C OMPOSITIO M ATHEMATICA
J AMES C. B ECKER
D ANIEL H ENRY G OTTLIEB
Coverings of fibrations
Compositio Mathematica, tome 26, n
o2 (1973), p. 119-128
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119
COVERINGS OF FIBRATIONS by
James C. Becker and Daniel
Henry
Gottlieb *COMPOSITIO MATHEMATICA, Vol. 26, Fasc. 2, 1973, pag. 119-128 Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
Let E B be a fibration. Then a
covering
map Egives
rise to afibration B. If the fibre of this fibration is
connected,
then the fibreis a
covering
space of the fibre of E ~ B. In this case, we shallcall Ê
a
covering fibration of
thefibration
F ~ E Bextending
thecovering 11: : F -+ F.
This note takes up the
following questions.
A. Given a fibration F ~ E ~ B and a
covering F,
when can we find acovering E extending F?
B. Given a space F and a
covering
spaceF,
under what conditions can wealways
find acovering
fibrationextending F
for any fibration withfibre F?
For
F
a universalcovering
space, we can answerquestion
A(see
Theorem
1)
in terms of conditions on the fundamental groups of the fibration F - E ~ B. For oriented fibrations and universalcoverings F,
we find the answer to B
depends
uponGl(F), (see
Theorem2).
Someresults on
covering
fibrationsextending
non-universalcoverings F are
found
in §
3.The existence of
covering
fibrations leads to variousapplications.
Themost
striking
of them is theorem 15 whichgeneralizes
a theorem ofBorel’s
[ 1 ],
lemma 3.2.By fibration,
we shall mean Hurewicz fibration(i.e.
F ~ E B hasthe
homotopy covering property).
We assume that E and F arepath connected, locally path connected,
andsemi-locally
1-connected. Most of the resultsproved
willobviously
be true for other types of fibrations.In § 3,
we consider fibre bundles(locally homeomorphic
to aproduct
ofa
neighborhood
in the base and thefibre). By
an orientedfibration,
weshall mean the strongest
possible interpretation.
Thatis,
F - E ~ B isoriented if
11:1 (B)
operatestrivially
on03BE(F),
where03BE(F)
is the group ofhomotopy
classes of selfhomotopy equivalences. By 03BE0(F)
we shallmean the group of based
homotopy
classes of based selfhomotopy equivalences.
* Partially supported by a National Science Foundation grant.
Also
X
willalways
denote acovering
space of X.Covering
spaces arealways
assumed to bepath
connected.2.
Covering
fibrationsextending
universalcoverings
First we answer
question
A for universalcoverings.
THEOREM 1: Let
F
E .4 B be afibration
with connectedfibre
F andlet
F
be the universal coverof
F. Then there exists acovering fibration
É of
E P Bextending F if
andonly if
a) i* :
03C01(F) -
03C01(E)
isinjective
and
b)
p* :03C01(E) ~ 03C01(B)
has aright
inverse(which
is ahomomorphism).
PROOF:
Suppose
thatE
is acovering
fibration. Then we havewhere x denotes
covering
maps. From the exact ladder associated with thisdiagram,
we haveThus d is
trivial, hence 1* :
xi(F) - 03C01(E)
isinjective.
On the other hand(p03C0)* : 03C01() 03C01(B).
So(p03C0)*
has aninverse j : xi (B) - xi(É).
Then03C0*j : 03C01(B) ~ 7r 1 (E)
is therequired
inverse to p* :7r 1 (E)
~ nl(B).
Conversely,
supposea)
andb) hold, Let Ë
be thecovering
space of Ecorresponding
to asubgroup
Hof 03C01(E)
such that p * : H -03C01(B)
is anisomorphism.
Thenp’
= p03C0 :É - B
is a fibration with fibre F’ and we121
have the commutative
diagram
where ii is the restriction of 03C0.
Now we will show that F’ =
F.
Notice thatis the
composition
ofisomorphisms
and is therefore an
isomorphism.
Hence F’ ispath
connected. Since ii : F’ ~ F is a fibration with discretefiber,
it is acovering
space[9,
Theorem10,
p.78].
We will show now
that irl(F’)
= 0. If v~ 03C01(F’)
theni’*(v) =
0by
exactness. Hence 0 =
x*1§(v) = i*03C0*(v).
But03C0*
isinjective
since 03C0is a
covering and i*
isinjective by hypothesis a).
Therefore v = 0 as was to be shown.Next we turn to
question
B in the case of universalcoverings
andoriented fibrations. We need a few technical remarks.
First,
let us recall the definition of the first evaluationsubgroup G1(F)
cxi(F).
Ahomotopy ht :
F ~ F is called acyclic homotopy
ifho = hl = 1F.
Theloop
Tgiven by i(t) - h,(*) (where *
E F is the basepoint)
is called the trace of thehomotopy.
ThenG1(F)
consists of thesubgroups of 03C01(F,*)
whose éléments arerepresented by loops
which aretraces
of cyclic homotopies.
For a listof properties
ofG1 (F),
see[5].
We also need the fact
(see
theorem16.9, [3])
that for any fibre F, there exists a universal fibration F -Eoo -+ Boo.
Now thetransgression d~ : 03C02(B~)
-03C01(F)
is related toG’i(F) by
the fact thatd~(03C02(B~)) = G1(F). Now,
asalways,
we assume that F is connected. Then let.800
be the universal
covering
space ofB 00’
Let F - D -B 00
denote the fibration classifiedby
thecovering
map 03C0 :~ ~ B 00’
It is easy to seethat d(03C02(~)) = Gl(F),
where d isthe transgression d : TC2(Boo)
~03C01(F)
It also is true that every oriented fibration may be
regarded
as apullback
of F ~ D -
.800’
This follows since aclassifying map k :
B ~Boo
of an oriented fibration F - E - B mapsxi(B)
to theidentity of 03C01(B~).
Hence, there is a
lifting
of k to:B ~ Boo
such that k =nk.
Now we can state and prove
THEOREM 2: There exists a
covering fibration of
anyoriented fibration
with
connected fibre
Fextending
the universalcovering
spaceF if
andonly if G l (F)
is trivial.PROOF: Let
F
D~ be
the fibration mentioned above. We shall show that there exists acovering fibration ~ B 00
which extendsF
if and
only
ifG1 (F) -
0. Thepull
back of theclassifying
map for F ~ E ~ B, denotedby
k : B -Boo,
of~
D ~ B will be therequired covering
fibration. Thatis, = k*(D).
Now
03C01(~) is
trivial. Thus(b)
of theorem 1 is satisfied. On the otherhand, G1 (F)
= 0implies
that d :03C02(~) ~ 03C01(F)
is trivial. Hencei* : 03C01(F) ~ 03C01(D)
isinjective.
So condition(a)
is satisfied and so we mayapply
theorem 1.If we assume that
G1 (F) 1= 0,
thenby
theorem 1, we cannot find acovering
fibration of F ~ D ~Boo extending F.
COROLLARY 3: Let F be a compact
polyhedron
andX(F) 1=
0. Then there exists acovering fibration of
any orientedfibration extending
F, the universalcovering
F.PROOF: We know that
X(F) 1=
0implies
thatG1(F) =
0.COROLLARY 4: Let F be a compact
polyhedron
withX(F) 1=
0. Thenthere exists a cross-section over the two-skeleton
of
the base spaceof
anyoriented fibration with , fibre
F.PROOF:
By
thecorollary above,
we havewhere B is a 2-dimensional CW
complex.
The groups in which the obstructions to a cross-section
of É £ B
liemust vanish. Thus there is a cross-section c : B -
É.
Now xc is the re-quired
cross-section.Now we shall
study
those fibrations with fibresRP",
realprojective
space. We
begin by giving
alternativeproofs
to some results of Olum.123
LEMMA 5
(Olum [8]) :
PROOF:
Let 0(Sn)
denote the groupof homotopy
classes of basepoint preserving equivariant homotopy equivalences
of S’". We have an iso-morphism
defined
by sending f
to theunique
basepoint preserving map
thatcovers f. By
the method ofproof
of[2,
TH.2.5],
theoperation
of sus-pension
defines anisomorphism
Therefore,
we haveFinally,
COROLLARY 6.
PROOF: Let L denote the space of self
homotopy equivalences
of RP".Let
Lo
be thesubspace of maps
of L which preserve the basepoint.
NowG1(RP2n) =
0 andsee
[4], corollary
1.6 and theorem II.5. From the exact sequencearising
from the fibration
L0 ~ L RP",
andnoting
that theimage
of cv* isG1 (Rpn),
we see thatd:03C01(RPn) ~ 03C00(L0)
is zero if n is odd andinjective
if n is even.
Thus i*:03C00(L0) ~ 03C00(L)
isinjective
if n is odd and has kernelZ2
if n is even.But 03C00(L0) = 03BE0(RPn) ~ Z2 by
lemma 5. Alsoi*:03C00(L0) ~ 03C00(L)
must be onto since RPn is connected. Thusç(Rpn) = x o (L)
isZ2
when n is odd and 0 when n is even.COROLLARY 7.
Every fibration with fibre
Rp2n is orientable.PROOF:
Since ç(Rp2n)
istrivial,
the fundamental group of the base must acttrivially
on03BE(RP2n).
THEOREM 8.
Every fibration with fibre
Rp2n is coveredby
an S2nfibra-
tion with an involution.
PROOF: The conclusion means that we can
always
find acovering
fibration which extends the universal cover S2" of RP2". But this follows
immediately
from theorem 2 andcorollary
7. The extended total space must be a 2-foldcovering
of theoriginal
total space, and the deck transformation is the involution.REMARK: Theorem 8 is not true for Rp2n+l even when the fibration is orientable. Consider the "universal oriented
fibration",
Now
is trivial since
Thus in theorem 1 condition
a)
fails.The next
corollary
is anapplication
of theorem 8.COROLLARY 9. Let Rp2n -+ E ~ B be
a fibration.
Thenas
Z2-vector
spaces.PROOF: Let S2n ~ + ~ B be an
S2n-fibration
with an involution whichcovers RP2n ~ E - B. Let À :
É - Soo
be anequivariant
map and let 2 : E- RP~ denote thequotient
map. Then thecomposition RP2n E Â eRP’
sends the generator c EH1(RP~; Z2)
to the generatoré E
HI(Rp2n; Z2).
Acohomology
extension of the fiberis defined
by O( êt) = À *( ct),
t ~ 0. Thecorollary
now follows from theLeray-Hirsch
theorem[9,
p.257].
3.
Arbitrary coverings
We shall restrict our attention to
coverings
of fibre bundles. Ourtechnique
canprobably
be extended to Hurewiczfibrations,
or Doldfibrations,
but the neededpropositions
have not been written down yet.First we answer
question
B for fibre bundles. Then followapplications.
Let G be a group of
homeomorphisms
of F onto itself. Let G* be theself
homeomorphisms
of acovering
of F, denotedF,
which areliftings
of
homeomorphisms
in G. Let 0 : G* ~ G be the map which takes amap
f ~
G* to the induced mape(f)
E G. Then e is continuous and also is ahomomorphism.
THEOREM 10:
If
there exists a cross-section c : G ~ G* to 0 which is also ahomonl0rphism,
then there is acovering
bundleof
any G-bundlewith fibre
F which extendsF.
PROOF: Let F - E - B be a G-bundle and let G - E* - B be the associated
principal
G-bundle. Then consider the G-bundle~
E xGF -+
125
B where G acts on
F by
means of the cross-section c. ThenÉ
= E xGF
is the
required covering.
Theprojection
1t :~
E isgiven by
The next theorem
gives
conditions for Theorem 10 to hold. We shall letG*
be theidentity
component of G* andGe
theidentity
component of G.THEOREM 11: A
covering
bundleextending F always
existsif a)
G1 (F)
r-- n 1(F ),
andb)
there is ahomomorphism
which is a cross-section to the
homomorphism
induced
by
03A6.PROOF: Note that 0 : G* - G is a fibration with a discrete fibre. The fibre over
1F
consists of theliftings
of1F.
Assumethat ~ Gé
is alifting
of
1p.
Then thepath from 1 to 1F
induces acyclic homotopy ht :
F - F.The trace
of ht
must represent an element in1tl(F) by a).
Thus the tracelifts to a closed
path
inF. Thus j has
a fixedpoint
andhence
=1F .
This fact allows us to conclude that
Ge
andGé
areisomorphic.
Thuswe may construct the cross-section
required by
theorem 10 overGe,
and condition
b)
allows us to extend the cross-section over the other components.REMARK: If G is connected and
G1 (F) - 0,
we may extend any cover-ing F
to acovering
G-bundle of anyarbitrary
G-bundle.COROLLARY 12:
If F is a
compactpolyhedron
andX(F) :0 0,
thencan be extended to a
covering bundle for any fibre
bundle with connectedstructural group.
Let M be a closed
topological
manifold which is unorientable and let1B1
denote its oriented doublecovering.
Let0(M)
be thesubgroup
of1tl (M)
of elementsrepresented by
orientationpreserving loops.
Then0(M)
is thesubgroup
of1tl (M) corresponding
to the oriented doublecovering M.
From this
point
on, we shall consider fibre bundles with fibre M andstudy covering
bundles which extendM.
The end result willyield
themain
application
of thesetechniques,
Theorem 15.THEOREM 13:
For any.fibre
bundle withfibre
M, a closed unorientabletopological manifold,
the oriented doublecovering M
extends to acovering
bundle. In
addition,
the structural groupof
thiscovering
bundle preserves the orientationof M.
PROOF:
Every homeomorphism
h : M - M has twoliftings ~ M,
one of which preserves and the other reverses the orientation of
M.
Consider the
correspondence
i which sends every h E G to its orientationpreserving lifting
in G*. It is easy to see that i : G - G* iscontinuous,
ahomomorphism
of groups and a cross-section of 0 : G* - G. Thus theorem 10 is satisfied. The group of thecovering
fibre bundle isi(G),
so the second statement of the theorem is
clearly
true.If we compare theorem 13 with condition
a)
of theorem11,
we arelead to
conjecture
thatG1(M)
cO(M).
This is in fact true, as thefollowing
theorem shows.THEOREM 14:
PROOF: Let oc E
G,(M).
Then there is acyclic homotopy h, :
M ~ Mwhose trace represents a. Now
ht
lifts to ahomotopy ht : ~ M
and1
is alifting
of theidentity.
Since1
ishomotopic
to1, 1
preserves the orientation onM.
There areonly
twoliftings
of theidentity
and oneof them reverses orientation. So
1
=1.
Thus the trace ofht
is aloop
which covers the trace of
ht.
Hence oc must be inO(M).
THEOREM 15: Let M - E B be a
fibre
bundlewith fibre
a closedtopological n-manifold
and with structural group G.If ~(M) ~
0(mod p),
where p is a
prime,
thenis injective.
PROOF: First consider the case where
03C01(B)
operatestrivially
onHn(M; Z,).
Then the theorem is true if M is orientableor if p = 2 [6,
Theorem
12]. Suppose
now that M is unorientable and p =1= 2. LetM
denote the orientable doublecovering
of M.By
theorem13,
we haveand
03C01(B)
operatestrivially
onHn(; Zp).
Now~()
=2~(M) ~
0(mod p).
SinceM
is orientable it follows that * isinjective.
Thenby
commutativity
03C0* isinjective.
127
Now consider the case where
03C01(B)
operatesnon-trivially
onH"(M; Zp)
p odd. Let K c
03C01(B)
denote the normalsubgroup
of index 2consisting
of elements which operate
trivially
onH"(M; Z).
LetB .!4 B
denote the2-fold
covering
whichcorresponds
to K. We havewhere ~
is the bundle inducedby
q. Since 1tl(B)
operatestrivially
on
H"(M; Zp)
we have from thepreceding paragraph
that îl* isinjective.
To show that 1t* is
injective
it is now sufficient to show thatis
injective.
There is the transfer map[10, Chapter 5]
and
03C4q*, being multiplication by 2,
is anisomorphism.
Thereforeq *
isinjective.
Thiscompletes
theproof
of the theorem.This theorem was first noted
by
A. Borel in[1 ] with
the extrahypoth-
eses that the bundle is differentiable and oriented in some sense and M is oriented. In
[6],
the second author removed thedifferentiability hypothesis
and the above theorem removes theorientability hypotheses.
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ALBRECHT DOLD
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DANIEL HENRY GOTTLIEB
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DANIEL HENRY GOTTLIEB
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