Download Math 527 - Homotopy Theory Spring 2013 Homework 9 Solutions

Math 527 - Homotopy Theory
Spring 2013
Homework 9 Solutions
Problem 1. Consider the Hopf map η : S 3 → S 2 .
a. Describe the cofiber C(η). It is a familiar space.
Solution. Recall that η : S 3 CP 1 was defined as the quotient by the action of S 1 ⊂ C×
(via scalar multiplication) on the unit sphere S 3 ⊂ C2 \ {0}. Its cofiber C(η) is obtained by
attaching a 4-cell to CP 1 using η as attaching map. This CW structure on C(η) is the standard
CW structure on CP 2 .
b. Consider the canonical comparison map ϕ : F (η) → ΩC(η) from the homotopy fiber to the
loop space of the cofiber. Find the lowest dimension k such that πk F (η) is not isomorphic to
πk ΩC(η) (and thus ϕ cannot possibly induce an isomorphism on πk ).
Solution. Since η is a fibration, its homotopy fiber F (η) is homotopy equivalent to its strict
fiber η −1 (∗) = S 1 . Thus the comparison map ϕ can be viewed as a map
ϕ : S 1 → ΩCP 2 .
Recall the natural isomorphism πi (ΩX) ∼
= πi+1 (X) for all i ≥ 0. To study the homotopy groups
of CP 2 , note that the long exact sequence on homotopy of the fibration
q
S1 → S5 →
− CP 2
yields the isomorphisms
∼
=
∂ : π2 (CP 2 ) −
→ π1 (S 1 ) ∼
=Z
∼
=
q∗ : πi (S 5 ) −
→ πi (CP 2 )
for all i ≥ 3. In particular, we obtain:
π0 (ΩCP 2 ) ∼
= π1 (CP 2 ) = 0
π1 (ΩCP 2 ) ∼
= π2 (CP 2 ) ∼
=Z
π2 (ΩCP 2 ) ∼
= π3 (CP 2 ) ∼
= π3 (S 5 ) = 0
π3 (ΩCP 2 ) ∼
= π4 (CP 2 ) ∼
= π4 (S 5 ) = 0
π4 (ΩCP 2 ) ∼
= π5 (CP 2 ) ∼
= π5 (S 5 ) ∼
= Z.
Comparing with the homotopy groups of S 1 , we conclude πi (S 1 ) ' πi (ΩCP 2 ) for all i < 4
whereas π4 (S 1 ) 6' π4 (ΩCP 2 ).
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Problem 2. Show that a (strictly) commutative diagram
f0
/
W
g0
Y
g
/
X
Z
f
is homotopy k-Cartesian if and only if for all x ∈ X, the induced map on homotopy fibers
f 00 : Fx (g 0 ) → Ff (x) (g)
over the respective basepoints x ∈ X and f (x) ∈ Z is k-connected. Here we have k ≥ 0 or
k = ∞.
Solution. Consider the comparison map ϕ from W to the homotopy pullback
f0
W
ϕ
#
X
g0
pY
×hZ
Y
g
pX
(
/
Y
/
X
Z.
f
Because the homotopy pullback square induces a homotopy equivalence on homotopy fibers
'
F (pX ) −
→ F (g), the condition on homotopy fibers in the statement is equivalent to the following:
The map on homotopy fibers ϕ00 : F (g 0 ) → F (pX ) induced by the diagram
ϕ
/
W
g0
X ×hZ Y
pX
/
X
X
id
is k-connected. For any basepoint x ∈ X (which we omit from the notation), consider the
diagram
ϕ00
/
0
F (g )
ϕ
/
W
g0
F (pX )
X ×hZ Y
/
X
id
2
X
pX
(1)
where both columns are fiber sequences. Now consider the induced map of long exact sequences
on homotopy
/
/
·
...
'
πi+1 (ϕ)
...
/
·
/
/
·
/
·
πi (ϕ00 )
·
·
·
/
/
...
/
...
·
πi−1 (ϕ00 )
'
/
/
·
πi (ϕ)
/
/
·
/
·
·
Recall the four-lemma for epimorphisms, schematically summarized here:
/
·
/
·
/
·
·
/
·
∴epi
/
·
/
·
_
·
This implies the following:
1. If πi (ϕ) is an epimorphism, then πi (ϕ00 ) is an epimorphism.
2. If πi (ϕ00 ) is an epimorphism and πi−1 (ϕ00 ) is a monomorphism, then πi (ϕ) is an epimorphism.
Now recall the four-lemma for monomorphisms, schematically summarized here:
·
/
·
·
/
/
_
/
·
/
·
·
∴mono
/
·
·
_
This implies the following:
1. If πi (ϕ) is an monomorphism and πi+1 (ϕ) is an epimorphism, then πi (ϕ00 ) is a monomorphism.
2. If πi (ϕ00 ) is a monomorphism, then πi (ϕ) is a monomorphism.
From both statements 1., we deduce: If ϕ is k-connected, then ϕ00 is k-connected.
From both statements 2., we deduce: If ϕ00 is k-connected, then ϕ is k-connected.
Therefore, ϕ is k-connected if and only if ϕ00 is k-connected.
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Alternate solution. Taking horizontal fibers of the two bottom rows in (1) yields the diagram
ϕ00
/
0
F (g )
/
F (ϕ)
g 00
/
X ×hZ Y
/
∗ ' F (id)
ϕ
W
g0
F (pX )
pX
/
X
X.
id
It is a fact (c.f. Strom, Theorem 8.57) that filling in the top left corner of the 3-by-3 diagram
using either the vertical or the horizontal homotopy fiber will yield equivalent results:
F (g 00 ) ' F (ϕ00 )
ϕ00
/
/
F (ϕ)
g 00
/
∗ ' F (id)
F (pX )
ϕ
/
W
g0
/
F (g 0 )
X ×hZ Y
/
X
id
from which we deduce the equivalence F (ϕ00 ) ' F (g 00 ) ' F (ϕ).
To conclude, consider the equivalent conditions:
ϕ is k-connected
⇔ F (ϕ) is (k − 1)-connected
⇔ F (ϕ00 ) is (k − 1)-connected
⇔ ϕ00 is k-connected.
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X
pX
Problem 3. Let f : X → Y be an n-connected map between spaces, and assume X is mconnected.
a. Using Blakers-Massey, show that the canonical comparison map
ϕ : F (f ) → ΩC(f )
from the homotopy fiber to the loop space of the cofiber of f is (m + n)-connected.
Solution. Consider the homotopy pushout diagram
f
/
X
Y
ho
∗
p/ C(f ).
Since X is m-connected, the map X → ∗ is (m+1)-connected. By Blakers-Massey, the diagram
is (m+1+n−1 = m+n)-Cartesian. By Problem 2, the map induced on (horizontal) homotopy
fibers F (f ) → ΩC(f ) is (m + n)-connected.
b. Looking back (c.f. Problem 1) at the example of the Hopf map η : S 3 → S 2 , conclude that:
• The connectivity estimate m + n in part (a) cannot be improved in general;
• The map ϕη : F (η) → ΩC(η) does in fact induce isomorphisms on homotopy groups below
the least dimension k satisfying πk F (η) 6' πk ΩC(η).
Solution. The map η : S 3 → S 2 is 1-connected, and its source S 3 is 2-connected. By part
(a), the comparison map
ϕ : F (η) → ΩC(η)
is 3-connected. In particular, ϕ induces isomorphisms on homotopy groups πi for i < 4. In the
cases i = 0, 2, and 3, this is automatic since both groups are trivial. In the remaining case
i = 1, the fact that ϕ is 3-connected guarantees that it induces an isomorphism on π1 .
However, ϕ cannot be 4-connected, because the induced map on π4
ϕ∗
0 = π4 (S 1 ) −→ π4 (ΩCP 2 ) ∼
=Z
cannot be surjective.
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