Download Math 527 - Homotopy Theory Spring 2013 Homework 1 Solutions

Math 527 - Homotopy Theory
Spring 2013
Homework 1 Solutions
Problem 1. Show that the following conditions on a topological space X are equivalent.
1. X is contractible.
2. The identity map idX : X → X is null-homotopic.
3. For any space Y , every continuous map f : X → Y is null-homotopic.
4. For any space W , every continuous map f : W → X is null-homotopic.
5. For any space W , the projection map pW : W × X → W is a homotopy equivalence.
Solution. Throughout this problem, let cz denote any constant map with value z (with appropriate source and target).
(1 ⇔ 2) Consider the following equivalent statements.
X is contractible.
⇔ The unique map f : X → ∗ is a homotopy equivalence, i.e. has a homotopy inverse g : ∗ → X.
⇔ There is a point g(∗) = x0 ∈ X such that the constant map cx0 = g ◦f : X → X is homotopic
to the identity map idX .
⇔ The map idX is null-homotopic.
(3 ⇒ 2) Particular case Y = X and f = idX .
(2 ⇒ 3) Assume idX ' cx0 for some point x0 ∈ X. Writing f as the composite f = f ◦ idX , we
obtain
f = f ◦ idX ' f ◦ cx0 = cf (x0 )
so that f is null-homotopic.
(4 ⇒ 2) Particular case W = X and f = idX .
(2 ⇒ 4) Assume idX ' cx0 for some point x0 ∈ X. Writing f as the composite f = idX ◦ f , we
obtain
f = idX ◦ f ' cx0 ◦ f = cx0
so that f is null-homotopic.
(5 ⇒ 1) Particular case W = ∗.
(1 ⇒ 5) Assume idX ' cx0 for some point x0 ∈ X. Then the map
(idW , cx0 ) : W → W × X
is homotopy inverse to the projection pW : W × X → W . Indeed, one composite is already the
identity
pW ◦ (idW , cx0 ) = idW
while the other composite satisfies
(idW , cx0 ) ◦ pW = idW × cx0 ' idW × idX = idW ×X .
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Problem 2. Let C be a locally small category with finite products, including a terminal object.
Let G be a group object in C. Show that for any object X of C, the hom-set HomC (X, G) is
naturally a group.
In other words, the structure maps of G induce a group structure on HomC (X, G), and this
assignment
HomC (−, G) : C op → Gp
is a functor.
Solution. Since the functor HomC (X, −) : C → Set preserves limits (in particular finite products), the set HomC (X, G) inherits a group object structure in Set from G, so that HomC (X, G)
is a group. Explicitly, its multiplication map is
µ∗
HomC (X, G) × HomC (X, G) ∼
= HomC (X, G × G) −→ HomC (X, G)
and likewise for the unit and inverse structure maps.
To prove naturality, it suffices to prove that for any morphism f : X → Y in C, the induced
map of sets
f ∗ : HomC (Y, G) → HomC (X, G)
is a group homomorphism. This follows from commutativity of the diagram
HomC (X, G) × HomC (X, G) o
∼
=
µ∗
O
O
f ∗ ×f ∗
HomC (Y, G) × HomC (Y, G) o
/
HomC (X, G × G)
O
f∗
∼
=
f∗
µ∗
HomC (Y, G × G)
2
HomC (X, G)
/
HomC (X, G).
Problem 3. Consider S 1 as the unit circle in R2 with basepoint (1, 0), and consider the
“pinch” map
p : S 1 → S 1 /S 0 ∼
= S1 ∨ S1
which collapses the equator S 0 ⊂ S 1 , i.e. identifies the points (1, 0) and (−1, 0).
a. Show that the pinch map is (pointed) homotopy coassociative. More precisely, the diagram
p
/
S1
p
S1 ∨ S1
S1 ∨ S1
/
p∨id
S1 ∨ S1 ∨ S1
id∨p
commutes up to pointed homotopy.
Solution. Denote the two summand inclusions by ιi : S 1 → S 1 ∨ S 1 for i = 1, 2. Using the
model S 1 ∼
= I/∂I, the pinch map can be explicitly written as
(
ι1 (2s)
if 0 ≤ s ≤ 12
p(s) =
ι2 (2s − 1) if 12 ≤ s ≤ 1.
The composite via the top right is


if 0 ≤ s ≤ 14
ι1 (4s)
(p ∨ id) ◦ p(s) = ι2 (4s − 1) if 41 ≤ s ≤ 12


ι3 (2s − 1) if 21 ≤ s ≤ 1
whereas composite via the bottom left is


if 0 ≤ s ≤ 12
ι1 (2s)
(id ∨ p) ◦ p(s) = ι2 (4s − 2) if 12 ≤ s ≤ 34


ι3 (4s − 3) if 34 ≤ s ≤ 1.
The formula

4s

if 0 ≤ s ≤ 1+t
ι1 ( 1+t )
4
2+t
H(s, t) = ι2 (4s − 1 − t) if 1+t
≤
s
≤
4
4

 4s−2−t
ι3 ( 2−t )
if 2+t
≤
s
≤
1
4
defines a pointed homotopy H : S 1 × I → S 1 ∨ S 1 ∨ S 1 from (p ∨ id) ◦ p to (id ∨ p) ◦ p.
In fact, a very similar argument shows that S 1 is a homotopy cogroup object in Top∗ . (Do not
show this.) Comultiplication is the pinch map S 1 → S 1 ∨ S 1 , the counit is the constant map
S 1 → ∗, and the coinverse S 1 → S 1 reverses the last component (viewed in R2 ).
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b. Conclude that for any pointed space (X, x0 ), the set π1 (X, x0 ) is naturally a group.
More precisely, the structure maps of S 1 as homotopy cogroup object induce a group structure
on π1 (X, x0 ), and moreover this assignment defines a (covariant) functor π1 : Top∗ → Gp.
Solution. Since S 1 is a homotopy cogroup object in Top∗ , it becomes a cogroup object
in the homotopy category Ho(Top∗ ), and therefore a group object in the opposite category
Ho(Top∗ )op . We have
π1 (X, x0 ) = [S 1 , (X, x0 )]∗
= HomHo(Top∗ ) S 1 , (X, x0 )
= HomHo(Top∗ )op (X, x0 ), S 1
which is naturally a group, by Problem 2.
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Problem 4. For pointed spaces, show that the smash product distributes over the wedge.
More precisely, there is a natural isomorphism
X ∧ (Y ∨ Z) ∼
= (X ∧ Y ) ∨ (X ∧ Z).
of pointed spaces. Don’t forget to argue that the isomorphism is natural.
Solution. Recall that the wedge is the coproduct in Top∗ . Applying the functor X ∧ − to
the natural summand inclusions Y → Y ∨ Z and Z → Y ∨ Z produces maps
X ∧ Y → X ∧ (Y ∨ Z)
X ∧ Z → X ∧ (Y ∨ Z)
which together define a natural map
ϕ : (X ∧ Y ) ∨ (X ∧ Z) → X ∧ (Y ∨ Z)
of pointed spaces. It remains to show that ϕ is an homeomorphism, thus an isomorphism in
Top∗ .
To construct the inverse of ϕ, consider the map ψ in the commutative diagram
e
ψ
X ∧ (Y ∨ Z)
/
(X ∧ Y ) ∨ (X ∧ Z)
O
<
O
O
ψ
X × (Y ∨ Z)
(X ∧ Y ) q (X ∧ Z)
O
O
O
X × (Y q Z) o
∼
=
(X × Y ) q (X q Z)
in Top. The bottom is an isomorphism because the functor X × − : Top → Top preserves
(arbitrary) coproducts.
Now the map
X × (Y q Z) → X × (Y ∨ Z)
is a quotient map, hence so is the composite
X × (Y q Z) → X × (Y ∨ Z) → X ∧ (Y ∨ Z).
on the left-hand side of the diagram. Since ψ is constant on the equivalence classes defined by
this quotient map, it induces a unique continuous map
ψe : X ∧ (Y ∨ Z) → (X ∧ Y ) ∨ (X ∧ Z)
making the diagram commute.
By construction, ψe is clearly the set-theoretic inverse of ϕ.
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