Yet another Hopf Invariant

0.18 MB·English

by  Jean-Paul Doeraene

YET ANOTHER HOPF INVARIANT
5
1 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
0
2

g
u
A Abstract. TheclassicalHopfinvariantisdefinedforamapf:Sr →X. Here
wedefine‘hcat’whichissomekindofHopfinvariantbuiltwithaconstruction

7 in Ganea’s style, valid for maps not only on spheres but more generally on
2 a ‘relative suspension’ f: ΣAW → X. We study the relation between this
invariant and the sectional category and the relative category of a map. In
] particular,forιX:A→Xbeingthe‘restriction’offonA,wehaverelcatιX 6
T hcatf 6relcatιX+1andrelcatf 6hcatf.
A Our aim here is to make clearer the link between the Lusternik-Schnirelmann
. category (cat), more generally the ‘relative category’ (relcat), closely related to
h
t James’sectional category (secat), and the Hopf invariants. In order to do this, we
a
introduce a new integer, namely hcat, that combines the Iwaze’s version of Hopf
m
invariant [3], based on the difference up to homotopy between two maps defined
[
for a given section of a Ganea fibration, and the framework of the sectional and

4 relativecategories,searchingforthe least integer suchthatthe Ganeafibrationhas
v a section, possibly with additional conditions. To do this combination, we simply
2 define our invariant hcat, as the least integer such that the Ganea fibration has a
1
sectionσ with additionalconditionthat the correspondingtwo maps(f◦σ andω
7 n
in this paper) are homotopic.
3
0 It appears that for f: Sr → X or even for f: ΣW → X, we obtain an integer
. that can be either cat(X), or cat(X)+1. More generally, for any f: Σ W →X,
1 A
0 wehaverelcat(f◦θ)6hcat(f)6relcat(f◦θ)+1,whereθ: A→ΣAW isthe map
5 arising in the construction of ΣAW.
1 Insection2,westudythe influenceofhcatinahomotopypushout. Insection3,
:
v weintroducethe‘strong’versionofourinvariant,andweobtainanotherimportant
Xi inequality: for any f: ΣAW → X, we have relcat(f) 6 hcat(f). In section 4, we
give alternative equivalent conditions to get hcat. Applications and examples are
r
given.
a
1. The Hopf category
We work in the category of pointed topological spaces. All constructions are
made up to homotopy. A ‘homotopy commutative diagram’ has to be understood
in the sense of [4].
Recall the following construction:
2010 Mathematics Subject Classification. 55M30.
Key words and phrases. Ganeafibration,sectionalcategory, Hopfinvariant.
1
2 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
Definition 1. For any map ι : A → X, the Ganea construction of ι is the
X X
following sequence of homotopy commutative diagrams (i>0):
>>A❊
ηi ⑥⑥⑥⑥ ❊❊❊❊ ιX
⑥ ❊
Fi❅⑥β❅⑥❅i⑥❅❅❅❅❅ α③i③+γ③i1③③③③❊③❊G<<"" i+1 gi gi+1 77//''X
G
i
where the outside square is a homotopy pullback, the inside square is a homotopy
pushoutandthemapgi+1 =(gi,ιX): Gi+1 →X isthewhiskermapinducedbythis
homotopy pushout. The iteration starts with g0 =ιX: A→X. We set α0 =idA.
For any i > 0, there is a whisker map θ = (id ,α ): A → F induced by
i A i i
the homotopy pullback. Thus we have the sequence of maps A θi //Fi ηi //A
and θi is a homotopy section of ηi. Moreover we have γi ◦αi ≃ αi+1, thus also
αi+1 ≃γi◦γi−1◦···◦γ0.
We denote by γi,j: Gi →Gj the composite γj−1◦···◦γi+1◦γi (for i< j) and
set γ =id .
i,i Gi
Of course, everything in the Ganea construction depends on ι . We sometimes
X
denote G by G (ι ) to avoid ambiguity.
i i X
Definition 2. Let ι : A→X be any map.
X
1)Thesectionalcategory ofι istheleastintegernsuchthatthemapg : G (ι )→
X n n X
X has a homotopy section, i.e. there exists a map σ: X → G (ι ) such that
n X
g ◦σ ≃id .
n X
2)Therelativecategory ofι istheleastintegernsuchthatthemapg : G (ι )→
X n n X
X has a homotopy section σ and σ◦ι ≃α .
X n
3)The relative category of order k ofι is the leastintegern suchthatthe map
X
g : G (ι )→X has a homotopy section σ and σ◦g ≃γ .
n n X k k,n
Wedenotethesectionalcategorybysecat(ι ),therelativecategorybyrelcat(ι ),
X X
and the relative category of order k by relcat (ι ). If A = ∗, secat(ι ) =
k X X
relcat(ι ) and is denoted simply by cat(X); this is the ‘normalized’ version of
X
the Lusternik-Schnirelmann category.
Clearly, secat(ιX) 6 relcat(ιX). We have also relcat(ιX) 6 relcat1(ιX), see
Proposition 6 below.
In the sequel, we will consider a given homotopy pushout:
W η //A
β θ
(cid:15)(cid:15) (cid:15)(cid:15)
A //Σ W
A
θ
Inother words,the map θ is a mapsuchthat Pushcatθ ≤1 in the sense of[2]. We
call this homotopy pushout a ‘relative suspension’ because in some sense, A plays
the role of the point in the ordinary suspension.
We also consider any map f: Σ W →X, and set ι =f ◦θ.
A X
YET ANOTHER HOPF INVARIANT 3
We don’t assume η ≃ β in general. This is true, however, if θ is a homotopy
monomorphism,andin this casewe can ‘think’ ofι asthe ‘restriction’of f onA.
X
For n>1, consider the following homotopy commutative diagram:
W β //A
Ayyrrrηrrrr ✤✤ θ // Σ W vv♥θ♥♥♥♥♥♥♥ ❖❖❖❖❖θ❖❖''Σ W
✤ A A
(†) (cid:15)(cid:15)✤ ωn (cid:15)(cid:15) αn−1
Fn−1(ιX) //Gn−1(ιX) f
Azztttttt //G ((cid:15)(cid:15)ι )ww♦♦♦♦γ♦n♦−1 ◆◆◆g◆n◆−◆◆1//&&X(cid:15)(cid:15)
αn n X gn
where the mapW →Fn−1 is induced by the bottom outer homotopypullback and
the mapω : Σ W →G is induced by the top inner homotopypushout. We have
n A n
f ≃g ◦ω by the ‘Whiskersmapsinside a cube’lemma (see [2], Lemma49). Also
n n
notice that αn ≃ωn◦θ ≃γn−1◦αn−1; so ωn ≃(αn,αn) is the whisker map of two
copies of α induced by the homotopy pushout Σ W. Finally, for all k > 1, we
n A
can see that ω ≃γ ◦ω .
n k,n k
Definition3. TheHopfcategoryoff istheleastintegern>1suchthatg : G (ι )→
n n X
X has a homotopy section σ: X →G (ι ) such that σ◦f ≃ω .
n X n
We denote this integer by hcat(f).
Actually, speaking of ‘Hopf category of f’ is a misuse of language. We should
speak of ‘Hopf category of the datas η, β and f’.
Example 4. Let X =Σ W and f ≃id . Then, asmight be expected, hcat(f)=
A X
1. Indeed, in this case, as g1 ◦ω1 ≃ f ≃ idX, ω1 is a homotopy section of g1.
Moreover, ω1◦f ≃ω1◦idX ≃ω1, so hcat(f)=1.
Example 5. Let X 6≃ ∗ and W = A∨A, β ≃ pr : A∨A → A and η ≃ pr :
1 2
A∨A → A the obvious maps. Then Σ W ≃ ∗ and we have no choice for f that
A
must be the null map f: ∗ → X. In this case the condition σ◦f ≃ ω is always
n
satisfied, so hcat(f)=secat(ι )=cat(X).
X
Notice that relcat is a particularcaseofhcat: WhenW =A, η ≃β ≃id , then
A
ιX ≃ f, ωn ≃ αn and hcat(f) = relcat(ιX). Also relcat1 is a particular case of
hcat: When W = F0, then ΣAW ≃ G1, θ ≃ γ0 ≃ α1, and if, moreover, f ≃ g1,
then ωn ≃γ1,n and hcat(f)=relcat1(ιX).
Thefollowingpropositionshowsthattheseparticularcasesareinfactlowerand
upper bounds for hcat(f).
Proposition 6. Whatever can be f (and ι =f ◦θ), we have
X
secat(f)6relcat(ιX)6hcat(f)6relcat1(ιX)6relcat(ιX)+1.
Proof. Consider the following homotopy commutative diagram (n>1):
αn 00:: Gn
A θ //ΣAW✈✈✈ω✈✈n gn
ιX ■■■f■■..$$ X(cid:10)(cid:10)
4 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
We see that if there is a map σ: X → G such that ω ≃ σ◦f then α ≃ σ◦ι
n n n X
and this proves the second inequality.
Now consider the following homotopy commutative diagram (n>1):
ωn 00==Gn
ΣAW ω1 //G1③③③γ③1,n gn
❉❉g1
❉❉"" (cid:10)(cid:10)
f ..X
We see that if there is a map σ: X →Gn such that γ1,n ≃σ◦g1 then ωn ≃ σ◦f
and this proves the third inequality.
The first inequality comes from secat(f) 6 secat(ι ) 6 relcat(ι ), the first of
X X
these two inequalities comes from [2], Proposition 29.
Finally, the fourth inequality is proved in [1]. (cid:3)
So hcat(f) establishes a ‘dichotomy’ between maps f: Σ W →X:
A
– Either hcat(f)=relcat(ι ) andwe haveaσ suchthat f◦σ ≃ω already
X n
for n=secat(ι );
X
– either hcat(f)=relcat(ι )+1 andwe haveaσ suchthatf◦σ≃ω only
X n
for n>secat(ι )
X
Our last example of the section shows that the inequalities of Proposition6 can
be strict, and even that two may be strict at the same time:
Example 7. Let X = ∗, A 6≃ ∗ and consider ι∗: A → ∗. We have Gi(ι∗) ≃ A ⊲⊳
... ⊲⊳ A, the join of i+1 copies of A. For any k, γ ≃ id, so it cannot factorize
k,k
through ∗; but γk,k+1 is homotopic to the null map, so relcatk(ι∗) = k+1. Now
consider f ≃ g1(ι∗): A ⊲⊳ A → ∗. As said before, in this case we have hcat(f) =
relcat1(ιX). So we get secat(f)=0<relcat(ι∗)=1<hcat(f)=relcat1(ι∗)=2.
2. Hopf invariant and homotopy pushout
Let us consider any homotopy commutative square:
Σ W ρ // B
A
(‡) f κY
(cid:15)(cid:15) (cid:15)(cid:15)
X //Y
χ
Proposition 8. The homotopy commutative square above can be splitted into the
following homotopy commutative diagram:
f
ΣAW //G1(ιX) // Gn(ιX) ++//X
ρ χ
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
B //G1(κY) // Gn(κY) 33//Y
κY
YET ANOTHER HOPF INVARIANT 5
Proof. Setφ=ρ◦θ. Since θ◦η ≃θ◦β, alsoφ◦η ≃φ◦β. Firstnotice thatwe can
insert the original homotopy square inside the following homotopy commutative
diagram:
W η //A
β~~⑥⑥⑥⑥⑥ {{✇✇✇✇✇ ❅❅❅❅ιX
A // Σ W // X
A
f
φ
(cid:15)(cid:15) ρ (cid:15)(cid:15)
φ B B χ
B(cid:15)(cid:15) ⑤⑤⑤⑤⑤⑤⑤⑤ B(cid:15)(cid:15) ✈✈✈✈✈✈✈✈✈✈ ❆❆❆❆// Y(cid:15)(cid:15)
κY
Byinductiononn>1,startingfromthe outsidecube oftheabovediagramand
φ0 =φ, we can build a homotopy diagram:
W ❑❑❑❑❑❑%% vv♥♥♥♥♥//♥Fn−1✤✤(ιX) {{✈✈✈✈✈✈//A❂❂❂❂❂ιX(cid:30)(cid:30)
Gn−1(ιX) ✤ //Gn(ιX) //X
(cid:15)(cid:15) φn−1 (cid:15)(cid:15)✤ φn (cid:15)(cid:15)φ
B ❑❑❑❑❑❑%% (cid:15)(cid:15) vv♥♥♥♥♥//♥Fn−1(κY) (cid:15)(cid:15) {{✈✈✈✈✈✈//BκY❂❂❂❂❂(cid:30)(cid:30) (cid:15)(cid:15)χ
Gn−1(κY) //Gn(κY) gn // Y
wherethedashedanddottedmapsareinducedbythehomotopypullbackFn−1(κY)
and the homotopy pushout G (ι ) respectively.
n X
So we obtain a homotopy commutative diagram:
W // A
~~⑦⑦⑦⑦⑦ zz✉✉✉✉✉ ❄❄❄❄ιX(cid:31)(cid:31)
A // G (ι ) //X
n X gn
(cid:15)(cid:15) (cid:15)(cid:15)
B B χ
(cid:15)(cid:15) ⑦⑦⑦⑦⑦⑦⑦⑦⑦⑦ (cid:15)(cid:15) zz✉✉✉✉✉ ❄❄❄❄(cid:31)(cid:31) (cid:15)(cid:15)
B //G (κ ) //Y
n Y gn
Finally take the homotopy pushout inside the upper and lower lefter squares to
get the homotopy commutative diagram:
Σ W // G (ι ) //X
A ωn n X
ρ χ
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
B //G (κ ) //Y
n Y
and this gives the required splitting of the original square. (cid:3)
Proposition 9. If the square ‡ is a homotopy pushout, then
relcat(κ )6hcat(f).
Y
Asaparticularcase,whenB ≃∗,Y isthehomotopycofibreoff,andrelcat(κ )=
Y
cat(Y). So the Proposition asserts that hcat(f)>cat(Y).
6 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
Proof. Let hcat(f) 6 n, so we have a homotopy section σ of g (ι ) such that
n X
σ◦f ≃ω . First apply the ‘Whisker maps inside a cube’ lemma to the outer part
n
of the following homotopy commutative diagram:
Σ W //B
ωxxrrnrrr A a(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) ■■■■α■n$$
G (ι ) // S //G (κ )
n X n Y
b
c
X(cid:15)(cid:15) xxqqqqqqqΣAW // Y(cid:15)(cid:15) (cid:127)(cid:127)⑦⑦⑦⑦// B ❏❏❏❏❏❏$$Y(cid:15)(cid:15)gn
where the inner horizontal squares are homotopy pushouts, and c and b are the
whisker maps induced by the homotopy pushout S. Next build the following ho-
motopy commutative diagram:
Σ W //B
A
X xxqqqfqqqq //YκY(cid:127)(cid:127)⑦⑦⑦⑦
d
σ (cid:15)(cid:15) ωxxrrnrrrΣAW (cid:15)(cid:15) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)a(cid:0)// B ■■■■α■n$$
G (ι ) // S // G (κ )
n X n Y
b
where d is the whisker map induced by the homotopy pushout Y. Let σ′ = b◦d.
We haveg ◦σ′ ≃g ◦b◦d≃c◦d≃id and σ′◦κ ≃b◦d◦κ ≃b◦a≃α . (cid:3)
n n C Y Y n
Corollary 10. In the diagram ‡, if relcat(κ ) = relcat(ι )+1, then hcat(f) =
Y X
relcat(ι )+1.
X
Proof. ByProposition9,thehypothesisimpliesthathcat(f)>relcat(ι )+1. But
X
by Proposition 6, we have hcat(f)6relcat(ι )+1. So we have the equality. (cid:3)
X
It is now easy to exhibit examples of maps f with hcat(f) = relcat(ι ) +
X
1. Indeed there are plenty examples of homotopy pushouts where relcat(κ ) =
Y
relcat(ι )+1:
X
Example 11. Let A=B =∗ andf: Sr →Sn be any ofthe Hopf maps S3 →S2,
S7 →S4 or S15 →S8. So here relcat(ι )=cat(Sn)=1. On the other hand it is
X
well known that those maps have a homotopy cofibre Sn/Sr of category 2, so here
relcat(κ )=cat(Sn/Sr)=2. By Corollary 10, we have hcat(f)=2.
Y
Example 12. Let f be the map u in the homotopy cofibration
Z ⊲⊳Z //ΣZ∨ΣZ // ΣZ×ΣZ
u t1
where Z ⊲⊳Z ≃Σ(Z ∧Z) is the join of two copies of Z and is also the suspension
of the smash product of two copies of Z. Let A = B = ∗, ΣZ 6≃ ∗. We have
relcat(ι ) = cat(ΣZ ∨ ΣZ) = 1 and relcat(κ ) = cat(ΣZ × ΣZ) = 2, so by
X Y
Corollary 10 again, we have hcat(u)=2.
YET ANOTHER HOPF INVARIANT 7
Example 13. For i>1, let f be the map β in the Ganea construction:
i
A θi // F ηi // A
❅ i
❅
❅
❅
αi❅❅❅❅(cid:31)(cid:31) (cid:15)(cid:15)βi (cid:15)(cid:15)αi+1
Gi γi // Gi+1
Actually Fi is a join over A of i+1 copies of F0, and also a relative suspension
Σ W where W is a relative smash product. For any i 6 relcat(ι ), we have
A X
relcat(α )=i, see [2], Proposition 23. So by Corollary 10 again, if i<relcat(ι ),
i X
we have hcat(β )=relcat(α )+1=i+1.
i i
3. The Strong Hopf category
In[2],weintroducedthestrongversionofrelcat,namelyRelcat. Inthissection,
we introduce the strong version of hcat, namely Hcat. This gives an alternative
way, sometimes usefull, to see if a map has a Hopf category less or equal to n.
Also this will lead to a new inequality: hcat(f) > relcat(f). Consequently, if
relcat(f)>relcat(ι ), then hcat(f)=relcat(ι )+1.
X X
Definition 14. The strong Hopf category of a map f: Σ W → X is the least
A
integer n>1 such that:
– there are maps ι0: A → X0 and a homotopy inverse λ: X0 → A, i.e.
ι0◦λ≃idX0 and λ◦ι0 ≃idA;
– for each i, 06i<n, there is a homotopy commutative cube:
W β //A
η~~⑥⑥⑥⑥ zz✈✈✈✈✈
A // ΣAW ιi
(♮)
Z(cid:15)(cid:15) zi ζi+1 // X(cid:15)(cid:15)
i i
A(cid:127)(cid:127)⑧⑧⑧⑧ιi+1 //Xi(cid:15)(cid:15)+1{{✇✇✇χ✇✇i
where the bottom square is a homotopy pushout.
– X =X and ζ ≃f.
n n
We denote this integer by Hcat(f).
Noticethatιi+1 ≃ζi+1◦θ ≃χi◦ιi. Inparticular,thismeansthatPushcat(ιi)6i
in the sense of [2], Definition 3.
For 0 6 i 6 n, define the sequence of maps ξ : X → X with the relation
i i
ξi = ξi+1 ◦ χi (when i < n), starting with ξn = idX. We have ξn ◦ ιn ≃ ιX
and ξi ◦ ιi = ξi+1 ◦ χi ◦ ιi ≃ ξi+1 ◦ ιi+1 ≃ ιX by decreasing induction. Also
ιX◦λ≃ξ0◦ι0◦λ≃ξ0. Moreover,for 0<i6n wehavewe haveξi◦ζi ≃f by the
8 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
‘Whiskermaps inside acube lemma’. Sowe havethe followinghomotopydiagram:
W η //A
}}④β④④④④ {{✇✇✇✇✇ ●●●●θ●##
A // Σ W Σ W
A A
(cid:15)(cid:15) ζi+1
ιi (cid:15)(cid:15) ~~⑤⑤⑤⑤Zi (cid:15)(cid:15) ||①①①ι①i①+//1A❋❋❋ι❋X❋❋"" (cid:15)(cid:15)f
Xi χi //Xi+1 ξi+1 // X
We say that a map g: B → Y is ‘relatively dominated’ by a map f: B → X if
there is a map ϕ: X →Y with a homotopy section σ: Y →X such that ϕ◦f ≃g
and σ◦g ≃f.
Proposition 15. A map g: Σ W → Y has hcat(g) 6 n iff g est relatively domi-
A
nated by a map f: Σ W →X with Hcat(f)6n.
A
Proof. Consider the map ω : Σ W → G (ι ) as in diagram † and notice that
n A n Y
Hcat(ω )6n. If hcat(f)6n, then f is relatively dominated by ω .
n n
Forthereversedirection,byhypothesis,wehaveamapϕandahomotopysection
σ such that ϕ◦f ≃ g and σ◦g ≃ f; composing with θ, we have also ϕ◦ι ≃ ι
X Y
andσ◦ι ≃ι . Fromthe hypothesisHcat(f)6n,wegetasequenceofhomotopy
Y X
commutative diagrams, for 0 6 i < n, which gives the top part of the following
diagram.
Weshowbyinductionthatthemapϕ◦ξ : X →Y factorsthroughg : G (ι )→
i i i i Y
Y up to homotopy. This is true for i = 0 since we have ξ0 ≃ ιX ◦λ, so ϕ◦ξ0 ≃
ϕ◦ιX ◦λ ≃ ιY ◦λ = g0◦λ. Suppose now that we have a map λi: Xi → Gi(ιY)
such that g ◦λ ≃ϕ◦ξ . Then we construct a homotopy commutative diagram
i i i
Z //A
zz✈z✈i✈✈✈✈ ✤✤i ιiyyt+t1tttt ❃❃❃❃(cid:31)(cid:31)
Xi ✤ //Xi+1 //X
ξi+1
(cid:15)(cid:15)✤ λi+1
λi (cid:15)(cid:15) {{✈✈✈✈✈Fi (cid:15)(cid:15) yytttαtit+t1//A❃❃❃❃❃(cid:30)(cid:30) (cid:15)(cid:15)ϕ
Gi(ιY) // Gi+1(ιY) gi+1 //Y
where Z ❴ ❴ ❴//F is the whisker map induced by the bottom homotopy pullback
i i
and λi+1: Xi+1 //Gi+1(ιY) is the whisker map induced by the top homotopy
pushout. The composite gi+1◦λi+1 is homotopic to ϕ◦ξi+1. Hence the inductive
step is proven.
Atthe endofthe induction, wehaveg ◦λ ≃ϕ◦ξ =ϕ◦id =ϕ. Aswe have
n n n X
a homotopy section σ: Y →X =X of ϕ, we get a homotopy section λ ◦σ of g .
n n n
Moreover,we have (λ ◦σ)◦g ≃λ ◦f ≃λ ◦ζ ≃ω . (cid:3)
n n n n n
Example 16. If we consider any relative suspension Σ f: Σ W → Σ Z (and
A A A
in particular, of course, when A = ∗, any suspension Σf: ΣW → ΣZ), we have
Hcat(Σ f) = 1. And so, any map g that is relatively dominated by a (relative)
A
suspension has hcat(g)=1.
YET ANOTHER HOPF INVARIANT 9
In fact, by definition, a map g has Hcat(g) = 1 if and only if g is a (relative)
suspension. There are maps forwhich the strong Hopfcategoryis greaterthanthe
Hopfcategory: Forinstance,considerthe nullmapf: ∗→X ofExample5; ifX is
a space with cat(X)=1 that is not a suspension, then f cannot be a suspension,
so Hcat(f)>hcat(f)=1.
Proposition 17. In the diagram ♮, we have
Relcat(ζ )6i
i
As ω is a particular case of ζ , this implies Relcat(ω )6i.
i i i
Proof. Fori>0,letbuildthefollowinghomotopydiagramwherethethreesquares
are homotopy pushouts:
η
W //Zi−1 ((//A
β zi−1 θ
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A //Ci−1 //ΣAW ιi
ci−1 ζi
ιi−1 '' (cid:15)(cid:15) (cid:3)(cid:3) (cid:15)(cid:15) (cid:3)(cid:3)
Xi−1 //Xi
andwherethemapci−1 =(ιi−1,zi−1)isthewhiskermapinducedbythehomotopy
pushout.
Wehavesecat(ιi−1)6Pushcat(ιi−1)6i−1by[2],Theorem18. Sosecat(ci−1)6
i−1 by [2], Proposition 29. So Relcat(ci−1)6(i−1)+1=i by [2], Theorem 18.
And this implies Relcat(ζ )6i by [2], Lemma 11. (cid:3)
i
Theorem 18. For any f: Σ W →X, we have
A
Relcat(f)6Hcat(f) and relcat(f)6hcat(f)
Proof. If Hcat(f) = n, then we have f ≃ ζ in ♮. So Relcat(f) = Relcat(ζ ) 6 n
n n
by Proposition 17.
If hcat(f) = n, then f is relatively dominated by ω . As Relcat(ω ) 6 n, we
n n
have relcat(f)6n by [2], Proposition 10. (cid:3)
As a corollary, we get an indirect proof of Proposition 9 because relcat(κ ) 6
Y
relcat(f)by[2],Lemma 11,thatassertsthata homotopypushoutdoesn’tincrease
the relative category.
It is not difficult to find an example where these inequalities are strict:
Example 19. Let f be the map t1 in the homotopy cofibration
Z ⊲⊳Z // ΣZ∨ΣZ // ΣZ×ΣZ
u t1
LetA=∗,ΣZ 6≃∗. Ast1 isahomotopycofibre,wehaverelcat(t1)6Relcat(t1)6
1, see [2], Proposition 9. On the other hand, we have Hcat(t1) > hcat(t1) >
relcat(ι )=cat(ΣZ×ΣZ)=2 by Proposition 6.
X
10 JEAN-PAULDOERAENEANDMOHAMMEDELHAOUARI
4. Equivalent conditions to get the Hopf category
Let be given any map f: Σ W → X with secat(ι ) 6 n and any homotopy
A X
section σ: X →G of g : G →X. Consider the following homotopy pullbacks:
n n n
Q π //Σ W
π′ (cid:15)(cid:15) θnWA(cid:15)(cid:15) ❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍
Σ W //H // Σ W
A σ¯ n ηW A
n
f fn f
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
X // G //X
σ n gn
where θnW = (ωn,idΣAX) is the whisker map induced by the homotopy pullback
H . By the ‘Prism lemma’ (see [2], Lemma 46, for instance ), we know that the
n
hnoomticoetotphyatpπul≃lbaπc′ksionfcσe πan≃dηfWn i◦sθiWnd◦eeπd≃ΣAηWW,◦aσ¯n◦dπt′h≃atπη′nW. ◦σ¯ ≃ idΣAW. Also
n n n
Proposition20. Letbegiven anymap f: Σ W →X with secat(ι )6n andany
A X
homotopy section σ: X →G (ι ) of g : G (ι )→X. With the same definitions
n X n n X
and notations as above, the following conditions are equivalent:
(i) σ◦f ≃ω .
n
(ii) π has a homotopy section.
(iii) π is a homotopy epimorphism.
(iv) θW ≃σ¯.
n
Proof. We have the following sequence of implications:
(i) =⇒ (ii): Since σ ◦f ≃ ωn ≃ fn ◦θnW ◦idΣAW, we have a whisker map
(f,idΣAW): ΣAW →Q induced by the homotopy pullback Q which is a homotopy
section of π.
(ii) =⇒ (iii): Obvious.
(iii) =⇒ (iv): We have θW ◦π ≃σ¯◦π since π ≃π′. Thus θW ≃σ¯ since π is a
n n
homotopy epimorphism.
(iv) =⇒ (i): We have σ◦f ≃f ◦σ¯ ≃f ◦θW ≃ω . (cid:3)
n n n n
Theorem 21. Let be a (q−1)-connected map ι : A → X with secatι 6 n. If
X X
Σ W is a CW-complex with dimΣ W < (n+1)q −1 then σ ◦f ≃ ω for any
A A n
homotopy section σ of g .
n
Proof. Recall that g is the (i+1)-fold join of ι . Thus by [4], Theorem 47, we
i X
obtain that, for each i > 0, g : G → X is (i+1)q−1-connected. As g and ηW
i i i i
have the same homotopy fibre, the Five lemma implies that ηW: H → Σ W is
i i A
(i+1)q −1-connected, too. By [5], Theorem IV.7.16, this means that for every
CW-complexK withdimK <(i+1)q−1,ηW inducesaone-to-onecorrespondence
i
[K,H ] → [K,Σ W]. Apply this to K = Σ W and i = n: Since θW and σ¯ are
i A A n
both homotopy sections of ηW, we obtain θW ≃σ¯, and Proposition 20 implies the
n n
desired result. (cid:3)
Example 22. Let A = ∗ and W = Sr−1, so Σ W = Sr, and X = Sm. In this
A
case secatι = catSm = 1. Hence Theorem 21 means that if r < 2m−1, we
X
have σ◦f ≃ω1, whatever can be f and σ: X →G1(ιX), so hcatf =1 and we get