 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/21f/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Q

give

H_{i} (X; Q) ≅ H_{i} (X; Z) \otimes_{Z} Q .

Thus, if

H_{i} (X; Z) ≅ Z^{b_{i}} \oplus T_{i}

with

T_{i}

torsion, then

H_{i} (X; Q) ≅ Q^{b_{i}}

For a finite triangulable space, its Euler characteristic is

χ (X) = \sum_{i} (- 1)^{i} dim_{Q} H_{i} (X; Q) .

K

is a finite triangulation and

f_{i}

is its number of

i

-simplices, then the Euler-Poincare formula is

χ (X) = \sum_{i} (- 1)^{i} f_{i} .

To prove it over

Q

, write

Z_{i} = ker \partial_{i}

and

B_{i} = im \partial_{i + 1}

. Rank-nullity and

H_{i} = Z_{i} / B_{i}

give

dim C_{i} = dim Z_{i} + dim B_{i - 1} = dim H_{i} + dim B_{i} + dim B_{i - 1} .

After multiplying by

(- 1)^{i}

and summing, the two boundary sums cancel, leaving the claimed equality.

Δ^{3}

is contractible, so

H_{i} (Δ^{3}) ≅ {Z, 0, i = 0, i > 0.

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