Simplicial Structures in Topology

III.4 Relative Homology 127
such that � ℓq = g. To prove that � ℓℓ and ℓ � ℓ are homotopic to their respective identity
functions, we construct the following homotopies:
1. (∀[x,t] ∈ CA) , H1([x,t],s)=[x,ts],
2. (∀x ∈ X) , H1(x,s)=G(x,1 − s)
and
H1 : (X ∪CA) × I −→ X ∪CA
H2 : X/A × I −→ X/A
H2(q(x),t)=ℓG(x,t) , x ∈ X � A .
We ask the reader to verify that these homotopies are well defined and to complete
the proof. �
We now turn to Theorem (III.4.1). It is sufficient to notice the following facts:
(a) |L| is a closed subspace of |K|; (b) the pair (|K|,|L|) has the Homotopy Extension
Property (see Theorem (III.1.7)), (c) |K ∪CL| ∼ = |K|∪|CL| is a pushout space;
finally, we apply Theorem (III.4.2).
Theorem (II.4.9) in Sect. II.4 has a corresponding version in the category of polyhedra:
let {|Ki||i = 1,...,p} be a finite set of based polyhedra, each with base point
given by a vertex x i 0 ∈ Ki; we then take the wedge product
∨ p
i=1 |Ki| := ∪ n i=1 ({x1 p
0 }×...×|Ki|×...×{x0 }) .
(III.4.3) Theorem. For every q ≥ 1,
We consider the topological space
Hq(∨ p
i=1 |Ki|,Z) ∼ = ⊕ p
i=1 Hq(|Ki|,Z) .
X = S 2 ∨ (S 1 1 ∨ S 1 2)
which is the wedge product of a two-dimensional sphere and two circles. The space
X is clearly triangulable and therefore, by Theorem (III.4.3), its homology is as
follows:
Hq(X;Z) ∼ ⎧
⎪⎨
Z q = 0
Z ⊕ Z q = 1
=
⎪⎩
Z q = 2
0 q �= 0,1,2 .
Therefore, X and the torus T 2 have the same homology groups. But these spaces
are not homeomorphic. To prove that X and T 2 are not homeomorphic, we recall
Remark (I.1.17). We suppose f : X → T 2 to be a homeomorphism and let x0 be the
identification point of spheres S 2 , S 1 1 ,andS1 2 ; then, X � {x0} and T 2 � { f (x0)} are
homeomorphic; but X � {x0} has three connected components while T 2 � { f (x0)}
is connected, which leads to a contradiction.