Simplicial Structures in Topology

68 II Simplicial Complexes
2. λn is well defined. We must verify that λn is independent from both the choice
of the cycle z representing the homology class and the chain ˜z mapped into z.
Let z ′ ∈ C ′′ be a cycle such that [z]=[z ′ ],andlet˜z ′ ∈ C ′ n be such that gn(˜z ′ )=z ′ ;
moreover, take a cycle c ′ in C ′ n−1 satisfying the property fn−1(c ′ )=∂ ′ n (˜z′ ).The
definition of homology classes implies that there exists a chain b ∈ C ′′
n+1 such
that ∂ ′′
n+1 (b) =z − z′ . Since gn+1 is an epimorphism, we can find a ˜b ∈ C ′ n+1
such that gn+1(˜b)=b. Hence,
gn(˜z − ˜z ′ − ∂ ′ n+1 (˜b)) = z − z ′ − ∂ ′′
n+1 (b)=0
and thus, there exists a ∈ Cn such that
At this point, we have that
fn(a)=˜z − ˜z ′ − ∂ ′ n+1 (˜b).
fn−1(∂n(a)) = ∂ ′ n(˜z − ˜z ′ − ∂ ′ n+1(˜b)) = fn−1(c − c ′ )
and because fn−1 is injective, we conclude that c − c ′ = ∂n(a). Therefore, c and
c ′ represent the same homology class in Hn(C).
(II.3.2) Remark. The previous items 1. and 2. are typical examples of the
so-called “diagram chasing” technique. We suggest the reader to draw the
diagrams indicating the maps without their indices which, although necessary
for precision, are sometimes difficult to read; all this will help in following up
the arguments.
3. The sequence is exact. To prove the exactness of the sequence of homology
groups, we must show the following:
a. imHn( f )=kerHn(g);
b. imHn(g)=kerλn;
c. imλn = kerHn−1( f ).
We shall only prove (b), leaving the proof of the other cases to the reader. We
pick a class [z] ∈ Hn(C ′′ ) and compute λnHn(g)([z]) = λn[gn(z)]. Since we can
take any (!) element of C ′ n which is projected onto gn(z), we choose z itself;
given that ∂ ′′
n (z)=0, we conclude that λn[gn(z)] = 0, that is to say, imHn(g) ⊆
kerλn. Conversely,let[z] be a homology class of Hn(C ′′ ) such that λn[z] =0;
the definition of λn implies that there exist ˜z ∈ C ′ n and a cycle c ∈ Cn−1 such that
gn(˜z)=z and Cn−1( f )(c)=∂ ′ n(˜z).
Because λn[z]=0, there exists ˜c ∈ Cn such that c = ∂n( ˜c). Notice that
∂ ′ n (Cn( f )( ˜c) − ˜z)=0
and moreover, Hn(g)(Cn(g)( ˜c) − ˜z)=z; hence, kerλn ⊆ imHn(g).
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