Simplicial Structures in Topology
Simplicial Structures in Topology
Simplicial Structures in Topology
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V.3 Po<strong>in</strong>caré Duality 191<br />
(V.3.3) Theorem. For every p-simplex σ ∈ Φ, the block eσ is a (n − p)-block of<br />
K (1) ; besides, the set def<strong>in</strong>ed by<br />
is a block triangulation of K (1) .<br />
Proof. We first have to prove that<br />
e(K (1) )={eσ |σ ∈ Φ}<br />
Hr(eσ , • eσ ;Z) ∼ =<br />
� Z if r = n − p<br />
0 ifr �= n − p .<br />
As we have already seen, if σ is an n-simplex, eσ is a simplicial subcomplex of<br />
K (1) hav<strong>in</strong>g only the vertex b(σ) and no other simplex; so, • eσ = /0 and the statement<br />
above is true. We then assume that dimσ ≤ n − 1. S<strong>in</strong>ce |K (1) | ∼ = |K| and<br />
dimK (1) = n, the relative homology of the simplicial subcomplex S(b(σ)) of K (1)<br />
(see the def<strong>in</strong>ition given <strong>in</strong> Theorem (II.2.10))is<br />
(see Lemma (V.1.4)).<br />
We now prove that<br />
•<br />
�Hr(S(b(σ));Z) ∼ =<br />
� Z if r = n − 1<br />
0 if r �= n − 1<br />
S(b(σ)) = | (σ) (1) ∗ • eσ |<br />
where (σ) (1) is the barycentric subdivision of the boundary of σ. In fact,<br />
|{σ 1 ,...,σ r }| ∈ S(b(σ)) ⇐⇒ {σ 1 ,...,σ r }∈Φ (1) with σ ⊂ σ 1 ⊂ ...⊂ σ r<br />
⇐⇒ (∃t) σ t ⊂ σ ⊂ σ t+1 with {σ 1 ,...,σ t }∈<br />
•<br />
On the other hand, | (σ) (1) | ∼ = Sp−1 and consequently<br />
S(b(σ)) ∼ = S p−1 ∗| • eσ| .<br />
•<br />
•<br />
(σ) (1) and {σ t+1 ,...,σ r }∈ •<br />
Bσ .<br />
Let us now go back a step to note that if L1,...,Lp are p simplicial complexes, each<br />
of them hav<strong>in</strong>g only two 0-simplexes (and no 1-simplex), then<br />
|L1 ∗ ...∗ Lp| ∼ = S p−1<br />
and we conclude that S(b(σ)) is homeomorphic to the pth suspension of | • eσ | (the<br />
suspension of an abstract simplicial complex K is def<strong>in</strong>ed on p. 47 – it is easily seen<br />
that the suspension of the sphere S n is homeomorphic to the sphere S n+1 ;thepth<br />
suspension of K is def<strong>in</strong>ed by iteration). By Exercise 5 of Sect. II.4, we conclude<br />
that<br />
�Hq(S(b(σ));Z) ∼ = �Hq−p( • eσ ;Z)