Simplicial Structures in Topology
Simplicial Structures in Topology
Simplicial Structures in Topology
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164 IV Cohomology<br />
S<strong>in</strong>ce g and π r are simplicial functions, H ∗ (g;Z) and H ∗ (π r ;Z) are r<strong>in</strong>g homomorphisms;<br />
however, s<strong>in</strong>ce H ∗ (π r ;Z) is an isomorphism, also H ∗ (π r ;Z) −1 preserves<br />
cup products. We conclude that if two polyhedra |K| and |L| are of the same<br />
homotopy type, then their cohomology r<strong>in</strong>gs are isomorphic (as r<strong>in</strong>gs).<br />
We now wish to prove that the equality of the (co)homology groups is a necessary<br />
but not sufficient condition for the cohomology r<strong>in</strong>gs to be isomorphic. To this end,<br />
let us return to the cohomology of the bi-dimensional torus T 2 and of the space<br />
X = S 2 ∨ (S 1 ∨ S 1 ). We have already proved (see Sect. III.4) that these two triangulable<br />
spaces have the same homology and, by the Universal Coefficient Theorem<br />
<strong>in</strong> Cohomology, also the same cohomology; moreover, we have already seen (see<br />
Sect. III.4 aga<strong>in</strong>) that T 2 and X are not homeomorphic. We now compute the cohomology<br />
of T 2 and X once more, by explicitly consider<strong>in</strong>g the r<strong>in</strong>g structure.<br />
We beg<strong>in</strong> with T 2 . Let us consider the oriented triangulation of T 2 used to<br />
compute H∗(T 2 ;Z) (see Fig. II.10 <strong>in</strong> Sect. II.2, p.63). We recall that if z1 1 and<br />
z2 1 are the two 1-cycles<br />
then<br />
z 1 1 = {0,3} + {3,4} + {4,0} and z 2 1 = {0,1} + {1,2} + {2,0},<br />
w 1 1<br />
= {1,6} + {6,5} + {5,8} + {8,7} + {7,2} + {2,1}<br />
is a 1-cycle homologous to z 1 1 (see how H∗(T 2 ,Z) was computed) and<br />
w 2 1 = {3,7} + {7,5} + {5,8} + {8,6} + {6,4} + {4,3}<br />
is a 1-cycle homologous to z 2 1 . For each 1-simplex {i, j} of T 2 ,let<br />
be the function such that<br />
{i, j} ∗ ∈ Hom(C1(T 2 ),Z)<br />
(∀{k,ℓ}∈T 2 ) {i, j} ∗ ({k,ℓ})=<br />
We now perform some calculations:<br />
�<br />
1 if {k,ℓ} = {i, j}<br />
0 if {k,ℓ} �= {i, j}.<br />
1. ∂ 1 {1,6} ∗ ({1,6,5})=({1,6} ∗ ∂2)({1,6,5})={1,6} ∗ (∂2{1,6,5})=1<br />
2. ∂ 1 {1,6} ∗ ({1,2,6})=−1<br />
we conclude that ∂ 1 {1,6} ∗ = {1,6,5} ∗ −{1,2,6} ∗ . Similarly, we f<strong>in</strong>d that<br />
∂ 1 {6,5} ∗ = {1,6,5} ∗ −{5,6,8} ∗ , ∂ 1 {5,8} ∗ = {5,8,7} ∗ −{5,6,8} ∗ ,<br />
∂ 1 {8,7} ∗ = {5,8,7} ∗ −{7,8,2} ∗ , ∂ 1 {7,2} ∗ = {7,2,1} ∗ −{7,8,2} ∗ ,<br />
∂ 1 {2,1} ∗ = {7,2,1} ∗ −{1,2,6} ∗ ,