Simplicial Structures in Topology

152 IV Cohomology
(IV.1.1) Theorem. Suppose that the sequence of Abelian groups
A
f
��
B
is exact. Then, for every Abelian group G, the sequence of Abelian groups
is exact.
0
��
Hom(C,G)
�g
g
��
C
��
Hom(B,G)
��
0
�f
��
Hom(A,G)
Proof. Let us prove that �g is injective. Take φ ∈ Hom(C,G) such that �g(φ) =0.
Then, for every b ∈ B, φ(g(b)) = 0. Since g is surjective, for every c ∈ C, there
exists b ∈ B such that c = g(b). Hence, for every c ∈ C, φ(c)=φ(g(b)) = 0; that is
to say, φ = 0.
We now prove the exactness at Hom(B,G). Foreveryφ ∈ Hom(C,G),
�f �g(φ)=φ(gf)=0
since gf = 0; hence, im �g ⊂ ker �f .Letψ ∈ Hom(B,G) be such that �f (ψ)=ψ f = 0.
Then the restriction of ψ to f (A) is null and so there exists a homomorphism
ψ ′ : B/ f (A) → G such that the composite function
B
q
��
B/ f (A)
(where q is the quotient homomorphism) coincides with ψ. On the other hand,
since im f = kerg, there exists an isomorphism g ′ : B/ f (A) ∼ = C for which g ′ q = g.
The homomorphism φ = ψ ′ (g ′ ) −1 : C → G is such that �g(φ) =ψ and therefore,
ker �f ⊂ im �g. �
We note that if A is the direct sum of the Abelian group of the integers Z with
itself n times (A = Z × ...× Z = Z n ), then
Hom(Z × ...× Z,G)
� �� �
n times
∼ = Hom(Z,G) × ...× Hom(Z,G) ∼= G × ...× G
� �� � �
;
�� �
n times
n times
indeed, the function
ψ ′
��
G
φ : Hom(Z × ...× Z,G) −→ G × ...× G
(defined by φ( f )=(f (1,0,...,0),..., f (0,...,1)) for every f ∈ Hom(Z × ...×
Z,G)) is an isomorphism.
We use Theorem (IV.1.1) for computing Hom(Z2,G). In view of the exact sequence
0
��
Z
2 ��
Z
��
Z2
��
0,