Math 527 - Homotopy Theory
Spring 2013
Homework 5 Solutions
Problem 1. Find a space X such that no choice of basepoint will make it well-pointed.
Solution. Consider the space Q of rational numbers, with its standard metric topology. For
∼
=
any point q ∈ Q, there is a homeomorphism Q −
→ Q sending q to 0, e.g. translation by −q.
Therefore it suffices to show that the inclusion of some basepoint, say, i : {7} ,→ Q is not a
cofibration.
Consider the mapping cyclinder M (i) ∼
= Q × {0} ∪ {7} × I ⊂ Q × I, with the natural maps
f : Q → M (i) and H : {7} × I → M (i). Recall that Q is totally disconnected (i.e. connected
components are all singletons), and the same argument shows that singletons {(q, 0)} ⊂ M (i)
with q 6= 7 are connected components of M (i). In particular, any path in M (i) starting at (q, 0)
must be constant. Therefore, any homotopy G from f to a map g : Q → M (i) must satisfy
G(q, t) = G(q, 0) = f (q) = (q, 0)
for all t ∈ I and all q 6= 7. At time t = 1, the resulting map g : Q → M (i) satisfies g(q) = (q, 0)
for all q 6= 7. By continuity, it also satisfies
g(7) = (7, 0) 6= (7, 1) = H(7, 1).
e: Q×I →
Therefore the homotopy H : {7} × I → M (i) cannot be extended to a homotopy H
M (i).
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Problem 2. (May § 9.4 Lemma) Show that for all n ≥ 0, the functor πn : Top∗ → Set∗ preserves products. In other words, for all pointed spaces X and Y , there is a natural isomorphism
πn (X × Y ) ∼
= πn (X) × πn (Y ).
Solution. Recall that the Cartesian product is the product in Top∗
(X, x0 ) × (Y, y0 ) = (X × Y, (x0 , y0 ))
as well as in the homotopy category Ho(Top∗ ). Therefore the natural map
∼
=
πn (X × Y ) = [S n , X × Y ]∗ −
→ [S n , X]∗ × [S n , Y ]∗ = πn (X) × πn (Y )
is an isomorphism (in Set∗ ).
Remark. This argument works for arbitrary products, not just finite:
!
Y
Y
∼
=
→
πn (Xα ).
ϕ : πn
Xα −
α
α
Moreover, the isomorphism ϕ is an isomorphism of groups, since each of its coordinates
!
Y
Y
ϕ
πn (Xα ) πn (Xβ )
πn
Xα −
→
α
α
is a group homomorphism, namely πn (pβ ) induced by the projection pβ :
2
Q
α
Xα Xβ .
Problem 3. (May § 9.6 Problem 1) Let X and Y be pointed spaces, and n ≥ 2.
a. Show that the map j∗ : πn (X × Y ) → πn (X × Y, X ∨ Y ) is zero.
Solution. Consider the natural isomorphisms
(pX∗ ,pY ∗ )
πn (X × Y )
/
∼
=
πn (X) × πn (Y ) o
∼
=
πn (X) ⊕ πn (Y )
where the last step comes from the fact that πn (X) and πn (Y ) are abelian groups (n ≥ 2). One
readily checks that the inverse isomorphism is
(ιX∗ ,ιY ∗ )
πn (X) ⊕ πn (Y )
/
∼
=
πn (X × Y )
where ιX : X → X × Y is the “slice inclusion” ιX (x) = (x, y0 ) and likewise for ιY . Therefore,
any element θ ∈ πn (X × Y ) can be (uniquely) written as a sum θ = θX + θY with θX ∈ im ιX∗
and θY ∈ im ιY ∗ .
Any element θX ∈ im ιX∗ is represented by a map Dn → X × Y whose image is contained in
X × {y0 } ⊆ X ∨ Y , which implies j∗ (θX ) = 0 ∈ πn (X × Y, X ∨ Y ), and likewise j∗ (θY ) = 0. We
conclude
j∗ (θ) = j∗ (θX + θY )
= j∗ (θX ) + j∗ (θY )
=0+0
=0
using the fact that πn (X ×Y, X ∨Y ) is a group (as n ≥ 2) and j∗ is a group homomorphism.
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b. Show that there is an isomorphism
πn (X ∨ Y ) ' πn (X) ⊕ πn (Y ) ⊕ πn+1 (X × Y, X ∨ Y ).
Solution. Note that the slice inclusions ιX and ιY factor through the subspace X ∨Y ⊆ X ×Y :

X ιX

X ∨Y /
i
/
7
X × Y.
ιX
This provides a section of the map i∗ as follows:
i∗
πn (X ∨ Y )
/
g
πn (X × Y )
O
∼
=
(ιX∗ ,ιY ∗ )
(ιX∗ ,ιY ∗ )
πn (X) ⊕ πn (Y ).
Hence the long exact sequence
j∗
πn+1 (X × Y )
/ πn+1 (X × Y, X ∨ Y )
∂
0
i∗
/ πn (X ∨ Y )
c
/ πn (X × Y )
j∗
/ πn (X × Y, X ∨ Y )
0
breaks down to a split short exact sequence
0
/
∂
πn+1 (X × Y, X ∨ Y )
/
i∗
πn (X ∨ Y )
e
/
πn (X × Y )
which yields the isomorphism
πn (X ∨ Y ) ∼
= πn (X × Y ) ⊕ πn+1 (X × Y, X ∨ Y )
∼
= πn (X) ⊕ πn (Y ) ⊕ πn+1 (X × Y, X ∨ Y ).
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/
0
Problem 4. (May § 9.4 Lemma) Let n ≥ 2 and consider the n-dimensional real projective
space RP n . Show that the following holds: π1 (RP n ) ' Z/2 and πk (RP n ) ' πk (S n ) for all
k ≥ 2.
Solution. Recall that RP n is obtained as the quotient
RP n = S n /O(1)
of the sphere by the free action of the group O(1) = {−1, 1} ⊂ R× . The quotient map
p : S n RP n is thus a two-sheeted covering, and is in fact the universal cover of RP n , since
S n is simply-connected for n ≥ 2. Therefore π1 (RP n ) is isomorphic to O(1) ' Z/2.
'
For k ≥ 2, the covering map p induces an isomorphism p∗ : πk (S n ) −
→ πk (RP n ).
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Problem 5. (May § 9.6 Problem 2) Let n ≥ 3.
a. Compute the group πn (RP n , RP n−1 ).
Solution. The standard inclusion i : RP n−1 → RP n lifts to universal covers to the standard
inclusion ei : S n−1 → S n , as illustrated in the commutative diagram
S n−1
p
RP n−1
ei
i
/
Sn
/
p
RP n .
Note that ei : S n−1 → S n is null-homotopic, as πn−1 (S n ) = 0. For any k ≥ 2, applying πk to this
diagram yields
πk (S
n−1
p∗
'
ei∗ =0
/
)
πk (S n )
'
πk (RP n−1 )
i∗
/
p∗
πk (RP n )
from which we conclude i∗ = 0. Hence the long exact sequence
i∗
πn (RP n−1 )
/ πn (RP n )
j∗
/ πn (RP n , RP n−1 )
∂
/ πn−1 (RP n−1 )
0
i∗
/ πn−1 (RP n )
0
breaks down to a short exact sequence
0
/
j∗
n
πn (RP )
/
πn (RP n , RP n−1 )
∂
/
πn−1 (RP n−1 )
/
0
which is automatically split, since πn−1 (RP n−1 ) ' Z is a projective Z-module. This yields the
isomorphism
πn (RP n , RP n−1 ) ' πn (RP n ) ⊕ πn−1 (RP n−1 )
∼
= πn (S n ) ⊕ πn−1 (S n−1 )
∼
= Z ⊕ Z.
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b. Deduce that the quotient map of pairs
q : (RP n , RP n−1 ) → (RP n /RP n−1 , ∗)
does not induce an isomorphism on homotopy groups.
Solution. Recall that the standard CW-structure on RP n has one cell in each dimension
0, 1, . . . , n and RP n−1 is the (n − 1)-skeleton of RP n . Therefore we have a homeomorphism
RP n /RP n−1 ∼
= S n.
The quotient map q cannot induce an isomorphism on the nth relative homotopy groups, as the
two groups are non-isomorphic:
πn (RP n , RP n−1 ) ' Z ⊕ Z
πn (RP n /RP n−1 , ∗) ∼
= πn (S n , ∗) ∼
= Z 6' Z ⊕ Z.
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