C AHIERS DE
TOPOLOGIE ET GÉOMÉTRIE DIFFÉRENTIELLE CATÉGORIQUES
V ERA ˇ T RNKOVÁ
Simultaneous representations by metric spaces
Cahiers de topologie et géométrie différentielle catégoriques, tome 29, n
o3 (1988), p. 217-239
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
SIMULTANEOUS
REPRESENTATIONS BY METRIC SPACESby
V~raTRNKOVÁ
CAHIERS DE TOPOLOGIE ET
GÉOMÉTRIE DIFFÉRENTIELLE
CATÉGORIQUES
Vol. XXIX-3 (1988)
Dedicated to the memory of
Evelyn
NELSONRESUME,
Etant donn6 trois monoïdesMI c M2 C M3,
il existe unespace
mdtrique complet
P tel que toutes lesapplications
non-constantes de P dans lui-m6me
qui
sontnon-dilatantes forment un monoide =
M,
uniformément continues f orment un monoide =
M2
continues forment un monoide =
M3-
Des r6sultats
plus g6n6raux
etplus
forts sontprouvds.
Onétudie aussi les foncteurs de
completion.
I, SIMULTANEOUS
REPRESENTATIONS AND THE MAINTHEOREM,
1.
By [2],
every group can berepresented
as the group of allautohomeomorphisms
of atopological
space. This result wasstrengthen-
ed in the
following
two ways:(i) every monoid can be
represented
as the monoid of all nonconstant continuous maps of a metric space intoitself, by [5].
(Let us notice
explicitly
that all the nonconstant continuousendomaps
of a space need not form a
monoid; however, given
a monoidM,
thereexists a metric space P such that all the nonconstant continuous
endomaps
of P do form a monoid and this monoid isisomorphic
to M.)(ii) For two
arbitrary
groups G c H there exists a metric space P such that the group of all isometries of P isisomorphic
to G andthe group of all
autohomeomorphisms
isisomorphic
toH, by [3].
The result mentioned in the abstract
strengthens
them both. This result is aconsequence
of the Main Theorem below.We deal with almost full
embeddings
ofcategories.
Let us recall(see
[4])
that a functor F of acategory
K into a concretecategory
H is called an almost full
embedding
if it is faithful anda) every
morphism
of K ismapped by
F onto a nonconstantmorph-
ism of H
and b) if h: F (a) -> F (b) is a nonconstant
morphism
ofH,
then thereexists a
morphism
k: a e b in If’ such that h = F (k).By [5],
every smallcategory
admits an almost fullembedding
into thp
category
of all metricspaces
and all continuous maps. If the embeddedcategory
hasprecisely
oneobject,
we obtain therepresent-
ation of the
morphism-monoid
mentioned in (i) above.2.
Here,
weinvestigate
simultaneousrepresentations:
let D be adiagram scheme,
let C and D bediagrams
over D such that:a) for every
object
a ofD,
C(o-) and D CQ) arecategories;
B)
for everymorphism
m ofD,
C(m) and D(m) are functors.We say that a natural transformation i>: C -> D is a simultaneous
representation
of C in D if(i)
mo :
C (a ) A 0 (a) is an almost fullembedding
whenever 0 (a)is a concrete
category
in which all constants aremorphisms,
(ii)
I>o-:
C(a) e D(a) is a faithful and full functor (= fullembedding)
else.LEMMA. Let C and D be
diagrams
ofcategories
and functors over ascheme
D,
let I>: C -4 D be a simultaneousrepresentation
of C in D.If,
for amorphism
m ofD,
D(m) is a faithfulfunctor,
then C (m) is also a faithful functor.PROOF. If 4l: C 4 D is a simultaneous
representation
of C inD, then,
for every
morphism
m: u -4 a’ ofD,
we have:and both
I>o-, I>o-
are faithful functors. If D(m) issupposed
to befaithful,
thenD (m)I>o-
is alsofaithful,
so that C(m) must be faith- ful.3. Let us denote
by
Metr the
category
of all metric spaces of diameter 1 and all theirnonexpanding
mapsi.e.,
f is amorphism
ifffor every x, y of the domain of f
),
Metrc its full
subcategory generated by
allcomplete
spaces,Unif the
category
of all uniform spaces and alluniformly
continuousmaps,
Unifc its full
subcategory generated by
allcomplete
spaces,Top
thecategory
of alltopological
spaces and all continuous maps.In the Main Theorem
below,
weinvestigate
simultaneous repre- sentations in thefollowing diagram
M:In the
diagram M,
thecompletion
functors are as follows:c,: Metr -+ Metrc is the metric
completion
of metricspaces,
c,,: Unif - Unife is the uniformcompletion
of uniformspaces,
the letter Falways
denotes theforgetful
functor:F"(M)
is theuniform space
underlying
the metric spaceM,
andFu (U)
is thetopo- logical
spaceunderlying
the uniform space U.Clearly,
the square in thediagram /1 commutes, i.e., F.oc. = cuoF..
Moreover,
the square has thefollowing property:
ifM,, M2
are metric spaces (of diameter ( 1) andf:
C"(M1)
AC"(M2)
is amorphism
in Metrland g:
F. (M,
) ->F. (M2)
is amorphism
in Unifsuch that
F"
(f) = Cu(g),
then there exists aunique morphism
h:M, d M2
in Metr such that c. (h) = f and
F.
(h) = g.(In
fact,
f is anonexpanding
map of thecompletion c.(M,)
into thecompletion c..(M2);
the conditionF"(f) = cu(g)
says that f(x) =g (x)
for every
point
x EM, ,
so that f mapsM,
intoM2 ;
since f is non-expanding,
itsdomain-range-restriction
h:M, 4 M2
is alsononexpand- ing,
hence amorphism
ofMetr ; then, clearly,
f =c"(h) and g = F" (h).>
The last
property
says that "the square is apullback
onmorphisms".
(The square in M is also a"pullback
onobjects"
but thisplays
no r61e in ourinvestigations.)
Let us call any commutative square with thisproperty
asubpullback.
4. Let us denote
by
Cat thecategory
of all smallcategories
and allfucntors.
OBSERVATION. Let
be a commutative square in Cat. Then it is a
subpullback iff, forming
a
pullback
the
unique
functorG: ko ->
h such thatHioG = Gi
for I =1,2,
is a fullembedding.
5. LEMMA. If
C, D
bediagrams
ofcategories
and functors over ascheme
D,
let I>: C -> D be a simultaneousrepresentation
of C in D.Let
CY
be a commutative square in D such that its
D-image
is asubpullback.
Suppose
that eithera)
i>o-.
is a fullembedding
or b)
I>o-o, I>o-,
,I>o-2
are almost fullembeddings
and D(m,),
I =1,2,
preserve constantmorphisms.
Then the
C-image
of the square above is also asubpullback.
PROOF is
quite straightforward.
LetA,, A2
beobjects
of thecategory C(0o),
letbe
morphisms of C(o-i),
i =1,2,
such thatThe
morphisms gi = 4l,; t’ff)
(which arenonconstant,
in the case b) ful- fillso that there exists a
(unique!)
(Moreover, t
isnonconstant,
in the case b.) We find a(unique!)
mor-phism
h:A,
AA2
inC (uo)
such thatI>o-o (h)=
1.Then, clearly,
so that the
C-image
of the square above isreally a subpullback.
06. In the
terminology
of simultaneousrepresentations,
the result mentioned in the abstract says thefollowing:
everydiagram
where
k1, k2, k3,
areone-object categories
andG,, G2
are one-to-onefunctors,
has a simultaneousrepresentation
in thediagram
As a
corollary
of the Main Theorem (see Remark b in 7) we obtain thefollowing
assertion:C,
has a simultaneousrepresentation
inD1,
whenever
k, , k2, k3
are smallcategories
andG,, G2
are faithful, functors. This is an essentialgeneralization.
If we choose e.g. thecategories k,, k2, k3
discrete (= theironly morphisms
are the ident-ities) and such that
the
cardinality
ofobj k3
isequal
to m3,G2
sends m2objects of k2
on eachobject
of&, G, sends m, objects of k,
on eachobject
ofk2,
where m, , /)2’ m3 are
given cardinals,
we obtain "arigid
tree" of metric spaces: there is a set ofcardinality
n)3 of metrizable spaces without nonconstant nonidentical continuous maps; each of these spacescan be uniformized
by m2
metrizable uniformities such’ that there is nononconstant nonidentical
uniformly
continuous map in the obtained setof uniform
spaces;
and each of these uniformspaces
can be metrizedby
m, metrics such that there are no nonconstant nonidenticalnonexpanding
maps in this set of metric spaces.7. Let S be the
following
scheme:where
p1 om1 = p20m2, Clearly,
thediagram
M in 3 is adiagram
over S.MAIN THEOREM. Let C be a
diagram
in Cat over S. Then C has a simul-taneous
representation
in M iff all the functorsC (q), C(Pi), C Cmi ),
i -
1,2,
are faithful and theC-image
of the square in S Is asubpull-
back.
REMARKS. a) The
necessity
of the conditions in the Main Theorem is almost evident: since all the functors CM, c",F., Fu
in tl arefaithful,
all the functorsC(q), C (mi), C(pi)
must be alsofaithful, by
Lemma2;
and the
C-image
of the square of S must be asubpullback, by
Lemma 5.The
parts
II and III of thepresent
paper are devoted to theproof
that the above conditions are also sufficient. In the
part IV,
wepresent
somestrengthenings
aboutrepresentation
of groups and Brandtgroupoids.
b) Let us mention
explicitly
how the result in 6 isimplied by
the Main Theorem: ifis a
diagram
in Ca t such thatG, , G2
arefaithful,
choose C: S -> Ca tas follows:
k
Then C fulfills the conditions of the Main Theorem so that there is a
simultaneous
representation
I>: C -> M. Then{I>o-1,I>o-3,I>o-4)
form asimultaneous
representation
ofC,
inD1.
I I ,
METRIC AND TOPOLOGICALCONSTRUCTIONS.
1. Let us recall that the
category
Metr has allcoproducts:
if((M,,d,)
I 1. EI)
is a collection ofobjects
ofMetr,
itscoproduct IIicI,(Mi,di)
is the space(M,d)
withM = UicIMix{i}
andIf there is no
confusion,
we suppose that the setsM,
aredisjoint
andwe omit the
multiplication by
theone-point set {i} making
these setsdisjoint. Hence,
weput simply
M =UicIMi
and d is an extension of all the metricsd, by
the ruled (x,y) =
1 if x, y are in distinctM/s.
Werecall that Metr has also
quotients:
let(M,d)
be anobject
of Metr and q: M 4Q
be asurjective
map; define c onQ by
where the infimum is taken over all sequences xo, yo,..., xn, yn of
elements
of M such that
then c is a
pseudometric
onQ and, identif.ying
thepoints
x, y ofQ
with
cex,y)
=0,
we obtain thequotient
ofCM,d?
inMetr,
determinedby
the map q.
In the construction
below,
we use the above constructions in Metr.However,
all ourquotients
will be sosimple
that thepseudo-
metric c
given by
the above formula willalready
be a metric.2. Let C be a Cook
continuum, i.e.,
acompact
connected metricspace, nondegenerate (i.e.,
with more than onepoint)
and such- that:if K is a subcontinuum of C and f: K e C is a continuous map, then either f is constant or f (x) = x for all x E K.
A continuum with these
properties
was constructedby
H. Cook in[1].
Amore detailed version of the construction is contained in
Appendix
ALet J be the set of all
integers,
letbe
systems
ofnondegenerate
subcontina of C such that thesystem
is
pairwise disjoint.
Hence(4) if
X,Y
cX ,
K is a subcontinuum of X and f: K -> Y is acontinuous map, then either f is constant or X = Y and f (x) = x
for all x E K.
We
may
suppose(by
a suitablemultiplication
of metrics) thatChoose,
in each member ofX,
twopoints
with the distanceequal
tothe diameter and denote them
by
Let us denote
by
P thespace
obtained from thecoproduct
(in Metr) of all members of Xby
thefollowing
identifications (thequotient
inMetr):
the obtained
point
is denotedby
ao;for i> 1
the
obtainedpoint
isfor i -1 denoted
by
a;;the obtained
point
is denotedby bn,o;
for i
1}
the obtainedpoint
isfor i ( -1
}
denotedby bn, j;
the obtained
point
is denotedby
Cn,O;for i >
1}
the obtainedpoint
isfor i -1 denoted
by
Cn, iMoreover, identify,
foreach
EJ, j ? 0,
The space P is indicated
by
thefollowing figure
To obtain a
completion
cP ofP,
we have to add fivepoints
toP, namely
3. In what
follows,
weinvestigate
thesubspace Q = cPB {83}
of cP.Let us denote
by
d the metric of cP. We consider thefollowing
threemetrics on
Q:
OBSERVATIONS. a) The
identity
map ofQ
isnonexpanding
as the map(Q,p3) -> Q,P2)
and(Q, p2 ) -> (Q,p1), uniformly
continuous but notnonexpanding
as(Q,p1)
A(QIP2),
continuous but not
uniformly
continuous as(Q,P2) -> (Q,p3).
b)
Q
with each of the metrics p1, P:z: P3 is Al"omp1ptA space,
thecompletion
ofQB {82} in
each of these metrics isQ again.
Let usdenote
QB {82} by Q-
and the metric p; restricted toQ+ by
p;again.
4. All the metrics p,, P2, P3 are
equivalent,
let us denotesimply by Q
thecorresponding topological
space.CONVENTION. To
simplify
thenotation,
wesuppose
that the continuaA;, Bn. i, C.,., ABn, it BCn. i, CAn,j
in thefamily
X aresubspaces
ofQ, homeomorphic
to theprevious
ones,i.e., homeomorphic
to adisjoint family
of subcontinua of the Cook continuum C.LEMMA. Let Y be a
topological
spacecontaining Q
as a closedsubspace
and such that the closure
YBQ c (a+,a,e1, ) U (YBQ).
Let X be a continu-um in
X ,
let f: X -> Y be a nonconstant continuous map. Then either f(X) cQ
and f is the inclusion(i.e:,
f (x) = x for all x E X) or f(X) cYBQ.
PROOF. Put
Let us suppose that f(X) intersects
QBS.
Then 0 = f-I(QBS)
isnonempty
and open.
a) If XBO =
0,
thenf maps
the whole X intoQBS;
since f(X) isconnected,
it must be contained in some member K ofX,
distinct fromX,
so that f must be constant. This is a contradiction.b) Let us suppose that X B0 # 0. Choose x E 0 and denote
by
C the
component
of 0containing
x. Since the closureC
of C intersects theboundary
of0,
hencef(C)
intersects theboundary
ofQBS.
Find themember K of X such that F(X) c K. Since f(C) c
K,
then alsof(C) c
K.But
C
is a subcontinuum of X and K is distinct member ofX,
hence f is constant onC by
(f).Consequently f maps
the wholeC
on thepoint
f(X) E
QBS,
which is a contradiction.We conclude that
(X)n (QBS) =
0. Since f(X) is anondegenerate
con-nected space,
necessarily
either f(X) c X (and then f(x) = x for allx E X) or f(X) c
Y BQ.
5. PROPOSITION. L e t Y be a
topological
spacecontaining Q
as aclosed
subspace
such that theYBQ c {a+,a,e1} U (Y-Q).
Let f:Q+
e Ybe a continuous map. Then ei ther
f(Q+)
cQ
and f is the inclusion(i. e.,
f (x) = x for all x EQ+)
orf(Q+) C YBQ
or f its a constant map.PROOF.
By II.4, Lemma,
f restricted to any member X of X is either a constant or f(X) cYBQ
or f(X) = X and f (x) = x for all x E X. Let ussuppose that f restricted to some X in X is
constant,
say f(X) ={Xo}.
Let S be as in the
proof
ofII.4,
Lemma.a) If xo E
QBS,
then every member of X which intersects X has to bemapped by
f on xo. We can continue to the next members of X .Finally,
we obtain that f maps the wholeQ+
on xo.b) Let us suppose that
b1) Xo $
X: In the definition of P (see11.2),
the identifications ofpoints
in the members of X are chosen such that for every K in X notcontaining Xo,
there exists a chainXo = X, X1 ,
...,Xk -
K of members of X such that none of themcontains Y,
andX;
intersectX;+, for J
=0,..., k-1. By
II. 4Lemma,
fhas
t o mapXo = X, X1 ,
...,Xk -
K on t o 4.Hence f maps any K in X not
containing xo
onto X0Consequently
any member K of X whichcontains 4
contains also apoint
x distinct from xo with f (x) = v.0.By
11.4Lemma, f maps
K onto )(0. We conclude that f maps the wholeQ+
onto xo.b2) Xo
E X:let h
be a member ofX ,
which intersects X in apoint
distinct from xo;then Xo$X
(see 11.2).By
11.4Lemma, f maps 5;:
onto )(0.
Now,
use the case bl for2.
We conclude that if
f(Q+)
intersectsQB{a+,a-,e,,82},
then either f isconstant or F(x) = x for all x E
Q+. Moreover,
iff(Q+)
contains e2, then it isnecessarily
constant. (Infact, f(Q+)
is connectedand e2
isan isolated
point
ofIn the
remaining
case,f(Q+) C YBQ.
III.
THE PROOF OF THE MAINTHEOREM.
1. Let D =
(D,4)
be aposet
(=partially
ordered set) with alargest
element t. For
every 4
ED,
denoteby GD (do)
thefollowing category:
objects
are allpairs (X, {Ro-
I d>4 }),
where X is aset, Rd C
XxXfor every d E
D,
d >d,,
anda) the directed
graph (X,Ri)
is connected(i.e.,
for every x, y E X (notnecessarily
distinct) there exist Xo = x, x, , ..., Xn = y in X such that(xi-1,xi)
ERfuR,-1
for I =1,..., n>,
contains noloops (i.e.,
never (1(.:V) E
Ri)
and rardY 1 2;
b)
if cg d1 d2,
thenRo-1 c Rd2.
A map f; X -> X’ is a
morphism
of
GD (Ct) if f,
for every d>do, (x,y)
ERd implies Cf Cx), f Cy))
ER’d (Le.,
f isRdR’d-compatible
for every d ;do).
If
do ( d, ,
there is a naturalforgetful
functornamely
We consider the
poset
D as acategory:
ifdo ( d, ,
denote theunique morphism from do
.tod, by m(do,d1).
Weinvestigate
thediagram Go
overD
consisting
ofGD(do)
andIn
[6],
thefollowing auxiliary
lemma isproved.
AUXILIARY LEMMA. Let D be a
poset
with a last element t. Let C:D -4 Cat be a
diagram
suchthat,
for everymorphism
m ofD,
the functor C (m) is faithful. Then there exists a simultaneousrepresent-
ation m of C in
Go. Moreover,
ifdo, d1,d2
ED, d, = d1^d2
in D and the squareis a
subpullback
inCat,
thencan be chosen such
that,
for everyobject
of CCdo),
its(Dd -image
(X, {Ro-
I d>do})
fulfillsR,,
=Rd " Rd .
REMARK. Since constants need not be
morphisms
ofGO(d)
the functorsI>d:
C (d) ->Go
are fullembeddings
(notonly
almost full as it is in the case ofMetr, Unif, Top,
see the definition of the simultaneousrepresentation
in 1.2).2. In the
proof
of the MainTheorem,
weapply
theauxiliary
lemmato the
diagram
scheme S in 1.7. Forshortness,
let us write I instead of a; so that S is thefollowing
scheme(p1 om1 = pom2)
Let C: S -4 Cat be a
diagram
suchthat,
for everymorphism
m ofS,
thefunctor C (m) is
faithful;
moreover, let the squarebe a
subpullback
in Cat (since the functor C(q)
isfaithful,
it isequivalent
to the fact thatis a
subpullback
in Cat).By
theauxiliary lemma,
there exists a simul- taneousrepresentation
I>: C ->Gs,
whereGs(i)
is as in III.1 for I =1,2,3,4,
andGs (0)
is the fullsubcategory
ofGs (0)
in III.1generated by
allthe
(X, { Ro . R1 , R2 ,R3, R4 })
withRo = R, n R2,
and,
for everymorphism
m ofS, Gg (m)
is as inIII.1, i.e.,
the corres-ponding forgetful
functor E.3. A
composition
of a fullembedding
and an almost fullembedding
is an almost full
embedding again. Hence,
to prove the MainTheorem,
it is sufficient to find a simultaneous
representation
’l1:Gg
-> M. Then W0I> is a simultaneousrepresentation
of C in M.4.
First,
we define the functorWo: Gs (0 )
e Metr. Let o- =(X, {Ro,R1,R2,R3,R4})
be anobject
ofGs(O), i.e.,
card X) 2 andR,
is aconnected
binary
relation on X (hence in every x E X either an arrow starts or terminates) withoutloops,
Let us denote
by
a,., ar-> er, thepoints
a+, a-, e, in the copy Z’. In thecategory Metr,
we form thecoproduct IIrER4
Z r and then thefollowing
identifications:The obtained metric space is
To
(a).Let
be a
morphism
ofGs (0), i.e.,
fxf mapsRd
intoRd’
for d =0,1,2,3,4.
Wedefine Wo (f) = g
suchthat g
maps apoint z
in a copy Z r to the samepoint z
in the copy Z" with r’ = (fxf)(r). Since f isRdR’d-compatible,
the identifications (**) are
preserved. Moreover,
since f isR2R’2- compatible,
thepoint
e 2 hasreally
itsimage er’2
in Z r’ whenever r ER2. Consequently
the map g:Wo (0-) -> Wo (o-’)
iscorrectly
defined.Since the
identity
mapsare
nonexpanding, g
is also anonexpanding
map. We conclude thatTO:
Gs(O) ->
Metr is acorrectly
defined functor. It isfaithful, obviously.
The fact that it is almost full will be
proved
in 9.5. Now we define the functor
’l1,: Gs(1)
-9 Metrl. Let a =(X, {R1,R3,R})
be an
object
ofGs (1), i. e.,
card X > 2 andR,
is a connected relation withoutloops, R, c R3 C R,),
For every r E
R, ,
let Z’ be a copy of(Q,p1).
For every r E
R3 BR1,
let Z r be a copy of(Q,P2)’
For every r E
R4 BR3,
let Z r be a copy of(Q,p3).
In the
category Metr,
we form thecoproduct IIeCR4 Zr
and then the identifications (**) as above. The obtained metricspace T,
(a) iscomplete,
hence it is anobject
of Metrc. Ifis a
morphism
ofGs (1),
we define the mapW1 (f) = g similarly
as in2, i.e.,
g maps z in a copyZ’’
to the samepoint
z in the copy Z r’ with r’ = (fxf) (r). Then g is anonexpanding
map.Clearly, ’II,: Gs
(1) -> Metrc is acorrectly
defined faithful functor.6. We define the functor
W2: Gs (2)
e Unif as follows: let ,a =(X, {R2,R3,R4})
be anobject
ofG s (2 (i.e., (X,R,)
is as in 3 andR2 c R3 c R4).
For every r E
R2,
let Z be a copy of(Q,P2)’
For every r E
R3BR2,
let Z r be a copy of(Q+,P2).
For every r E
R. BR3,
let Z be a copy of(Q+.P3).
In the
category Metr,
we form thecoproduct IIrCR4, Zr
and the identif-ications (**). The uniform space determined
by
the obtained metric space(i.e.,
theF"-image
of the obtained metricspace)
isW2 (o-).
If f: o- e a’ -
(Xc,{R’2,R’3,R’4})
is amorphism
ofGs (2),
we de f ineW2
(f) = g as in 2 or3, i.e.,
it maps apoint
z in a copy Z r to thesame
point
z in thecopy
Zr’. One can seeeasily that W2
is acorrectly
defined faithful functor.
7. The functor
W3: Gs(3)
e Unifc is defined as follows: let 0’ =(X,{R3,R4})
be anobject
ofGs(3).
For every r E
R3,
let Z be a copy of(Q,P2).
For every r E
R4 BR3,
let Z be a copy of(Q, P3)’
In the
category Metr,
we form thecoproduct IIrCR4 Zr
and the identifi- cations (**) as above. ThenT3(a)
is the uniform space(i.e.,
theF.- image)
of the obtainedcomplete
metric space. The’I13-images
of mor-phisms
f: a 4 o) ofG3 (3)
are defined as in 2 or3, i.e., W3
(f) sends every z in Z’ r to the samepoint
z in Z r’ with r’ = (fxf)(r). ThenW3
isa
correctly
defined faithful functor.The f unctor
W4: Gs
(4) ->Top
is def ined as follows: if o-(X, {R4 })
is an
object
ofGs(4),
denoteby
Z r a copy of(Q,p3);
in thecategory Metr,
form thecoproduct IIrcR
Z-r and the ident if icat ions (**) as above.Then
W4(o-)
is thetopological
space of the obtained metric space ti.e., theFuoF"-image
of the obtained metricspace).
If f: a -> o-’ is amorphism
ofG, (4), W4
(f) is def inedsimilarly
as in 2 or 3.PROPOSITION. W:
Gs
-> M is a natural transformation.PROOF. a) If a =
(X,{R3,R4})
is anobject
ofGs (3), Gs (q)
sends itto the
object (X,{R4}).
HoweverFu(W3((X,{R3,R4}))
is the sametopo- logical
space asW4,((X,{R4}))
because p, and P3 areequivalent
metricson
Q. Consequently
the squarecommutes.
b) The square
commutes because the metrics p, and P2 are
uniformly equivalent
onQ.
c) The square
commutes because
(Q,p2)
is acompletion
of(Q+P2)
and(Q,p,)
is acomplet ion
of(Q+,P3).
d) The square
commutes because
(Q,p,)
is acompletion
of(Q+,pi), i = 1,2,3.
e) The square
commutes because p, and P2 are
uniformly equivalent
metrics onQ.
8. PROPOSITION. The functor
W4: G,(4) -> Top
is almost full.PROOF. a) Let Q -
(,. {R4})
be anobject
ofGs
(4). If h:Q
->T4
(o-) isa continuous map, then either h is constant or there exists r E
R,
such that h sendsQ
to the copy Z’ r ofQ
such that each z EQ
is sentto the same z in ZI. This follows
immediately
from 11.5.b) Let a =
(X, i R4 1),
a’ =(X’,(R’,))
beobjects
ofGs(4),
leth:
’P, (0’)
->W4 (o-’)
be a continuous map. Let us suppose that there exists ro ER4
such that the restriction Z’- ->W4(o-’)
of h isconstant,
so that
Since
R’4
is a connectedbinary
relation onX’, necessarily
the restric- tion of h to anyZ’,
r ER"
isconstant,
so that h is a constant map.If h is
nonconstant, then,
for every r ER,,
there exists r’ ER’4
sucht hat h maps Z r onto Zr (and
hence,
it sends z in Zr on z inZr ’, by
a).Since
R,
is connected(hence,
in every EX,
an arrow either starts orterminates) and h
preserves
the identifications(**),
there exists acompatible
map f: a -> Q’ suchthat h = W4
(f),9. PROPOSITION. The functors
W3: Gs (3)
-4Unifl, W1 : Gs (1)
-4Metre, T2: Gs
(2) ->Unjf, q.r 0: Gs (0)
-> Metr are almost full.PROOF. a) Let o- =
(X,{R3,R4.})
o- =(x’,(R’3,R4’,})
beobjects
ofG5(3),
let h:’P3 (a)
-4W3(o-)
be a nonconstantuniformly
continuous map.Then
is a nonconstant continuous map so that there exists an
R4R’4- compatible
map f:(X, (R, ))
->(X’, {R’4})
suchthat 9 = T4 (f), by
111.8.Then
both fi and
h send each copy of Z r on the copy Z r’ with r’ = (fxf) (r). Since h isuniformly
continuous while theidentity
map(Q,P2)
->(Q,p3)
isnot, necessarily
r’ is inR’3
whenever r ER3,
hence fis also
R3R’3-compatible.
b) The
proof
thatWi
is almost full isanalogous.
Givenobjects
and a
nonexpanding
nonconstant map h:B11,
(a) iWi (a’),
we find anR,R’,-
compatible
mapThen f must also be
R3R’3-compat ible
andR, R’, --compat ible
because theidentity
maps(Q,P2)
4(Q,P3)
andCQ, p, )
->(Q,P2)
are notnonexpanding.
c) The
proof
that’V2
is almost f ull isanalogous.
It usesthe facts that the
identity
map(Q+,P2)
-1(Q+,P3)
is notuniformly
continuous and that there is no nonconstant
uniformly
continuous map(Q,P2)
->(Q+,P2)
d) The
proof
thatWo
is almost full is alsoanalogous, only
more facts are
used, namely
that:the
identity
maps(QIP2)
->(QIP3)
and(Q,p1)
A(Q,P2)
are notnonexpanding
(andanalogously
forQ*)
and there are no nonconstant
nonexpanding
maps(Q,P2) -> (Q+,P2)
and(Q,p1) -> (Q+,p1).
I V, REPRESENTATION
OF GROUPS ANDBRANDT
GROUPOIDS.
1. In 1.1 we
already
mentioned the result of[3]:
for every two groups G c H there exists a metric space P such that the group of allisometries of P is
isomorphic
to G and thegroup
of all autohomeomor-phisms
of P isisomorphic
to H. The result about three monoidsMI C M2 C M3, proved
here and mentioned in theabstract,
isstronger
even if we choose groups
M,
andM3, disregarding Ms.
In this case weobtain
that,
for twoarbitrary
groupsM,
cM3,
there exists a metric space P such that(***) every nonconstant
nonexpanding
map of P into itself isalready
an
isometry
and every nonconstant continuous map of P into itself isalready
anautohomeomorphism
and the isometries form a group
isomorphic
toM,
and the autohomeo-morphisms
form a groupisomorphic
toM3-
2. In
[3], embeddings
of agiven
metric spacePo
into a metric spaceP, representing
thegiven
groups G cH,
areinvestigated.
The authorproves there that for every metric space
Po
there exists a metric space Pcontaining Po
andrepresenting
thegiven
groups G c H in the above sense. This is not true ingener.al,
if werequire
also the val-idity
of (***). Infact,
if thegiven
spacePo
contains an arc, then everycompletely regular
space Pcontaining Po
admits many nonconstant continuous maps into this arc so that the nonconstant continuous maps cannotrepresent
the trivial group.However, disregarding (f..), stronger embedding
results can beproved by
thepresent methods,
see theproposition
below.3. Let us recall that a small
category
b is called a Brandtgroupoid
if each of its
morphisms
is anisomorphism.
If K is acategory,
let usdenote
by
iso K itssubcategory
formedby
allobjects
of K and allisomorphisms
of K. Let us denoteby F 1ft
andFu again
thedomain-range
restrictions of the
forgetful
functors. Weinvestigate
thefollowing diagram
Iso D:PROPOSITION. Let a metric
space Po
with diamPo
1 begiven.
Thenevery
diagram
where
b, , b2, 16
are Brandtgroupoids
andG, , G2
are faithfulfunctors,
has a simultaneous
representation
such that for every
object
a ofb,,
the spacePo
is a retract of theSpace I>1
B01 in Metr(i.e.,
there arenonexpanding
maps e:Po
-1I>1
(a) andr:
I>1
(a -iPo
such that roe =identity
onPo).
REMARKS. a) The
equation
roe =identity implies
that e preserves the metric ofPo
so thatPo
is asubspace
ofI>1
(o-). Since the choice of the Brandtgroupoids b1,b2, b3
is ratherfree,
we canalways
suppose thatso that
F*(Po)
is a retract in Unif of everyI>2(o-),
a Eobj b2,
andFu(F"Po>)
is a retract inTop
of everyI>3(o-),
a Eobj b3.
If each ofb1, b2, b3,
hasprecisely
oneobject,
we obtain the result aboutrepresent-
ation groups. But
b1,b2,b3,
can be chosen e.g., to be discrete and thenwe obtain the existence of an
"isomorphism-rigid tree", analogously
asin 1.6.
b) The restriction that diam
Po
1 is notessential;
it ispresumed
in theproposition
for the sake of asimple
formulation.Inspecting
theproof below,
one can see thisimmediately.
4. PROOF OF THE PROPOSITION. 1) Let
Po
be a metric space with diamPo ;
1 and adiagram
of Brandt
groupoids
and faithful functors begiven.
Let h be adiscrete
category
withobj
h =Po.
Weput ki = bi
IIh,
I =1,2,3
(wesuppose
that b;
and h aresubcategories
ofki)
and extend the functorsG,, G2
as identities on h (let us denote the extended functorsby G,
and
G2 again). By
1.7 (Remarkb),
there exists a simultaneous repre- sentation B =(B1,B2,B3)
of thediagram
into the
diagram
2) For every z E
obj h = Po,
choose apoint
az inB1
(z).Now,
wedefine a functor
I>: b,
-> iso Metr : for every o- Eobj b, , I>1 (o-)
isobtained from the
coproduct
(in Metr)by
the identification (in Metr ) of each z EPo
with a3 EB,
(z). Let L denote thesubspace
ofm,
(o-) obtained fromPo
uIIzEpo B,
(z) and let ussuppose that
Moreover,
let ussuppose
that(Clearly, I>1,
(a) can be retracted toPo; .
thenonexpanding
map rsending B, (z)
to 6z andB, (c)
to anarbitrary point
ofPo
fulfills roe =ident,
where e:
Po fl 0,
(Q) is the inclusionmap.)
If m: a -> a’ is amorphism
of
b, , i>1
(m) mapsB,
(a) ontoB,
(o-’) asB,
(m) and it is identical on L.Thus, 0,
is acorrectly
defined faithful functor.3) Since we may suppose that
the
commutat ivit y
conditionsalready
determine the functorsThey
arefaithful, obviously.
4) To finish the
proof,
we have to show that all the functors01, 01, I>3
are full. LetQ,a’
Eobj b,
and let f be ahomeomorphism
ofthe metric
space 0,
(a) =B,
(a) u L onto0,
(a’). We prove that f mapsB,
(a) ontoB,
(a’) and it is identical on L.a) The definition of h and B
implies that,
forevery
z EPo,
every continuousmap
ofB,
(z) intoB,
(o-’) isconstant;
since f is one-to-one, necessarily f(B,
(z)) c L for every z EP..
We conclude thatb) For every z E
Po,
let r.: L ->B,
(z) be the retractionsending
every .x EL B8, (z)
to a,. If z1, Z2 EPo, Z1,#
z, then the mapis constant (this follows from the fact that the
category
h is dis-crete). Since f is
one-to-one,
this constant map cannot be inB1 (z2)B {az2 )
becauserZ2 -1 (y)
={y}
for everyy E B, (Z2)B{ar2}
Con-sequently
rz¿(f(B1(z1))) = {ar2}’ If 4
ranges overPoB{z1},
we obtainthat
f(B,(z,)) c Po U B1(z1).
Let r:P,
uB, (z, )
->B1(z1)
be the retractionsending Po
to z,. Then ro f mapsB,(z1) continuously
intoB1(z1)
so it must be eitheridentity
or a constant. If it is a constant thennecessarily r(f(B, (z,))
={z1}
because r1(y)
={y}
for ally E
B1(z1)B{z1};
hencef(B, (z1)) c Po;
but this is a contradictioon becauseB, (z,)B {z1}
is open inI>1 (a),
f is ahomeomorphism
and nosubset of
Po
is open inI>1
(o-’). We conclude that ro f mapsB, (z1 )
onto itself as theidentity. Consequently
f maps L onto itself as the iden-tity.
Since f is one-to-one on0,
(a) =8,
(a) u L and it maps L ontoitself,
it mapsnecessarily B,
(a) intoB,
(a’). L,et us denoteby
g:B,
(o-) ->B,
(a’) thedomain-range
restriction of f. If f is anisometry, necessarily g = B1 (m)
for amorphism
m: a 4 or’ inby,
so that f =I)1
(m). If f is a uniformhomeomorphism
(orhomomorphism)
then it is02-image
(orØ3-image)
of m: Q 4 a’in b2
(or inb3) -
this is evidentnow.
REMARK. If the
given
spacePo
iscomplete,
then the constructed spaces1>1
(a) are alsocomplete
so that 0 is a simultaneousrepresentation
inthe
diagram
5) The
presented proof
can beeasily
modified to obtain thefollowing:
Let a metric spacePo
with diamPo ;
1 begiven.
Then everydiagram
over Swhere the square is a