C AHIERS DE
TOPOLOGIE ET GÉOMÉTRIE DIFFÉRENTIELLE CATÉGORIQUES
C. B. S PENCER Y. L. W ONG
Pullback and pushout squares in a special double category with connection
Cahiers de topologie et géométrie différentielle catégoriques, tome 24, n
o2 (1983), p. 161-192
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
PULLBACK AND PUSHOUT SQUARES IN A SPECIAL DOUBLE CATEGORY WITH CONNECTION
by
C. B. SPENCER and Y. L. WONG CAHIERS DE TOPOLOGIEET
GÉOMÉTRIE DIFFÉRENTIELLE
Vol. XXIV - 2
(1983 )
INTRODUCTION.
This paper continues the work of
[12]
in an attempt torephrase
classical
homotopy theory
in thesetting
of aspecial
double category with connection via the concepts ofpullback
andpushout
squares. In the pre-sent paper we
adopt
the more convenient notation of thin squaresemploy-
ed
initially by
R. Brown andP. J. Higgins
in their discussion of doublegroupoids
and otherhigher
dimensionalobjects [1, 3,
4 and5].
Most of the results obtained arose from an effort togeneralise
the «cube theorems) of[11]
to the presentsetting.
While it was found thatthey
are not true ingeneral,
the second author has obtained such results in the category of chaincomplexes [13].
The first two sections are devoted to basic ideas and definitions and in Section 3 we discusshomotopy equivalences
andhomotopy
commutative cubes. Anequivalence
relation on squares, similarto that
given
in[11],
is established.Next, pullback
andpushout
squaresare recalled and some of their further
properties
established. We obtain versions of the well-knownglueing
andcoglueing
theorems inhomotopy theory [2].
We includeexamples
ofpullback
andpushout
in thespecial
double category with connection of chain
complexes
CC . This will be dis- cussed further in[13].
1. DOUBLE CATEGORIES.
DEFINITION 1. 1. A double
category
D consists of setsDo, D1
andD2
(of points, edges
and squares,resp.) together
withboundary
mapssatisfying
the rules for a two-dimensional cubicalcomplex
such that(D I, D0, 600, 610, 1)
is a category(composition
in which is denotedby juxtaposition).
In addition we must havecomposition
laws +1 and+2 ,
called resp. horizontal and verticalcomposition,
defined onD2 satisfying
thefollowing
axioms :( i )
for eacha, b e- D2,
ai + b is defined whenever( iii )
for eachf E D1, =f (resp. | |f)
is theidentity
for thecomposi-
tion +1 (resp. +2). = f
and II f
are called resp. the horizontal and verticalidentity.
(iv)
for each x rD0, = 1 x =||1x
. This square is called the doubleidentity
and is denotedby ~x.
( vi ) (the interchange law )
To make the
picture
more clearare sometimes used to represent a square,
= f and I I f
inD2.
It is also convenient to use a matrix notation forcompositions
of squares.Thus,
ifa, b satisfy d’ a - 601 b ,
wewrite [a b]
forat b,
andif 612 a = 602 b,
we write
[ab] for a
+ b . Moregenerally,
for squaressatisfying
we write
for
The notion of double category was first introduced
by
C. Ehres-mannin[7].
In our
definition,
the horizontal and verticaledges
form the samecategory
D1
whereas in thegeneral
casethey
may be two different categ- ories. LetD 1
be a category, then the double category with squares cons-isting
of all those commutative squares inD1
is denotedby ~D1.
Cons-ider squares in D of the form
they
form a substructure of D which is refered to as the horizontal sub- doublecategory
h( D ) o f
D .Similarly
we have the concept ofvertical
sub-double category v (D) o f
D .They clearly
constitue a2-category
struc-ture under
f.
For more detailed discussions of doublecategories,
confer[6, 7,
8 and10].
DEFINITION 1.2. A double category D is said to be with connection if a
double
functors ~D1->
D is defined such thatA |(~D1)1
is the iden-tity
functor.Functions
-,- : D1-> D2
for which-f, -f
haveedges given
are determined
by restricting A
to squares of~ D1
of the formThe functions J and F are called connections and were first
introduced
in[3]
to deal with doublegroupoids.
Here we recall some of theproperties
of -.J and T described in
[12].
( i ) Transport
l aw :DEFINITION 1.3. A double category D is
special
i fh ( D )
forms a group-oid under
+1.
For each
a E h (D), - a
is used to denote its inverse. Here the wordspecial
has a differentmeaning
from theliterature,
e. g.[6, 12].
From now on, we shall restrict our attention to
special double
cat-egories
with connection.Throughout
this paper we shall use D to denote such anobject.
Asexamples,
we have thespecial
doublecategories
withconnection TT and CC of
topological pointed
spaces and chaincomplexes
resp. In T’T’ a square is the
homotopy
class of ahomotopy
from611 a 602 a
to
aJ a 601 a
and in CC, a is a chainhomotopy from 611 a 602 a
to612 a 601 a.
We recall from
[12]
that every2-category
determines aspecial
double category with connection. From
[12]
we also recall the process ofreflecting
squares. For eachin
D2 ,
letT (a) E h (D)
denote the squareThe function r :
D2 4 D2
definedby
is called the
reflection.
PROPOSITION 1. 1.
( i ) Let
th en
PROOF .
( i ) and ( ii )
follow from definitions.(iii)Since T(a) = T(b),
we haveand since a and b have the same
edges
we can compose both sides onthe left
by .
and on theright by resulting
inand hence a = b .
P ROP OSITION l. 2
(Theoreme
2.4of [1.2]).
(iv) r : h (D) -> v (D) is
anisornorphism (in
the senseof
a doublefunctor
or a2. functor).
2. THIN SQUARES.
DEFINITION 2.1. a E D is thin
(or degenerate
in[12])
if a has a de-composition consisting
of~, -a, lb , -c, Td for
someedges
a,b ,
c , d .The concept of thin squares was first introduced
by Brown ,
in theapplication
of doublegroupoids
toproblems
intopology,
see forexample [1, 3].
InTT ,
thin squaresactually correspond
to those with a constanthomotopy filling.
The essential feature of thin squares is described in thefollowing proposition.
PROPOSITION 2.2.
(i) at D2
is thiniff a = A (s) for some s E ~D1.
( ii )
Let s fE]Dl then A(s) is
theunique
thin squarehaving
theedges of
s .PROOF.
(i) Suppose a E D2
is thin so thatand
they
are in0 D 1 .
Thereforeand
they
arein ~ D1. By
the double functor propertyof A ,
which
means a = A (s)
forsome s E ~D1.
The converse is obvious.( ii ) Proposition
1.1 of[12].
Since any
composition
of thin squares is thin and every thin square is determinedby
itsedges
it follows that one canidentify
adecomposition
of thin squares
by simple noting
itsedges.
In each of the thin squaresD, H, l I
J orf
, theedges supplied
represent identities. We shall some- times omit thesubscript
of a thin square when it can be deduced from otherinformation,
forexample,
theedge
of aneighbouring
square. This notationwas introduced
by
Brown in[1].
3. HOMOTOPIES AND HOMOTOPY COMMUTATIVE CUBES.
As indicated in
[11],
the familiar modulohomotopy
category aris-ing
from a2-category
leads to ahomotopy theory
in aspecial
double cat-egory with connection
D ,
in whichf
and g inD 1
are said to be homo-topic
if there exists a Eh ( D )
such thatIn this context,
f
is ahomotopy equivalence
if there is ahomotopy
inversef
and squaresin
h (D) ;
andfurthermore, f
is astrong homotopy equivalence
ifPROPOSITION 3. 1.
f
is ahomotopy equivalence i f f f
is astrong
homo-topy equivalence.
PROOF.
Proposition
2.3 off 111 -
PROPOSITION 3.2. Let
rE E D1
andg Fzlb, gf2Jh ( h(D). If
and
f
is ahomotopy equivalence,
then a = b .PROPOSITION 3.3.
Suppose f g b fg
Eh (D)
withf
ahomotopy equival-
ence, then there exists a
unique
squarePROPOSITION 3.4.
( i )
Let a Eh (D ). Then 0-00,
a is ahomotopy equival-
ence
iff 611 a
is ahomotopy equivalence.
( ii )
Letf , g
fD1 . I f
any tuoedges from f ,
g,f g
arehomotopy equi-
valences,
then the third one is also ahomotopy equivalence.
Propositions 3.2, 3.3
and3.4
are direct consequences of the cor-responding
results in the modulohomotopy
category derived from the 2- categoryh (D). However,
we include below aproof
of3.2
as asimple
demonstration of how we may work
entirely
with squaresexploiting
whereappropriate
th e notion of thin square.PROOF of
Proposition
3.2. Letf
be thehomotopy
inverse off
with homo-topies
Composing
, we haveWhence
and so a = b .
Next we shall describe a process for
vertically inverting
squares of the formwhere
f
and g arehomotopy equivalences.
To carry out this process we shall need therespective homotopy
inversesf
and g andhomotopies
satisfying
( cf. Proposition 3 .1).
Th eresulting
inverted squareis defined
by
P ROP OSITION 3.5.
(i) 0(0( (a))
= a ,where è$
isde fined similarly to 0
wi th
f
andi,
g andg , d1
andd2, el and e2 ,
u and vinterchanged.
( iv ) 1 f
the square isatisfies
PROOF. We shall
only
prove the first part of(iv). Firstly, applying a 0
tothe first
equation,
we see that601 a f
=ïf. Since, by Proposition 3.3,
thereexists a
unique
square601 a e f in h (D)
such thate+2=f=ff,
e mustbe
-f,
so that601 a
=7. Similarly
we have611 a 9 -
That6a2 a
=6a2 O (a)
follows
immediately
from the fact that a + a- is defined and the first equa- tion.Therefore,
J andO (a)
have the sameedges.
Next weapply
T :Composing
both sides of the secondequality
at the topby - f
we haveComposing
both sides on the leftwith - dl +2- vf gives
and
finally, composing
on theright
withdi + -g u
we arrive atHence,
sincea and O (a)
have the sameedges, by Proposition
1.1( c ),
a - 0 ( a ) , completing
theproof.
Now,
for i = 1, 2, letbe a square with
homotopies
satisfying
the conditions forf i
and gi to be stronghomotopy equivalences.
We then have
corresponding inverted
squaresIf g1 = f 2,
thenis defined and the
homotopies d1a, e2a (a
=1,2) l
determine the invertedsquare
We then have
COROLLARY 3.6.
provided
PROOF. The result follows
directly
fromProposition
3.5( ii )
and( iv ) by considering
Similarly
ifu 1
=v2 ,
172
satisfy
the conditionsrequired
to showf2 fl
andg2 g1
are stronghomotopy equivalences.
Forexample,
we haveHence,
thehomotopies
D, Ea( a
= 1 ,2)
determine the inverted squareWe may now state a further
corollary
toProposition
3.5.PROOF.
1 1 1
Hence, by Proposition
3.5( iv ),
we have therequired
result.We shall devote the remainder of this section to a discussion of
homotopy
commutative cubes. The definition is modelled on the corres-ponding
concept intopology
as described in[11].
Theimportance
of cubesin the
setting
ofmultiple categories
was made evident in[8].
DEFINITION 3.8. The subset
{a0, a1, b1, b2, b3, b4}
ofD2 is
called acube if :
It is said to be
homotop y
commutative ifSuch a cube is
diagrammatically represented
as :PROPOSITION 3.9.
Let {a0, a1, b1, b2, b3 ’b 41 }
be acube,
then thefol-
lowing
statements areequival ent :
( i) ap, a,, bl , b2, b3, b4}
ishomotopy
commutative.PROOF. This follows from direct
simplifications
ofappropriate equations.
Let
{a0, allbi b2, b3, b4}
be a cube as shown in thefigure,
where
f..
i arehomotopy equivalences
withhomotopy
inversesf.
i,( i
= 1 ,.... ,
4 )
andhomotopies
satisfying
the conditions for thefi
’s to be stronghomotopy equivalences.
We may then use these
homotopies
to define the inverted squareswhere for
example
PROPOSITION 3. 10.
If {a0,
a 1,b1, b2, b3, b4}
is ahomotopy
commutative cube then so isPROOF. The condition of Definition 3.8
implies,
with thehelp
ofCorollary
fience, by Proposition
3.5 (ii )
and( iii ),
Rearrangement produces
theequation
vwhence, by
the conditions forf 1
andf4
to be stronghomotopy equival-
ences, we have
from which the
required
result followseasily.
The next
proposition
says that verticalcomposition
ofhomotopy
commutative cubes preserves
homotopy commutativity.
Horizontalcomposi-
tion will
similarly
preservehomotopy commutativity.
PROPOSITION 3.11.
I f
are
homotopy
commutativecubes,
thenis a
homotopy
commutative cube.Here
by homotopy commutativity
of the secondcube,
the four terms in thelower
right
hand corner of theexpression
may besimplified
togive :
which
by homotopy commutativity
of the first cube becomesDEFINITION 3.12. The square
al
is said to beequivalent
to the squarea2 ,
insymbols a1= a2 ,
if there exist squaresb1, b2, b3 and b4
whosevertical
edges
arehomotopy equivalences
and such thatis a
homotopy
commutative cube.PROPOSITION
3.13.= is
anequivalence
relation.PROOF. This follows from the above two
propositions
and the obvious fact that the trivial cube witha1 = a0
and thin vertical faces ishomotopy
commutative.
4. PULLBACK AND PUSHOUT SQUARES.
DEFINITION 4.1.
( i )
Apullback
square is a square a inD2
such that for any squareb in D with
611 b = a 1 a
anda2 b = a2 a ,
there existsquares c 1
andc2
withsuch that
and,
inaddition,
ifthen there exists
( ii )
Apushout
square is a square a inD2
such that for any square b inD2
with602 b = 602 a
and601 b= 601
a, there exist squarescl , c2
withsuch that
and,
inaddition,
ifthen there exists
such that
Here 0 and 0 are called the induced
morphisms.
At this
point,
we shallgive
twoexamples
ofpullback
andpushout
squares and
they
will be our sole concern.EXAMPLE 1. Consider the
special
double category with connection TT.f a )
LetA ,
B , X betopological
spacesand f : A ->
X , g : B -> X con-tinuous maps. Then the space
determines a
pullback
in TT.( b )
LetA ,
B , X betopological
spaces andf :
X->A ,
g : X 4 B con-tinuous maps. Then the double
mapping cylinder P = AU f (X x I)UgB
together
with the inclusions i : B 4P , j :
A - P andhomotopy
determines a
pushout
square in TT.EXAMPLE 2. Consider the
special
double category with connection C C .(a)
Letf :
A -> X , g : B -> X be chain maps. Define the chaincomplex
Let
p 1:
P ->A, p2:
P 4 B be therespective projections
of chain com-plexes
and T :f p 1 - gp2 : P -> X
the chainhomotopy
definedby
One then verifies for each n ,
The T then constitutes a
pullback
square in C C .Suppose
r : G 4A ,
s : G - B are chain maps such that H :f r -
g s . Letwe then have that 0 is a chain map,
Suppose 0 :
G 4 P is another chain map and U : r -p10, V : s-p2 0
arehomotopies
such that[-gV+T0+fU]=[H]. (Here [H]
denotes th eequivalence
class of the chainhomotopy
H . Two chainhomotopies
H’and H are
equivalent
if for each n , there is a grouphomomorphism
such that
It is thus necessary to obtain a
homotopy
For
each g
inGn ,
letso
that R ( g )
isin P n+1 and 6R + R6 = 0 - 0. Therefore R : 0- 0.
It is also clear that[p1R] = [U] and [p2R] = [V].
( b)
Letf :
X ->A ,
g : X 4 B be chain maps. Define the chaincomplex
Q = (Qn, 6)
wh ere withHere
[a, b,
x,ac]
denotes an element inQn
withrepresentative (a,
b, x,x).
Let
i1 : A -> Q, i 2: B - Q
be therespective
inclusions of chaincomplexes
and T :
i1f- i 2g :
X +Q
the chainhomotopy
definedby
One then verifies for each n ,
The T then constitutes a
pushout
square in CC .Suppose r : A -> G,
s: B 4 G are chain maps such thatH : r f -
s g . Let0 : Q + G be
We then have
9
is a chain map,0i 1
= r,0i 2 = s
and 0T = H .Suppose 0: Q 4
G is another chain map and U : r -ii 1,
V : s-0i2
arehomotopies
such that[-Vg+0T+Uf]= [H].
It is thus necessary to obtain ahomotopy
For each
(a, b, x, c] in Q ,
letwhere L is the
homomorphism
such thatWe then have
The
following
threepropositions
from[12]
are recalled.PROPOSITION 4.1
(Proposition
3.4of [ 12 ]). I f
a is apullbork (pushout)
square, then so is
r( a)
apullback (pushout)
square.P R OP OSITION 4.2
(Proposition
3. 6of [12]).
Let a be a square such thatone
pair of opposite edges
arebomotopy equivalences.
Then a is botha
pullback
and apushout
square.PROPOSITION 4.3
(Proposition
3.7of [12]). lf
a and b arepullback (pushout)
squares, th en a +h is apullback (pushout)
s quare(
i = 1,2 ) .
The nextproposition
includes the converse part ofProposition 3.2
in
[12].
PROPOSITION 4.4. Let a be a
pullback
square and a’ a square witha1i
a =62i
a’( i
= 1,2)
and let 0 be the inducedmorphism.
Then a’ is apullback
squareiff
0 is ahomotopy equivalence.
PROOF. Vle shall
only
prove the «if» part of theproposition.
Since a isa
pullback
square, there are squaresc, and c2
such thatFrom
Proposition 4.2, -0, c1
andc2
arepullback
squares, so that fromProposition 4.1, -0, c1
andr ( c2 )
arepullback
squares.Finally,
fromProposition 4.3,
we see that a is apullback
square. Of course, there should be a similar result forpushout
squares ; we will not write it down.PROPOSITION 4.5. Let the cube shown in
Figure
1 behomotopy
commu-tative and
suppose a1
is apullback
square. Then,i f
there are squaresis
homotopy
commutative, there existsPROOF. We first have
Secondly,
since the cubesare
homotopy commutative,
and
so that from the above
equation,
we havehence
by uniqueness
of thepullback
squarea r ,
there existswhich means
d1+bi = bi (i=1, 2).
PROPOSITION 4.6. Given squares
such
that a1
is apullback
square, there exist squaressuch that
{a0, a1, b1, b2 , b3, b4}
is ahomotopy
commutative cube.Furthermore,
supposeare squares. Then
{a0, a1, b1, b2, b3, b4 }
is alsohomotopy
commuta-tive iff there exists
PROOFo Consider the square
Since
al
is apullback
square, there are squaressuch that
We shall refer to this as
equation ( a ).
Next letOne
easily
checks with thehelp
of( a )
that{a0, a1, bl , b2 , b3, b4} is a
homotopy
commutative square. To prove the latter part of theproposition,
we
proceed
as follows:Suppose I ao a1, b1, b2, b3 , b4}
is ahomotopy
commutative cube. Therefore
so that
by equation ( a),
it isequal
toHence
by uniqueness
of thepullback
property ofal ,
there iswhich, by
definition ofb 1
andb2
isequivalent
to theequation
Conversely,
suppose there is a squaresuch that
As
I ao a1 , bi b2 , b3 , b4}
is ahomotopy
commutativecube,
we havethe
equal
squaresthe latter of which can be subdivided into
and
consequently
isequal
toHence
{a1, a2, b1, b2 , b3 , b4}
is ahomotopy
commutative cube.These two
propositions
express the fact that in thehomotopy
com-mutative cube
if the bottom square
al
is apullback
square, then each{f4}
determinesa
unique {f1},
where the bracket denotes thehomotopy
class of a mor-phism.
The dual result is that ifa0
is apushout
square, then each{f1}
determines a
unique {f4}.
Hencewhen a0
is apushout
square andal
isa
pullback
square, there is abijection
between the sets of the classes{f1}
and{f4}.
Proposition
4.6 leads to a number ofinteresting
results.COROLLARY 4.7. Let
pullback
squares, then there exist squaressuch that
f,
is ahomotopy equivalence and {a0, a1, b1, b2’ b3’ b4 I is
a
homotopy
commutative cube.PROOF.
By Proposition 4.6,
there are squaressuch that
are
homotopy
commutative cubes. Consider thehomotopy
commutative cubeswh ere
are
homotopies. By Proposition 4.5,
there is ahomotopy
such that
Similarly, by considering
the otherpair
ofhomotopy
commutativecubes,
the
homotopy
is obtained.
COROL L AR Y 4.8. Let
lao, all b1, b2 , b3 , b 4 }
be ahomotopy
commu-tative cube as shown in
Figure
1.If ao
anda1
arepull back
squares andf2’ f3’ f4
arehomotopy equivalences,
thenf1
is also ahomotopy equi-
valence.
PROOF. This follows
directly
fromCorollary
4.7 andProposition
4.6.COROLLARY 4.9. Let a
and 7
be squares such that a isequivalent
toä. 1f X
is apullback (pushout)
square, then a is apullback (pushout)
square.
PROOF. Here we have to assume
pullback
orpushout
squares exist in D.Let
b1, b2, b3 and b4
be squares with verticaledges homotopy equival-
ences. We then have the
homotopy
commutative cube{a, 7, b1, b2, b3, b4}
below.
Let
a1
be thepullback
square of611 a
andd 1 a
so thatfor some squares c1 and
c2 . By Corollary 4.7,
there are squaresb 1, b2
such that
601 b1 = a1 b2 = g (say)
is ahomotopy equivalence
andis a
homotopy
commutative cube. One then verifies thatthe square
so that
by Proposition 3.4, 0
is ahomotopy equivalence. Therefore, by Proposition 4.4, a
is apullback
square.The
following proposition
followseasily
from Definition 3.8.PROPOSITION 4.10. Let a be a
pullback
square andlet
be the
homotopy
cubeof Figure
1.Suppose
are
homotopi es.
Then the cubeis
homotopy
commutative.Now we obtain from
Proposition
4.6 that for the cube inFigure 1,
when
f2
andf3
arereplaced by homotopic edges,
the inducededge
fromao a1 aO
to8§ 601 a1
ishomotopic
to/1 .
Hence we have the
following
refinement ofProposition
4.5.COROLLARY 4.11
(re fer
toFigure 1).
Leta 1
be apullback
square anda0, b3 , b4 given
squares. Then there are squaresbi
andb2
such thatthe cube
{a0, a1, bi b2’ b3’ b4} is hornotopy
commutative.Furthermore.
i f f2, f3
andf¢
arereplaced by homotopic edges,
the inducededge
isbomotopic
to/1.
Finally,
we consider results converse toProposition 3.6
of[12].
Proposition
4.12 is well known forordinary pullbacks,
i. e.for pullback
squares
in DD,
wheneverf
is a fibration.Proposition 4.13 provides
anecessary and sufficient condition for the converse of
Proposition
4.12.By
this result andPropositions 3.7
and 3.8 of[12]
in any category satis-fying
the condition(for example, according
to[11],
in the category of CW-complexes),
thealgebra
ofpullback
andpushout
squares takes on a par-ticularly simple
form.PROPOSITION 4. 12. L et
be a
pullback
square.I f f
is ahomotopy equivalence,
then g is a horno-topy equivalence.
PROO F. Let
f
be thehomotopy
inverse off
with stronghomotopies
Consider the square
Since a is a
pullback
square, there exist squaressuch that
Therefore, - T ( c 1)
is ahomotopy
fromg g
to 1 . To obtain the other ho- motopy, letThen
However,
we haveHence, by uniqueness
of thepullback
property of a , there existssuch that
Therefore,
s is thehomotopy
fromg g
to 1 .PROPOSITION 4.13. The
following
two statements areequivalent:
(
1 ) For anypullback
squareif
g is ahomotopy equivalence
thenf
is ahomotopy equivalence.
(2)
For any squares a, b ,if
a and a + b arepullback
squares then b is apullback
square.PROOF.
(1)
=>(2). Suppose
a, b are squares such that a, a + b are 1pullback
squares.Let b’ be the
pullback
square of611 b
and612 b. Therefore,
there aresquares
c 1
andc2
such that is a homo-topy commutative cube. Let a’ be the
pullback
squareof 9 1 a
and601 b’.
Again
there are squaresd 1
andd2
such thatis a
homotopy
commutative cube.Composing
the twocubes,
we see thatis also a
homotopy
commutative cube. Since both a + b and a’ + b’ are1 7
pullback
squares,Corollary
4.8implies p
is a homo-topy
equivalence.
On the otherhand,
since isa
homotopy
commutativecube,
we obtainAs a is a
pullback
square andcomposition
ofpullback
squares is apull-
back square, we thus have
d2+2 a’
is apullback
square.By Proposition
3.8
of[12], d2
is apullback
square.By assumption,
becausep
is a ho-motopy
equivalence, q = 611 d2
is also ahomotopy equivalence. Finally,
we see that b is
equivalent
tob’ ,
so that b is also apullback
square.( 2 )
=>(1). Suppose a
is apullback
square such that g is a homo- topyequivalence.
Since the verticaledges
of( a
+-i
arehomotopy equi- valences, Proposition 3.6
of[12] implies
that it is apullback
square.Therefore, by assumption, J f
is apullback
square.Finally, by Proposi-
tion
4.12,
we havef
is ahomotopy equivalence.
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Strawberry
HillTWICKENHAM TW1 4SX U.-K.
Y. L. WONG: Department of Mathematics
University
ofHong Kong
HONG KONG