Qu 1:
Five Lemma:
Suppose given a commutative diagram of abelian groups and homomorphisms
A0
α0→ A1 α1→ A2 α2→ A3 α3→ A4
↓ f0 ↓ f1 ↓ f2 ↓ f3 ↓ f4
B0
β0→ B1 β1→ B2 β2→ B3 β3→ B4
in which both rows are exact. If f0, f1,f3,f4 are isomorphisms then f2 is also an
isomorphism.
In the following commutative diagram
i) both rows are exact and
ii) Hn(f0) Hn(f+)⊕Hn(f−), Hn−1(f0) and Hn−1(f+)⊕Hn−1(f−). are isomorphisms.
Hn(X
′
+ ∩X ′−) i∗→ Hn(X ′+)⊕Hn(X ′−) ν→ Hn(X ′) δ→ Hn−1(X ′+ ∩X ′−) i∗→ Hn−1(X ′+)⊕Hn−1(X ′−)
↓ Hn(f0) ↓ Hn(f+)⊕Hn(f+) ↓ Hn(f) ↓ Hn−1(f0) ↓ Hn−1(f+)⊕Hn−1(f+)
Hn(X+ ∩X−) i∗→ Hn(X+)⊕Hn(X−) ν→ Hn(X) δ→ Hn−1(X+ ∩X−) i∗→ Hn−1(X+)⊕Hn−1(X−)
Hence Hn(f) is an isomorphism by the Five Lemma.
If σ is a principal simplex of X let X+ be the subcomplex of X determined by σ and
let X− be the subcomplex of X determined by the set of principal simplices τ of X
where τ 6= σ. Then
X = X+ ∪X− where X+ ∩X− ⊂ ∂σ.
The subdivision of X at σ is the complex
X ′ = X ′+ ∪X ′−
where X ′+ = C(∂σ), X
′
− = X− so that X
′
+ ∩X ′− = X+ ∩X−.
Let Sq : X ′+ = C(∂σ) → X+ be a ‘squash map’ ; that is, Sq is the identity on ∂σ
and maps the cone point to a boundary vertex of ∂σ. Define f : X ′ → X by
f|X+ = Sq ; f|X− = Id
1
and consider the induced diagram of homology groups for n ≥ 1.
Hn(X
′
+ ∩X ′−) i∗→ Hn(X ′+)⊕Hn(X ′−) ν→ Hn(X ′) δ→ Hn−1(X ′+ ∩X ′−) i∗→ Hn−1(X ′+)⊕Hn−1(X ′−)
↓ Id ↓ Hn(Sq)⊕ Id ↓ Hn(f) ↓ Id ↓ Hn−1(Sq)⊕ Id
Hn(X+ ∩X−) i∗→ Hn(X+)⊕Hn(X−) ν→ Hn(X) δ→ Hn−1(X+ ∩X−) i∗→ Hn−1(X+)⊕Hn−1(X−)
Then H∗(Sq) is an isomorphism as both X ′+ and X+ are cones. Hence
Hn(f) : Hn(X
′
+)
'−→ Hn(X+)
is an isomorphism for n ≥ 1. As X ′, X are both connected then H0(X ′+) '−→ H0(X+).
2
Qu 2
χ(A∗) =
∑
r≥0
(−1)rdim(Ar).
χ(A∗) is well defined provided
i) there exists N ∈ N such that Ar = 0 for r ≥ N ;
ii) dim(Ar) is finite for 0 ≤ r ≤ N .
χ(B∗) = χ(A∗) + χ(C∗)
Proposition : χ(X) = χ(X+) + χ(X−)− χ(X+ ∩X−)
Proposition : χ(X × Y ) = χ(X)χ(Y ).
Proof : Let P(n) be the statement :
P(n) : χ(X × Y ) = χ(X)χ(Y ) when Y is a finite complex of dimension ≤ n.
Likewise let P(n,m) be the statement
P(n,m) : χ(X × Y ) = χ(X)χ(Y ) when Y is a finite complex of dimension ≤ n
having exactly m simplices of dimension n.
We are allowed to assume that P(n, 1) is true for each n. As
∧
m≥1
P(n,m) = P(n) = P(n+ 1, 0)
it suffices to prove that P(n− 1)∧P(n,m) =⇒ P(n,m+ 1).
Thus suppose Y is a finite n-dimensional complex having exactly m + 1 n-simplces
σ1, . . . , σm+1 and put
Y0 = Y
(n−1) ∪ σ1 ∪ . . . ∪ σm
so that Y = Y0 ∪ σm+1 and Y0 ∩ σm+1 ⊂ Y (n−1).Then
i) χ(X × Y0) = χ(X)χ(Y0) by P(n,m);
ii) χ(X × σm+1) = χ(X)χ(σm+1) by assumption;
iii) χ(X × (Y0 ∩ σm+1)) = χ(X)χ(Y0 ∩ σm+1) by P(n− 1).
As X × Y = X × Y0 ∪X × σm+1 then
3
χ(X × Y ) = χ(X × Y0) + χ(X × σm+1) − χ(X × (Y0 ∩ σm+1))
= χ(X)χ(Y0) + χ(X)χ(σm+1) − χ(X)χ(Y0 ∩ σm+1)
= χ(X){χ(Y0) + χ(σm+1) − χ(Y0 ∩ σm+1)}
= χ(X)χ(Y )
2
χ(S2n × Y ) = 2χ(Y )
χ(S2n × S2n × Y ) = 4χ(Y )
4
Qu 3
Define Σ0 = Σ − {2− simplex}; then
M#N = M0
⋃
∂=∂
N0
Write h1(M) = dimF2H1(M ;F2) so that χ(M) = 2 − h1(M).
Then χ(M#N) = χ(M) + χ(N) − 2. Hence
2 − h1(M#N) = [2 − h1(M)] + [2 − h1(N)] − 2
and so
h1(M#N) = h1(M) + h1(N)
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6
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8
Qu 4 : If A = (aij)1≤i,j≤n is an n× n matrix over a field F then
Tr(A) =
n∑
i=1
aii ∈ F.
We note that Tr(AB) = Tr(BA). Consequently if P = (pij)1≤i,j≤n is an invertible
n× n matrix over F then
Tr(PAP−1) = Tr(AP−1P ) = Tr(A)
Hence if f : V → V is a linear endomorphism of a finite dimensional vector space
then the trace Tr(f) of f may be defined unambiguously as Tr(f) Tr(A) where A is
any representing matrix for f .
If K is a finite simplicial complex and f : K → K be a simplicial map the Lefschetz
number λ(f) ∈ Q is defined by
λ(f) =
∑
r≥0
(−1)r TrHr(f)
where Hr(f) : Hr(K,Q)→ Hr(K,Q) is the induced map on rational homology. The
Lefschetz Fixed Simplex then states that
Theorem : Let K be a finite simplicial complex and f : K → K be a simplicial map.
If λ(f) 6= 0 then there exists a simplex σ in K such that (ignoring local orientation
on σ)
f(σ) = σ
Proof : The Lefschetz number λ(f) may also be computed as
λ(f) =
∑
r≥0
(−1)r TrCr(f)
where Cr(f) : Cr(K,Q) → Cr(K,Q) is the linear map induced by f on the space
of simplicial r-chains of K. Let (σi)1≤i≤Nr be an indexing of the natural basis of
Cr(K,Q) consisting of oriented r-simplices and let µ
r = (µrij)1≤i,j≤Nr be the matrix
of Cr(f) with respect to this basis. Then µ
r has at most one non-zero entry (±1 ) in
each column, and f fixes σi up to orientation if and only if µ
r
i,i = ±1.
Consequently if f fixes no r-simplex then the entire diagonal of µr is zero and
hence TrCr(f) = 0. Thus if f fixes no simplex of any dimension then TrCr(f) = 0
for each r and so λ(f) = 0.
In the contrapositive, if λ(f) 6= 0 then f fixes at least one simplex. 2
9
λ(f) = Tr(H0(f))− Tr(H3(f))− Tr(H9(f)) + Tr(H12(f)).
Clearly Tr(H0(f)) = 1 and Tr(H3(f)) = Tr(H9(f)) = −1. Hence
λ(f) = 3 + α
where α = Tr(H12(f)). As f
7 = Id then α7 = 1. As α ∈ Q then α = 1. Hence
λ(f) = 4.
Thus there exists a simplex σ of K such that f(σ) = σ.
If f 7k = Id then α7k = 1. If k is even then one has α = ±1. If k is odd then one
still has α = 1. Consequently
λ(f) =

2 or 4 k even
4 k odd
Either way, the conclusion is unaffected. In either case there exists a simplex σ of K
such that f(σ) = σ.