MATH 5713
Section: 001
Topology II
MWF 11:50 – 12:40
KIMP 308
Spring 2015
Prof. Matthew Clay
Homework 6 – Solution
1. Let A ⊆ X.
a. Show that H0 (X, A) = 0 if and only if A meets every path component of X.
b. Show that H1 (X) → H1 (X, A) is surjective if and only if every path component of X contains
at most one path component of A.
Proof. Considering the long exact sequence of the pair (X, A) we have:
· · · → H1 (X) → H1 (X, A) → H0 (A) → H0 (X) → H0 (X, A) → 0.
Hence H0 (X, A) = 0 if and only if H0 (A) → H0 (X) is surjective. Also, H1 (X) → H1 (X, A) is
surjective if and only if H0 (A) → H0 (X) is injective.
The homology group H0 (X), respectively H0 (A), is a free abelian group with basis the path
components of X, respectively A, and the map H0 (A) → H0 (X) is induced by inclusion of path
components.
Therefore H0 (A) → H0 (X) is surjective if and only if A meets every path component of X. This
proves a.
Further H0 (A) → H0 (X) is injective if and only if every path component of X meets at most
one path component of A. This proves b.
e n (X) ∼
e n+1 (SX) where SX = X × I/(X × {0}) ∪ (X × {1}) is the suspension of
2. Prove that H
=H
X.
Proof. Consider the long exact sequence for the pair (CX, X) where CX = X × I/X × {1} and we
identify X with the subspace X × {0} ⊂ CX:
e n+1 (CX) → Hn+1 (CX, X) → H
e n (X) → H
e n (CX) → · · ·
··· → H
e n (CX) = 0 for all n and hence Hn+1 (CX, X) ∼
e n (X). Since
As CX is contractible we have that H
=H
(CX, X) is a good pair (i.e., X is a deformation retract of an open set in CX, specifically the set
e n+1 (SX) as CX/X = SX.
X × [0, 1/2) ⊂ CX), we have that Hn+1 (CX, X) ∼
=H
3. Suppose X =
_
Sα2 and A ⊂ X is a finite set of points. Compute Hn (X, A).
α
1
e n (X) = ⊕α H
e n (S 2 ) hence:
Proof. We have that H
α
(
e n (X) = ⊕α Z if n = 2,
H
0
else
As X is path connected and A 6= ∅, we have H0 (X, A) = 0 by Problem 1. The long exact sequence
for the pair (X, A) contains the exact sequence:
e 1 (X) → H1 (X, A) → H
e 0 (A) → H
e 0 (X) = 0
0=H
e 0 (A) = Zd−1 where A contains d points. For n > 1 the long exact sequence
Therefore H1 (X, A) ∼
=H
contains the exact sequences:
e n (A) → H
e n (X) → Hn (X, A) → H
e n−1 (A) = 0
0=H
e n (X) for n > 1. Thus we find:
Therefore Hn (X, A) ∼
=H

d−1

if n = 1,
Z
Hn (X, A) = ⊕α Z if n = 2,


0
else
4. Prove that if nonempty open sets U ⊆ Rm and V ⊆ Rn are homeomorphic, then m = n.
Proof. For any x ∈ U , using X = Rm , A = Rm − {x} and Z = Rm − U , by excision we have
Hk (U, U − {x}) = Hk (X − Z, A − Z) ∼
= Hk (X, A) = Hk (Rm , Rm − {x}). Considering the long
e k−1 (Rm − {x}) as
exact sequence for the pair (Rm , Rm − {x}) we see that Hk (Rm , Rm − {x}) ∼
=H
m
m
m−1
e k (R ) = 0 for all k. Since R − {x} deformation retracts to S
H
we have Hk (U, U − {x}) ∼
=
m
m
Hk (R , R −{x}) = Z if k = m and equals 0 otherwise. Similarly, for any y ∈ V , Hk (V, V −{y}) = Z
if k = n and equals 0 otherwise.
If h : U → V is a homeomorphism and x ∈ U , then h∗ : Hk (U, U − {x}) → Hk (V, V − {h(x)})
is an isomorphism for all k and thus m = n.
5. If X is an n dimensional CW–complex show that Hk (X) = 0 for k > n and that Hn (X) is free.
Proof. If X is a 0 dimensional CW–complex then:
(
⊕x∈X Z if n = 0,
Hn (X) = ⊕x∈X Hn (x) =
0
else
Thus the statement holds in this case. Let n > 0 and assume that the statement holds for CW–
complexes of dimension n − 1. We consider the long exact sequence of the pair (X, X (n−1 ). This
contains the exact sequences:
Hk (X (n−1) ) → Hk (X) → Hk (X, X (n−1) ) → Hk−1 (X (n−1) )
2
e k (X/X (n−1) ). The quotient X/X (n−1) is a
As (X, X (n−1) ) is a good pair we have Hk (X, X (n−1) ) = H
e k (X/X (n−1) ) = ⊕α Z if k = n and equals 0 otherwise.
wedge of n dimensional spheres ∨α Sαn and so H
When k > n, we have Hk (X (n−1) ) = Hk−1 (X (n−1) ) = 0 by hypothesis and thus Hk (X) =
Hk (X, X (n−1) ) = Hk (∨α Sαn ) = 0.
When k = n we have that the map Hk (X) → Hk (X, X (n−1) ) is injective. As any subgroup of a
free abelian group is free abelian, the group Hn (X) is free abelian.
Thus by induction the statement holds for all n.
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