Homework 8 Solution
MATH 5713
Section: 001
Topology II
MWF 11:50 – 12:40
KIMP 308
Spring 2015
Prof. Matthew Clay
Homework 8 – Solution
1. Use the Meyer–Vietoris sequence to compute the homology of the Klein bottle RP2 #RP2 .
Proof. We decompose the Klein bottle RP2 #RP2 = A ∪ B as the union of two M¨obius bands glued
together along their boundary circles. As the M¨obius band deformation retracts onto its central
circle, H2 (A) = H2 (B) = H2 (S 1 ) = 0. Thus using reduced homology, the relevant part of the
Mayer–Vietoris sequence is:
ϕ
ψ
0 → H2 (RP2 #RP2 ) → H1 (A ∩ B)→H1 (A) ⊕ H1 (B)→H1 (RP2 #RP2 ) → 0
The group H1 (A ∩ B) is isomorphic to Z and H1 (A) ⊕ H1 (B) is isomorphic to Z ⊕ Z. Using these
identifications we have ϕ(n) = (2n, −2n). In particular ϕ is injective and so H2 (RP2 #RP2 ) = 0.
Thus H1 (RP2 #RP2 ) is isomorphic to H1 (A) ⊕ H1 (B) modulo the image of ϕ. Since {(1, 0), (1, −1)}
is a basis for Z ⊕ Z, we see that H1 (RP2 #RP2 ) is isomorphic to Z ⊕ Z2 . As RP2 #RP2 is pathconnected, H0 (RP2 #RP2 ) = 0.
2. Let X be the quotient of the unit cube obtained by identifying opposite faces via a one-quarter
twist. (See Exercise 14 in Section 1.2). Compute the homology groups Hn (X).
Proof. Refer to Figure 1 for the labeling of the cells in X. Additionally, we label the front/back
2–cell by A, the top/bottom 2–cell by B and the left/right 2–cell by C.
c
d
b
u
v
a
Figure 1: The space X in Problem 2.
Denote the cellular i–chain group by Ki (X). Thus K0 (X) = Z2 , K1 (X) = Z4 , K2 (X) = Z3
and K3 (X) = Z. Order the bases of Ki (X) using alphabetical order. With these conventions, the
cellular boundary maps are represented by:
1 1
1
0
1 −1 1
−1 1 −1 1
d3 = 0
d2 =
d1 =
1 −1 −1
1 −1 1 −1
0
1 1 −1
1
All of the following claims are justified by using integer row and column operations on these matrices.
The image of d1 is spanned by v −u and hence H0 (X) = Z. As a basis for the kernel of d1 we can use
{a + b + c + d, b + c, c + d}; as a basis for the image of d2 we can use {a + b + c + d, 2(b + c), 2(c + d)}.
Therefore H1 (X) = Z2 ⊕ Z2 . Since d2 is injective, H2 (X) = 0. Finally as d3 is the zero map,
H3 (X) = Z.
3. If a finite CW complex X is the union of two subcomplexes A and B show that χ(X) =
χ(A) + χ(B) − χ(A ∩ B).
Proof. First observe that if:
0 → Cn → Cn−1 → · · · C0 → 0
P
is an exact sequence of abelian groups then (−1)i rank Ci = 0. Indeed, this follows by induction.
For the case n = 0, if 0 → C0 → 0 is exact then C0 = 0 and so rank(C0 ) = 0. Applying induction
to the exact sequence of length n:
0 → Cn−1 im Cn → Cn−2 → · · · C0 → 0
we have that:
0=
=
=
n−2
X
i=0
n−2
X
i=0
n−2
X
(−1)i rank Ci + (−1)n−1 rank Cn−1 im Cn
(−1)i rank Ci + (−1)n−1 (rank Cn−1 − rank im Cn )
(−1)i rank Ci + (−1)n−1 (rank Cn−1 − rank Cn )
i=0
=
n
X
(−1)i rank Ci
i=0
as claimed. The equality between the second and third line comes as the map Cn → Cn−1 is
injective.
Applying this observation to the Mayer–Vietoris sequence for X = A ∪ B we have:
X
0=
(−1)i (rank Hi (A ∩ B) − rank Hi (A) ⊕ Hi (B) + rank Hi (X))
i
X
X
X
=
(−1)i rank Hi (A ∩ B) −
(−1)i (rank Hi (A) + rank Hi (B)) +
(−1)i rank Hi (X)
i
i
i
= χ(A ∩ B) − (χ(A) + χ(B)) + χ(X)
from which the statement follows.
4. For finite CW complexes X and Y show that χ(X × Y ) = χ(X)χ(Y ).
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Proof. The product X × Y has a product CW structure consisting of m–cells of the form eiα × em−i
β
where eiα ranges over the i–cells of X and em−i
ranges
over
the
(m
−
i)–cells
of
Y
,
0
≤
i
≤
m.
Thus:
β
χ(X × Y ) =
X
=
X
(−1)m (#m–cells in X × Y )
m
m
(−1)
m
=
(#i–cells in X)(#(m − i)–cells in Y )
i=0
m
XX
m
m
X
(−1)i (#i–cells in X) · (−1)m−i (#(m − i)–cells in Y )
i=0
!
=
X
(−1)m (#m–cells in X)
m
·
!
X
(−1)m (#m–cells in Y )
m
= χ(X)χ(Y ).
e → X an n–sheeted covering space, show that χ(X)
e =
5. For X a finite CW complex and p : X
nχ(X). Conclude that if Sg covers Sh then g = n(h − 1) + 1 for some n ∈ N. (Previously we have
shown that if g = n(h − 1) + 1 then there exists an n–sheeted cover Sg → Sh . See Example 1.41 in
Section 1.3.)
e → X has a CW structure consisting of m–cells em where em
Proof. The covering space p : X
α
α,i
` m
.
Specifically,
the
characteristic
map
e
)
=
ranges over the m–cells of X, 1 ≤ i ≤ n and p−1 (em
α
i α,i
m
e is the lift of Φα : (Dαm , 0) → (X, Φα (0)) to the covering space p : (X,
e Φi (0)) →
Φα,i : Dα,i
→ X
α
e is n
(X, Φα (0)) where p−1 (Φα (0)) = {Φ1α (0), . . . , Φnα (0)}. Therefore the number of m–cells in X
e = nχ(X).
times the number of m–cells in X. It follows that χ(X)
Hence if Sg covers Sh then for some n ∈ N, we have 2 − 2g = χ(Sg ) = nχ(Sh ) = 2n − 2nh.
Dividing by 2 and rearranging we have g = n(h − 1) + 1.
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