Simplicial Structures in Topology
Simplicial Structures in Topology
Simplicial Structures in Topology
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90 II <strong>Simplicial</strong> Complexes<br />
Hn(C;G)=ker(∂n ⊗ 1G)/im(∂n+1 ⊗ 1G);<br />
the graded Abelian group H∗(C;G) is the graded homology group of C with coefficients<br />
<strong>in</strong> G. In particular, if (C,∂ )=(C(K),∂), the cha<strong>in</strong> complex of the oriented<br />
complex K, thenH∗(C(K);G) – simply denoted by H∗(K;G) – is the homology of<br />
K with coefficients <strong>in</strong> G.<br />
We recall that the cha<strong>in</strong> complex (C(K),∂) is positive, free, and has an augmentation<br />
homomorphism ε : C0(K) → Z. To cont<strong>in</strong>ue with our work, we only need<br />
one of these properties, namely, that (C,∂ ) be free.<br />
For every free cha<strong>in</strong> complex (C,∂) and for each n ∈ Z, we have a short exact<br />
sequence of free Abelian groups<br />
Zn(C) ��<br />
��<br />
Cn<br />
∂n ��<br />
��<br />
Bn−1(C) .<br />
The ma<strong>in</strong> po<strong>in</strong>t is that, by tak<strong>in</strong>g the tensor product of each component of this exact<br />
sequence with G, we obta<strong>in</strong> aga<strong>in</strong> a short exact sequence.<br />
(II.5.1) Lemma. If<br />
A ��<br />
f<br />
��<br />
B<br />
is a short exact sequence of free Abelian groups and G is an Abelian group, then<br />
also the sequence<br />
is exact.<br />
g<br />
��<br />
��<br />
C<br />
A ⊗ G ��<br />
f ⊗ 1G g ⊗ 1G ��<br />
B ⊗ G ��<br />
��<br />
C ⊗ G<br />
Proof. We beg<strong>in</strong> by not<strong>in</strong>g that the group C is free; therefore, we may def<strong>in</strong>e a map<br />
s: C → B simply by choos<strong>in</strong>g for each element of a basis of C an element of its antiimage<br />
under g, and by extend<strong>in</strong>g this operation l<strong>in</strong>early; through this procedure, we<br />
obta<strong>in</strong> a homomorphism of Abelian groups<br />
s: C −→ B<br />
such that gs = 1C, the identity homomorphism of C onto itself. It follows that g(1B −<br />
sg)=g −(gs)g = 0, <strong>in</strong> other words, the image of 1B − sg is conta<strong>in</strong>ed <strong>in</strong> the kerg =<br />
im f ; for this reason, we may def<strong>in</strong>e the map r := f −1 (1B − sg): B → A that also<br />
satisfies rf = f −1 (1B − sg) f = 1A. We thus obta<strong>in</strong> the relations<br />
rf = 1A , gs = 1C , and fr+ sg = 1B.<br />
We know that the tensor product by G is a functor and that it transforms sums of homomorphisms<br />
<strong>in</strong>to sums of transformed homomorphisms; consequently, tensorization<br />
gives us the relations