =A ={ae:R11 ('tc) IS_(~6 )nA 1 ,}, where ~(a,d ={xe R1 f 1x-a1 < :::}

Let a e A and let there be given an a.rb:Ltra.ry e. Sin­

ce RJ_ (O:,e)n A :Ls not empty, there ex:Lsts an x in dom fne-1(1),

such that f(x) :Ls in K1 (a;e)f\A. It fol1ows that we can find an open

neighborhood WE of x wh:Lch :Ls contained in dom f' f\ e-1(1) and such

that f'(W )C K1 (a, e). e

Let v = w~ u (x-iie). We define a mapping e' of v in-

to :al in the following way: { 1 if xe WE,

e' {x) = o if xe x-we.

Now it is obvious that e• 2 = e' and that ee' .... e'. Moreover we have

l{f-~)e•l<! on dom fnV. For if zcdom fnV, then either zeWe

or z ~W€ and, in the first case e' {z)=O, while in the second we ha­

ve e'(z)=l, which implies that j(f{z)~{z))e'(z) 1= jf(z)- a jand

this is smaller than c. , since f(z)e RJ. (~, e) as z e W rl

Conversely, if t3 belongs to the r:Lght hand s:Lde, then,

g:Lven an e 1 there ex:Lsts an idempotent e' 1 .Q, with ee' ""e' and

such that 1 (t'-[z)e'l~.! on a dense open subset V of x.

·/7

Nov sin ce dom f, dom e, dom e' , V, are aJ.l dense o-

pen subsets and sin ce e' is not Q, we can find an x which is contai­

ned in the intersection of all these demains and such that e•(x) = 1.

It follows that lr(x) -f31 ~e. But f(x)e :a.{ef) since xe dom f

and- X.€ e.-1(1) as e' {x)=1 and ee'- e'. Therefore

COROLLARY 1

R~{er)!={ ae:al f CV..§)(3etro)(3he) [ e'2=e' & ee' _ e' & ~2-((r-~)e')2- ~ 1}

COROLLARY 2

R)et) ={ aeal 1 ( 'l(s)*) (3 (er)·*f(Q)*) (3 (h€ )1<J

[ {e1 f 2=(e' )* & (ee' )* = {e• )'*' &

(~)it2 - ( (f)*-(~)* ) 2 {e' )~ = {he)~]}

In the followil::\g we will always denote by

s( <2>*, (e)*, (r)*)

the algebraic sentence

(\1 <!_)*) (3 {e• )*f (Q)* ) (3(hE )*) [{e' )*2=(e• )* &

(ee' )*=(e' )* & (!.)*2 - ( (f)*-(g:) *)2 (e' )*2 = (hE )*2 ] ,

so that

~(ef) ={a e al 1 s( (g)*, (e)*, (r)*) }·

.j8

PROPOSITION 7

If V is an open dense subset of a topological space X,

then Q(X) = Q(V).

Proof'

We will denote by (:r)~ an element of' Q{X) and by (:r)v

an element of' Q(V). Let us def'ine a ma.pping <p : Q{X)- Q(V) in the

f'ollowing way. If' (:r)* € Q(X) and dom f' = u, then let x -,((:r)i) = (:rjvnu); •

We will verif'y that cp is a well def'ined isomorphism..

{a) If (:r)x = (g)x, where dom f = u and dom g = U',

it is clear that f/VnU and gjVnU' agree on vnunu•, which is

open and dense in v. Therefore (f/Vnu); = (gjvnu• >; and it f'ol­

lows that q> is vell defined.

* (b) If' (f)v eQ(V), it is evident that cp ((fJ;r) = (f)~

and so 1 cp is ont o.

(c) If q> ( (f);) =cp {(g)~p 1 where dom f = U and dom g=U',

then (fjvnu); = (gjVnU' )V' which implies that f and g agree on the

open dense set Vn U 1'\U'. It f'ollovs that f and g agree on Un U'.

Theref'ore (:r)~ ~ (g)~ and socp is one-one.

(d) q,((:r)i)+C){(g)i) = (:rjvnu); + (gjvnut >; = (f'/V nu + gJV n u• ); = (f+g/V f\ (Un ut)>;

= cp ((f+g)~ ::: qJ ((:r); + (g)~.

(e) The proof that q>((:r);)-cp((g);) = cp((:r);•(g);)

is similar to that of' (d).

·/9

COROLLARY

If X and Y are bath triangu].ab~e ma.nifo~ds of the sa-

me dimension, then Q(X) = Q(Y).

Proof

It is obvious that two triangulab~e manifo~ds of the

same dimension have homeomorphic dense open subsets, namely finite

unions of disjoint n-ce~lso Therefore if U and V are homeol!!D:rphic

dense open subsets of X and Y respectively, then

Q(X) = Q(U) = Q(V) = Q(Y).

The preceding corollary states that if X and Y are

triangulab~e manifolds of the sam.e dimension, then Q(X) = Q(Y).

The 1.l1ai.n abject of this work is to show that the con-

verse of this corollary ho~ds, i.e., if X and Y are triangulab~e ma­

nifo~ds such that Q(X) = Q(Y), then x:~d Y have the same dimension.

It Wi~ be sufficient, in arder ta get the re~t,

to prove that if m. and n are different natur~ numbers, then Q(tB)

:f Q(~) (where I is the open unit interv~ ]0,1( ), for if X and

Y are triangulab~e manifolds of dimension m. and n respectively, then

Q(X) = Q(tn) and Q(Y) = Q(fl).

The re:mainder of this work Wi~ be divided into two

part6. ile Wi~ prove that: (1) Q(fl) :f Q(I1 ) for every natur~

number n greater than 1; (2) Q(Im.) :f Q(In), where m. and n are dif-

ferent natur~ numbers greater than 1, asSUllling a certain plausible

conjecture.

.jlO

PART II

In the following, I (a, e) will denote the open n-cube n

of side 2~ whose center is a =(a1,a

2, ••• ,an). T.he set

{ x=(x1,x

2, •• ,x'l\.) 10 <x

1 -o;1 < E ,15 i:S n }

will be denoted by I+(a,e), and the set n

{ x=(xl,x2, •• ,xn)1 -t:< x1-ai <o, l::;i:;:n}

PROPOSITION 8

If (f) *..:: Q(:r'l) is such that R.;..,(ef) cFR+:·((l-e)f) for

every idempotent (e)* of Q(l") different from (0)* or (1)*, then f

is one-one. (Here, I still denotes the open unit interval.)

Pro of

Let x1 yÇ. dom f, with f{x) = f(y). T.hen, either

(a) f(x) or f(y) is a relative maximum or minimum of f; or

(b) neither f(x) nor ~(y) is a relative maximum or minimum of f.

Suppose (a) holds and that f(x) is a relative maxi.-

mum of f. Binee dom f' is open, there exi.sts an € such that

~(x,c)Cdom f and f(x)2:f(z), for all Z€ ~(x,€)•

It follows that r(r:(x,~)) and f(I;(x,€)) are closed intervals

.fil

of the form (a,f(x)) and (b,f(x)) respectively. Then a5b or

b< a. Let us suppose, here, that a:S b 1 i.e., f(ID.(x1 €))!;f(J:ri{x,t.::)).

Define a i'unction e to be l. on rn (x, €) and 0 on

tn - ~(x,e). Now R*(ef) = f(r:(x,e)) = f(~(x,€)) and

~((1-e)f) :2 R ((1-e)f/~(x,E:)) = f(Iri{x,e)}

= f(~(x,e)} 2 f(I~(x,~)) = R*{ef}.

Therefore there exists an idempotent e such that

Ri'.(ef) ç; R*((1-e)f) and this is a contradiction.

Now, suppose (b) ho1ds. Then, for every e and 8

such that lh(x,e) and In(y,ô) are contained in dom f, f(x) and

f(y) are interior points of f(~(x,e)) and f(~(y,ô)) respectively.

Take 8 such that ~(y,B) C dom f and x4f:~(y,8).

Let U = t(~(y1 8)). New there exista an €SUCh that t(~(x,E))5U,

In(x~) 5 dom f and ~(x,€ )n In(y,ô) = '/J• Let • s define an idempotent mapping e to be 1 on

In(x,G } and 0 on In - Izr(x,e ). It foll.ows that

R*(ef) = f(In(x,€)) 6 Ü

and R*{(1-e)f) 2 f(!n(Y, 8)) = Ü,

therefore

and this is a contradiction.

In Proposition 8, the condition

(V(e)~ r (Q)*, (;!,)'*) 'R)et}$R.xH1-e)r)

is obviously e~ivalent to the a.lgebra.:tc condition

(V(e)* r (Q)*, CtJ* )(3C~f) ("s((g)*,(eYk,(fP) &

not s((~)*,(l-e)*,(f)*)).

.f12

If an element (f)* satisfies this a.J.gebraic condition, we will say

that (f)* has the property s.

TBEOREM

For every n greater than 1, the ring Q(In) is not

isomorphic to Q(I).

Proof

The class (i)* of Q(I), where i is the identity

mapping on I, has obviously the property s. Therefore, if the rings

Q(I) and Q(In) were isomorphic, there would exist an element (f)*

in Q(F) having the property S {since the property S is that. of

satisfying an algebraic relation wbich is preserved by isomorphisme:

see Prop. 4). It would follow, by Proposition 8, that f is one-one.

Therefore the restriction of f to a solid closed sphere contained in

dom f' would be an homeomorphism of a n-dimensional subset of sn onto

a subset of the real line and this would contradict the Brouwer In-

variance of Ibmains Theorem.

Combining the preceding theorem with the remark made

r

in the third paragraph on page 8, we get the main result of part II.

THEO REM

If X and Y are triangul.abl.e mani:fol.ds SllCh that the

dimension of X is l. and Q(X) is isomorphic to Q(Y), then the di­

mension of Y i$ aJ.so J..

.fl4

PART III

Terminology f

In the following, if (f1

)*, ••• ,(fm)*are m elements

of Q(In), we 'Will denote by (f1

, ••• ,f ) the mapping into am, who--m.- m

se domain is .fl dom f. and who se value at a point x is the point ... ' ~ 1t'L

(f1

{x), ••• ,fm(x)). Clearly, this is a continuous mapping ofk() dom fi

into am. The range of the mapping (fl, ••• ,fm) 'Will be denoted by

R(f1 , ••• ,fm).

We 'Will use the notation Kn{a,r) for the n-dimen­

sional open solid ~here of center a and radius r, i.e., for the set

f = (xv···~~) l lx-a 1< r } •

Definition

If f = {f1, ••• ,fm), where (f1 )*, ••• ,(fm)*are ele­

ments of Q(In), and if {e)* is an idempotent of Q(In), then we de-

fine 1)71..

R.*"(efl, • •• ,efm) = R( f/ ndom f. 1'\ e -l(l) ). < ..c.:l ~

It is evident that R:+(ef1

, ••• ,efm) U (o, ••• ,o) == R(ef1 , ••• ,efm).

PROPOSITION 9

Rx(ef1 , ... ,efm) = { (a:l' ... ,anJ 1 (V.,t:~(3e'f.2)(Vi, l:Si:=;m)

e' 2=e' & ee'- e' & l(i -~)e' 1< i on some open dense set Vi}

Proo:f

Let us denote (f~1 .,.,.,,:fm) by :f and R*(e:r1 , ••• ,e:fm)

by A.. 'fb.en we have

·-R*{ef'l, .... ,e:fmJ = A = f~e am 1 eVE) Kin,(a,e)OA 1 ~} Let a= (c;_, ..... ,am) e A and let there be given an a.r­

bitrary E.. Since K (a,e)n A is not eq;:~ty, there exists an x in the rn m

set ndom :t'ir\ e-l(l) such that f'(x) is in K (a,E)nA .. :1=1 -ln

Nov we can :find an open neighborhood K (x,e) o:f x, m n

which is contained in ()dom :fit" e-1(1) such that f(Kn_(x,8))C Km(c~,-E) .. i=J,

Let V = K (x,8) V (In-K (x,8 )). We de:fine e': V~ Rl n n

in the following way: t 1

e' {x) = 0

if x eKn(x,8 ),

if x cfl-Kn (x, ô).

Now, it is obvious that e• 2=e' and that ee'N e'. Moreover we have m

1 (fi-.Œi)e' 1< ~ (l~ ~ m) on the open dense su.bset Qdom fin V. For

if Z€ [}dom fin V and Zf K:n_(x,ô), then e' {z}=O, while if z E: K:n_{x,8}

we have e' ( z )=l vhich ilqplies that

!(fi(z)-f_~(z))e•(z) 1 = ifi(z)-g1 l ~ lf(z)-a 1

and this is sma.ller t.han c, since f(z) a K (a, E), as z E:K (x, 5) .. m n

Conversely, if f3 = ( ~' ..... , fin) belongs to the right

hand side, then, given an €, there exists an ide:rqpotent e' r .Q vith

ee•- e' and such that ! (fi-'ê..i}e' 1< EJTm holds on some dense sub­

set Vi , for every i (1~ i~ m) o

Nov, since dom e, dom e', dom fi and v1 (1~ i~m)

are aJ.l. dense open subsets and since e 1 is not 0, ve ean f'ind an x

vhieh is eontained in the intersection of' all these domaine and sueh

that e•(x)=1o It f'ollows that l:ri(x)-~i!<E/~ Yhich implies l:r(x)-~J<~o

But ( (f'1 (x), ••• ,:rm(x)) ~ ~(ef'1, ••• ,ef'm), since xe {.1dom fi and

xe e-1(1) as e' {x)=1, XE dom e and ee•- e'. Therefore

Km(13,e:)n~(ef1, ••• ,ef'm) r ~ for all €

and so 13 = R:;(e:r1,e:r2, ... ,efm) •

COROLLARY 1

RJt(er1, ... ,efm) = {<a1, ... ,am) 1 {V.:_)(3e•f.Q){Vi, lSi:Sm)

(3h~) [ e• 2=e' & ee' -e• & .!2 - (<ri-~1 )e•)2 N~hi)2] }•

This is an immediate consequence of' the preceding pro-

position together Yith properties of' real valued f'Unctions.

COROLLARY 2

Rlf(ef'l, ... ,ef'm) ={<~, ... ,'in) t (V(~)j(3{e•)*f(.Q)?(vi,l$1.Sm) (3(hi)*) [{e 1 )*2=i:(e•)* & (ee•):. (e•)* &

(s.'f2 •_ ( (1'1)*- (5_) * ) 2 ( e') ,.2 = {h!).R l}

In the follo'W'ing, we Yill denote by

( * ·lt * * *) S (e) ; (f'l) , ••• ,{fm) ; (s_), .••• ,(Sn) ' . ' .

the a1gebraic sentence

.fl7

<'J(.s)*)(3 (e' )*f(p_)* )è'Ji, 1<; i( m)( .3 (h~)*)

{et)*2=(e• )* & (ee' )*:a (e• )* & (.f.)f-((:r1 )*-(s)~)2 (e• )*2 = (h~)*2 , so that

R,.(efl, ... ,efm) = {<o:l, ... ,am) 1 s((e) *; (fl.)*' .... , (:rmY; (9flh <>0 0' (gm~}

Definition

An ~tuple (f1 )*, ••• ,(fm)* of elements of Q(In)

will be said to have the property (P) if, for every idempotent (e)*J(~);

the set R-ii(ef1, ..... ,efm) has a non-void inter1or in aiD-, which can be

e2Pressed algebraically in the foll~-ng vay:

(\l{e)*f(Q.)*)(3 (çu);'" ••• ,{~)N)

M & (3(_i)*)(~(âJ.)*, ... ,(J1n)*) l ( 3(h1 )*, ••• ,(h:m_)*) N•R] ,

vhere M, N and R have the folloVing meaning:

M: s( ( e )'*; (fl)'*' • • .. ' (fm)"k; (~1):..;., • ••' (gm)"'") '

N: ~Y~2 -(<!2,1 )*- (~1 )*)2 = (hi )-;R for 1~ i~ m,

R: s( {e )*; (:r1 )"*, .... , (:r.m).*; (E1 )*, • • ·, (~m)*).

PROPOSITION 10

If the ~tuple (f1 )*, ••• ,(fm)* has the property (P),

then the interior points of R (e:f11 u.,efm) are dense in ito *

Pro of

.jlB

by A. Let a;= ( 'it••. pm) eX and let there be given an arbitrary c.

Since K (a,c)(1A is not empty, there exl.ste an x in 1'\aom f'1, e-1(1) -"ln i=l

auch that f(x} 1s in K (a,e:)nA. Nov we can t'ind a closed neighbor-m :a

hood xn (x,fl} of x, ccntained in n dom fi ne-1(1), S..Tl.d such that 1=1

f'(Kn (x18 })( Km(a, €}.

Let V = ~(x1 8) U (In-~(x,e1) and let e' be a real­

valued t'unction def'ined on V 1 tak.ing the value 1 on ~ (x, 8) and the

value 0 on I -Rh (x,5 j. Now

R {e•f'1, ••• ,e'fm) ={f'(x) 1 x ~a (\aom f'ine•-1 (1)},

~* 1~ and since 1 1 dom fin e'-1(1) = e•-1 (1) = K (x1 8) 1 we have

1~1 n R!!Te•f'

1, .... ,e'fm) = f'(Kb (x, tl)) = tÇ""""Kn,....,(..-x-,8~) }C Km(a, s).

On the other hand1 .il

f'(Kn(x,r)))CR.:;.(ef1, ... ,ef'm) as Kn{x1 (1)C {!dom fin e-1(1). J.;::Jj,

Nov the set r(xn_(x,s)) being equal to RJe•r11 ... ,e'f'm), has a non-

* * void interior in~ since the m-tuple (f'1 ) , ••• ,(fm) has the pro-

perty (P).

It follows that there exista an interior point of the

set R~(et1,ef'21 •• 1 efm) which belongs to Km{a1 €). This proves the

Proposition.

COROLLARY

* ·:+ Let t'= (t~ 1 ••• 1 f ) where (~) , ••• ,(f ) is an m-m m tup1e of el.ements of Q(In) hà.v'"ing the property {P). Then if' U is an

.jl9

open set conta.ined in (\ .iom fl.., the interior points of f(U) in rfl, i =l.

are dense in f(U).

Iefinition

An m-tup1e of elements of Q(X0) will be said to have

property (T), if it has property (P) and if

{V (e)*f(O )*, (1)-R-) R* (ef1 , • .. ,efm) () ~ ((1-e)fp ... , (1-e)fm) m

has a void interior in R , which can be e~ressed algebraically by the

folloW1ng relation:

( \' (e)"* f (Q)*, (1)-lf ) ( 'tJ<J:.>~· ... ,(!fJ"') (.A & B)=+ (V'(i)·~~")( 3(~}*, .... ,~)*)(3(~)*, ••• ,(hmV) C &-{D & E)

where A, B, c, D and E, in this sentence, have the following meaning:

A: S( (e)~;(t1)*, ••• ,(fm)*;~)·, ••• ,~)* ),

B: s{ (l-et'; (f1r\ ... , (fm):+; (al·)' ... , ••• , (cF) ... ) ,

c: (~i, l~ù(m) k>*2- (bi)*-(e .. :/')2 = (hi)*2,

D: S( (e}*;(f1 )•, ••• ,(fm}*;(fi:)*, ••• ,(~)* ),

E: s( (l-e)*;(f1 )*, ••• ,(tm)*;(e~r·, ••• ,(~)* ).

Definition

Let X be a topological space and f be a continuous

mapping of X onto a space Y. A point x of X is said to be a regu.lar

point if the imagt; under f 1 of every neighborhood of x is a. neighbor-

hood of f(x}.

.j20

PROPOSITION ll

If' f is a continuous mapping of an n-dimensional solid

closed sphere Kh (cr.,r) onto a subset of the m-dimensional Eu.clidean

space, such that for every open set U in Kn(~,r) , the set f(U) bas

a non-void interior in am, then the regu.lar points of K (o:,r) are n

dense in Kn(a,r).

Proof

It will be sufficient to prove that the set E of ail

points x for which f{x) is an interior point in 'Jif4 of the image of

every neighborhood of x, is dense in Y~(a,r).

Let VN = Kn(a,r - 1/N), N being a positive integer.

For every positive integer ~~N, we set

vi ={x ~SVN l (3~ < 1/p) f(x) ,zint r(!h(x,c >) in .am}. We will prove that: {a) Vi is dense in VN; (b) V~ is open.

{a) Let as: VN and let there be given an arbitrary f1 su<".h

that ~(a,s)c:.VN. Let us choose c such that r-:<t1 and c< lj2J). New,

as Int f(Kn(a,E)) f ~~ we can fL~d a z in f(Kn(a,r:)) and a positive

number e &uch that ~(z,e)c:f(Kn(a,e:)). Let U = f-~c~(z,e))nKn(a,~) and let y belong to U. It follows that

u cED_ (y ,2e) c K.o. (y ,lfp)

and therefore y€ V~ as 2'2<1/P• J«>reover yc Kb_(a.,ti) and ..

It follovs tha.t Vi is dense in VN.

(b) If a. € v:, there exist po si ti ve numbers t and "Y, such

.f21.

that 5<1./p, ~(f(a);r)c: f(Kn(a,s)). Let us choose real. numbers P,E

suchthat 'O<e<l./p, e.<e-5, t(Kn(a,c-:.))c~(t(a),'Y) andal.so

K:n_(a, e)c VN. Nov, if yt:. Krt(a, €), then

:r(l<h (y,l./p ));:, f(K"u (y ,e ))::>t(:r<n (a,!5 >);.Kln(t{a) ,'Y )=>t(y),

and there:f'ore y bel.ongs to Vft as e< 1/P• It :f'ol.lovs that we have

~(a, e:)CV~ , which ilz!plies that v: is open, since a "W"'as arbitrary.

Now, as VN ia l.ocally compact, it :f'ollows that Q. V~ is a dense subset of VN' that we will denote by Eu· Th.erefore if

Xë:' E.N- 1 then the image of every neighborhood of x is a neighborhood

of f(x). It follows that x is a regul.ar point of iÇ'(a,r) and that

~CE.

To complete the proot, it remains to show that the

set E is dense in KU (a:, r). This follows ilmnediately, sin ce, if U

is an open subset of K (a, r) 1 we can choose N l.arge enough so that n

Un xn (a:, r - 1./N) 1 ~ 1 vhich implies U 1\ ~ 1 rp. Th.ere:f'ore U ft E ~~

and this completes the :proof.

Definition

A continuous mapping :f of a space X o:nto a space Y

is said to be strongl.y irreducibl.e, if there is no proper closed sub-

set of X, whose image under f is a.ll. of Y.

.j22

It has been proved by G.T. Wbyburn ( [21 ), that a

continuous mapping of a compact spaee X onto a space Y is strongly

irred.ucible, if and only if the set {x 1 r·lr(x) = x} is dense in X.

He aJ..so proved that if X is compact and f(X)=Y ie continuous, then

the set {X 1 rlf(x) D x} is a a6

•

PROPOSITION 12

If f is a continuous mapping of an n-dimensionaJ..

solid closed sphere onto a subset of the ~dimensional Euclidea.n

space rf4 such that for every open set U in the sphere, the set f(U)

has a non void interior in ~~ then the folloWing are equivalent:

(a) f is strongly irreducible;

(b) for every regula.r point x, rl.t(x) = x;

(c) for every pair of open sets u, V, Un V=~ ._. Int f(U) n Int :t"{ V) = '/J•

Proof

Tl:l.j.s is an immediate consequence ot the preceding

devel.opment.

Wb.ether a contin.uous mapping satisf'ying the candi-

tions stated in the above proposition must fail to be strongly irre-

ducib.le if m< n, is an unsolved problem, but it seems qui te pl.ausi-

ble to conjecture it.

I:f m.< n and i:f t is a continuons ma.pping o:f an

n-dim.ensional solid closed sphere ~ onto a subset o:f the m.-dim.en­

sional Euclidean space am such that 1 tor every open set u in I(ll,

the set t(U) has a non void interior in Efl, then t is not strongly

irreducible.

Assuming the val.idity o:f the pl"eceding conjecture,

ve can prove the :following.

TBEOBEM

It m is smaller than n1 there is no m.-tuple o:f ele-

ments o:f Q(In) having property (T).

Proo:f

Let (:r1 )*, ••• ,(:rm.r· be an arbitra.ry m.-tupl.e ot Q(In)

"" having the property (P). We set t = (:r1 , ••• ,:rm.}• Let xE {Jdom. :fi

and let us denote bw g the restriction o:f t to ~(x,r}, r being eho­

sen so that K:n(x,r)c(\dom ti. Com.bining the corollary of' Proposi-.,..:::, tion 10 vith the preceding conjecture, it resulta that g is not strong-

ly irreducible. Th.eref'ore, according to part (b) o:f Proposition u,

there exist tvo points y and z in ~ (x,r), such that y is a regula.r

point and g(y) = g(z). This içlies that ve can f'ind open neighbor­

hoods U = K:n_(y,e:)nK:n{x,r} and V= Kb(z,B )nK:n(x,r) o:f y and z

respectively, such that unv = ~ and g(V)e g(U).

Let w = ~(z, ô)nKn(x,r) andE= w u(fl-w). Let us

define a mapping e': E ... R1 in the following vay:

{

1 if t(: w, e' (t) =

0 if te (In-v).

It follovs that

R}e•:r1 , ... ,e'fm) = g(V)cg(U)cR*((l-e' )r1, ... , (l-e' )fm)

and therefore (t1)*, ••• ,(fm)* does not have property (T).

THEO REM

Two tria.ngulabl.e manifol.ds X and Y have the sa.me di-

mension, if and only if Q(X) is iso~hic to Q(Y).

Proof

According to our previous discussion ( see Page 9),

it suff'ices to prove that1 if m and n are arbitra.ry different inte­

gers, Q(In) is not isomorphic to Q(Im).

Nov, let us suppose n>m. By the preeeding theorem,

there is no m-tuple of Q(In) having the property (T). On the othe!"

hand, the m-tuple {f1 )*, ••• ,(fm)* of Q(tm), where r1

, ••• ,fm are

mappings of Im into al such that t1 (~1 ••• ,Xm)~ x1

, for any 1=1, •• , m,

obviously has the property (T}. It follows that Q(In) is not iso­

mrphic to Q(Im}.

./25

APPENDIX

In the f'ollowing development 1 ve will f'ollow the no­

tation of' Gillman and Jerison.

T.b.e zero-set of' a continuous :f'unction fe C{X), i.e.,

the set of' its zeros, Will be denoted by Z{f). For the cozero set

of f' 1 we will vri te coz f :

coz f' = X - Z(f) = { x e X 1 f'(x)fO} •

For any ideal D in C{X}, ve define

Z{D) = () Z(f') 1 coz D = X- Z{D) ::: \ J coz f. f'ED fln

Obviously, Z{D) is a closed set and coz D is open.

L'EMMA 1

If D and D' are ideals of' C(X}, then

coz {Dn D' ) = coz Dl\ coz D' •

An ideal D in C(X) is dense - 1n the algebraic sense

def'ined on page 1 - if and on1y if coz D is dense (in the topological

sense).

ISOMORPHISM THEOREM

If' X is a completely regula.r topological space 1 then

the ring Q(C(X)), the maximal ring of' ~otients of C(X), is isomor-

.j26

phic vith Q(X), the ring of equivalence classes cf continuons :f'unc-

tions on dense open sets. (1)

Proof

A member o:f' Q(C(X)) is an equivalence class [~p] of

some <Pc Rom (D,c(x)), vhere D is a dense ideal in C(X).; on the other

hand, a member of Q(X) is an equivalence class (:rf of some fE c(v),

where V is a dense open set in X.

In order tO prove the theorem, we 'Will define a map )i

s: Vxom{D,c(x)) -+ \Jc(v), and a map i: q(c(x)) _., Q(X) by

s([q,J) = (sc,if; a.ud show that: (a) s is vell defined; (b) s is a:n

hom.omorphism; {c) s is one-one; (d) s is onto.

Let <p € Hom(D,C(X)) be given, where D is a dense i­

deal in C(X). According to lemma 2, coz D is a dense subset o:f' x. Let us define a mapping s{"P): coz D= lJ' coz d _,..aJ.,

by s{'il) = ~ on coz d. Then, (i) s(t?) is defined on .coz D1

sin ce d does not vanish on coz d; {ii) s ('P} is we~l. defined. For

Qis a module homomorphism, hence ve have, for any d D, d 1 12 D,

whicb. il:r.g;llies

lfl{d) .. d• = tp{dd•) = (Ç{d1 d) = q:t(dt )-d,

~ = Cl'(~ on coz dncoz d•. d d 1

Now, we define s: Q(c(x)}+Q(X) by ~<rl} = (s(q>)~ (a) This mapping is vell defined. For, if!'!'E: Rom(D,C(X)),

(1) The proof is a slight modification of the one given by Fine, Gilll:!lan and La:mbek, in an unpublished (:f'orthcoming) paper.

·/27

cp 'e Hom(D•,c(X)) and q,- cp', then cp/ DnD• = cp'/ DrtD'. There­

fore, if d € D n D' 1 we have 9 Ad) = <p Ad) on X, and successi v ely

~ = $ '{d) on coz dC::X, s(q>) = s{Q>') on coz d , s~ )=s(q>') d d

on coz (DnD•) = Ycoz d (deDnD•), and so s(cp)=s(q>') on the set

coz Dncoz D'. It follows that s(q>)""'s(cp'), i.e., (s(cp))*=(s(cp•))*.

(b) Let cp e Hom(D,c(x)) and Q>' e Hom(D' ,c(x)).

(i) If de:DnD•, then, on coz d, we have successively:

s(q>+ç>') = {ép+q>)(d) = q>(d)+q>'(d) = cp(d) + q,•(d) = s(cp) + s{<T>'). d d d d

Therefore, sW+Q>') = s(cp) + s(q>') on coz (Dt\ D'), and since

coz (DnD') = dom( s(cp+q>')) = dom( s(cp)+s{cp')) , it follows that

sw +cp')- s(cp)+s(cp' ), which implies s{ [cp] + [<p'] } = s {[cp]} + s{ [Q>']}.

(ii) In order tha.t s(cpcp 1 )- (s(cp)) • (s(q>' >), it suffices that

s (cp cp' ) and s (cp) • s (cp 1 ) agree on coz DD' , sin ce dom( s (cp cp' ) ) =

coz DD•c:coz (nnn•) = dom~(cp)•s(cp•)). Let us suppose deD, d' eD'.

If x € coz dd', then we have (s(cpcp• >) (x) = { 'q>')(dd' >} {x) • (dd') (x)

On the other ha:nd, since dd '€ Df'\ D', we have coz dd 'c. coz D (\coz D',

conseqQently we get successively

( s(cp )•s(cp')) (x) = ( q> (dd') )Cx) • {dd1 ) (x)

= (cp(d)•d') {x) • d( x) • d' (x) ~=-r-""1"'"""'::-:-T---r--

= (cp (d)){x)•d' (x) • (q> 1 (d' ))(x)•d{x) d(x)•d•(x) d(x)•d•(x)

= (cp(d) (x) • ('<J'' (d') (x) d x. ' x

= {(q(d))•(Q>'(d'))} (x)

(dd'} (x)

= {<r>(d•fd'))} (x) = (dd') (x)

= { ~ cp•) (dd' >} (x) (dd') {x)

./28

andso (s(q><p~))(x) = (s(cp)•s(cp•))(x) fora.ll xe:cozDD'.

It :f'ollows that s{ [ q>] • [q> J} = s{ [ q>]} • s{ [ q>'J} •

{c) If 8( [q>]) = s([ qf] ), then (s((J))'/ = (s(cp t >)* and so s(q>) = s(q>') on coz Dncoz D' = coz (Dn D'). If de DnD~

then coz d (coz (DnD') and we have successively: s(q>) = s(q>') ;

~ = (!) '(d) and q> (d) = q> 1 (d). New, in order to prove that d d

cp (d) = (J)' (d) on x, ve will show that cp (d) = q>' (d) on coz (Dn D') 1

vhich is dense in X.

If d'fd and d' e: (D nD'), then q> (d? =cp' (d') on

coz d'. Nov, Q>(d' )•d =cp (d)•d' and cp' (d' )•d = (!)' {d)•d' on x. Ren­

ee, on coz d', <r>(d)•d' =c:.''(d)•d' and {cp(d)- q>'(d)}d• = 0 and

cp {d) = (1) 1 (d). Since d' vas an arbitrary element of Df'\D', ve get

<P{d) = cpr(d) on coz (Dl' D'), and consequ.ently cp(d) =q>'{d) on x.

It follows that q> and (!' 1 agree on DnD• 1 which izçlies ( q:~] =[cp'] •

{d) Given an element {:r)* c Q(X), ve Yish to find an

element cp € U Hom(D,C(X)) such that s(( cp] ) = (f}ll-. Let Df denote

the set {he: C(X~ r f • h/V has an extension on X} , where V = dom f.

We de:f'ine q>: Df..C(X), by cp(d) = the extension of (d/V)•f. Nov, we

will prove that: ( i ) D:r i s dense; {ii) cp is a :f'un.ction su ch that

s(cp) ""f.

.j29

(1) If Xc c V, then there e:xists a. function g.:: C(X), which

vanishes on an open neighborhood G of X-V and such tha.t g(:xo) :f O.

Let us d.efine a. mapping h: x .. al, by

h(x) = t~;f. x eG,

f) {x) if xc v.

Clea.rl.y 1 h is a. continuous extension of gjV • f. Therefore 1 gE Df

and, since g{x0 ) :f 0 1 Xc belongs to coz Df. It follows tha.t V is

a. subset of Df and so D:f' is dense.

(11) If x c V, then xe coz d, for some d.: D :f'• Therefore 1

(~(q>}) (x) = ( q>(d)) (x) = ( d/V • f) (x) = f(x}1

and conse-. d(x) d(x)

qu.entl.y,

BIBLIOORAPHY

( l J R. Fine & L. Gillman & J. La.mbek: "Rings of Qu.otients of Rings of Continuous Functions11 (unpublished)

( 2 J G.T. Whyburn: Article in the "America.n Journal of Mathe­matics", Vol. 61 (19;9), PP• 820-822

G.T. Whyburn: n Ana.lytical Topology", Publ. of the Ameri­ca.n Mathematical Society, 1942

L. Gillman & M. Jerison: "Rings of Continuous Functions", Van Nos­tra.nd Co. Inc., 1960

W. Hurewicz & H. Wallman: "Dimension Theory", 1941