Simplicial Structures in Topology

VI.4 Obstruction Theory 233
c n+1
f : Cn+1(K;Z) −→ πn(W)
that takes each ∑i miσ i ∈ Cn+1(K;Z) into ∑i mi[ f n
σ i]. We notice that if σ ∈ Ψ, then
by Lemma (VI.3.4), [ f n σ ]=0: in fact, f is defined in |L| and f n can be therefore
extended to |σn+1|. This allows us to conclude that c n+1
f ∈ Cn+1 (K,L;πn(W)), that
is to say, c n+1
f is a cochain.
The next lemma is useful for proving that c n+1
f is a cocycle.
(VI.4.1) Lemma. Let W be an n-simple space and S n+1 be the sphere viewed as an
(n + 1)-manifold, which is triangulated by a simplicial complex K =(X,Φ), asin
Theorem (V.1.5);letσ i , with i = 1,2,...,s, be the (n + 1)-simplexes of K; finally, let
K n =(X,Φ n ) be the simplicial n-dimensional subcomplex of K, where
Then, for any map f : |K n |→W,
Φ n = {σ ∈ Φ | dimσ ≤ n}.
s
∑
i=1
[ f i σ ]=0.
Proof. By Theorem (V.1.5), everyn-simplex of K is a face of exactly two (n + 1)simplexes;
besides, by Definition (V.3.1), each n-simplex inherits opposite orientations
from its two adjacent (n + 1)-simplexes. The spaces | •
σ i | are homeomorphic
to the sphere Sn whose elements may be considered as t ∧x, with x ∈ Sn−1 ;thisway
of viewing the elements of Sn gives us an idea of the orientation of the sphere; in
other words, if we take the elements in the format t ∧ x, we travel the sphere with a
“positive” orientation but, if we take them in the format (1−t)∧x,wetravelSnwith the opposite orientation, that is to say, we give Sn a “negative” orientation. With this
in mind, we observe that the function f is applied twice on each | •
σ i |: once, viewed
as the function f (t ∧ x) and once, as the function f ((1 − t) ∧ x); on the other hand,
the product by νn
× of the homotopy classes of these functions is the trivial class of
πn(W ) (see Theorem (VI.3.3)).
We note that the base point of each homotopy class is irrelevant because W is
n-simple. �
(VI.4.2) Theorem. The cochain c n+1
f
is a cocycle.
Proof. It is necessary to prove that, for every (n + 2)-simplex σn+2 of K n ∪ L,
d n+1 (c n+1
f )(σn+2)=c n+1
f (dn+2(σn+2)) = 0.