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Simplicial Structures in Topology

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74 II <strong>Simplicial</strong> Complexes<br />

(II.3.10) Corollary. Let (C,∂) and (C ′ ,∂ ′ ) be given positive augmented cha<strong>in</strong> complexes<br />

and let (C,∂ ) be free. Then any cha<strong>in</strong> homomorphism f : (C,∂) → (C ′ ,∂ ′ ),<br />

extend<strong>in</strong>g the trivial homomorphism 0: Z → Z and hav<strong>in</strong>g an acyclic carrier S, is<br />

null-homotopic.<br />

We prove now an important result known as Five Lemma.<br />

(II.3.11) Lemma. Let the diagram of Abelian groups and homomorphisms<br />

α<br />

A<br />

��<br />

A ′<br />

f<br />

f ′<br />

��<br />

β<br />

B<br />

��<br />

��<br />

′<br />

B<br />

g<br />

g ′<br />

��<br />

C<br />

γ<br />

��<br />

��<br />

′<br />

C<br />

h ��<br />

h ′<br />

D<br />

δ<br />

��<br />

��<br />

′<br />

D<br />

k ��<br />

be commutative and with exact l<strong>in</strong>es. If the homomorphisms α, β, δ, and ε are<br />

isomorphisms, so is γ.<br />

Proof. Let c ∈ C be such that γ(c) =0; then δ h(c) =h ′ γ(c) =0 and because δ<br />

is an isomorphism, h(c)=0. In view of the exactness condition, there is a b ∈ B<br />

with g(b)=c and g ′ β(b)=γg(b)=0; thus, there exists a ′ ∈ A ′ such that f ′ (a ′ )=<br />

β (b).But<br />

c = g(b)=gβ −1 f ′ (a ′ )=gfα −1 (a ′ )=0<br />

and so γ is <strong>in</strong>jective. For an arbitrary c ′ ∈ C ′ ,<br />

kδ −1 h ′ (c ′ )=ε −1 k ′ h ′ (c ′ )=0<br />

and, hence, there exists c ∈ C such that h(c)=δ −1 h ′ (c ′ ). Moreover,<br />

h ′ (c ′ − γ(c)) = h ′ (c ′ ) − δδ −1 h ′ (c ′ )=0<br />

k ′<br />

ε<br />

E<br />

��<br />

��<br />

′<br />

E<br />

and hence, there exists b ′ ∈ B ′ such that g ′ (b ′ )=c ′ − γ(c). It follows that<br />

γ(c + gβ −1 (b ′ )) = γ(c)+g ′ ββ −1 (b ′ )=c ′<br />

and so γ is also surjective. �<br />

Exercises<br />

1. A short exact sequence of Abelian groups<br />

Gn+1 ��<br />

fn+1 ��<br />

Gn<br />

fn ��<br />

��<br />

Gn−1

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