20F (Algebraic Topology)

Triangulate

X

and adjoin one new vertex joined to every simplex; the resulting simplicial complex realizes the cone

CX

. The cone contracts to its apex, so

H_{0} (CX) ≅ Z, H_{i} (CX) = 0 (i > 0) .

b

Write

X^{'} = X \cup_{A} C A

. After replacing the two pieces by small open neighborhoods, their intersection deformation retracts to

A

H_{i} (C A) = 0

for

i > 0

and

H_{0} (C A) = Z

, becomes

\dots \to H_{i + 1} (X^{'}) \to H_{i} (A) \to H_{i} (X) \to H_{i} (X^{'}) \to H_{i - 1} (A) \to \dots

and at degree zero ends as

\dots \to H_{1} (X^{'}) \to H_{0} (A) \to Z \oplus H_{0} (X) \to H_{0} (X^{'}) \to 0,

which is the required sequence.

Attaching the cone collapses the circle

A = S^{1} \times {x}

in the torus. The quotient has the homotopy type

X^{'} ≃ S^{1} \lor S^{2} .

Equivalently, the exact sequence in part (b) kills one of the two generators of

H_{1} (T^{2})

and preserves the fundamental two-cycle. Therefore

H_{i} (X^{'}) ≅ {Z, 0, i = 0, 1, 2, i > 2.

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