Simplicial Structures in Topology

IV.1 Cohomology with Coefficients in G 157
(IV.1.5) Lemma. ker(�in) ∼ = Hom(Hn(C),G).
Proof. We define
φ : ker(�in) → Hom(Hn(C),G)
so that, for every x+Bn(C) ∈ Hn(C), φ( f )(x+Bn(C)) = f (x). This function is well
defined and is a homomorphism. On the other hand, we define
ψ : Hom(Hn(C),G) → ker(�in)
by setting ψ(g) =gp,wherep: Zn(C) → Hn(C) is the quotient homomorphism.
Also this function is a homomorphism; in addition, the compositions satisfy ψφ =
1 ker(�in) and φψ = 1 Hom(Hn(C),G). �
This result shows that ker(�in) is independent from the groups Hom(Bn(C),G) and
Hom(Zn(C),G), and depends only on Hn(C), the cokernel of the monomorphism
in : Bn(C) → Zn(C), and on G. As in the case of the functor −⊗G, we wonder
whether the same is also true for coker(�in−1). In fact, we have the following result
which is dual to Proposition (II.5.4):
(IV.1.6) Proposition. Let H be the cokernel of the monomorphism i: B → Zbetween
free Abelian groups and any Abelian group G. Then, both the kernel and
the cokernel of the homomorphism �i: Hom(Z,G) → Hom(B,G) depend only on
H and G. Moreover, ker(�i) ∼ = Hom(H,G), while coker(�i) gives rise to a new contravariant
functor
Ext(−,G): Ab −→ Ab,
called extension product.
Proof. The method used in this proof is analogous to the one for proving
Proposition (II.5.4), that is to say, merely replacing the functor −⊗G with the
functor Hom(−,G) and keeping in mind the contravariance of the latter. We leave
the details to the reader. Nevertheless, we note that Ext(−,G) does not depend on
the free presentation of H. �
Let us compute Ext(Zn,G) for any Abelian group G. SinceExt(−,G) does not
depend on any particular free presentation of Zn, we choose the presentation
Z ��
n ��
Z
��
��
Z/n
where n is the multiplication by n. Then, we have the exact sequence
Hom(Zn,G)
��
Hom(Z,G)
�n ��
Hom(Z,G)
��
Ext(Zn,G)
and since Hom(Z,G) ∼ = G, we conclude that Ext(Zn,G) ∼ = G/nG. In particular,
Ext(Zn,Z) ∼ = Zn.
Due to Lemma (IV.1.4),ifH is free, Ext(H,G)=0.
��
0