Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 4 / i

Solution

Write

F = K (Z /4, 1)

E = K (Z, 2)

, and

B = K (Z, 2)

. In total degree at most two, the

E_{2}

page of the mod-two Serre spectral sequence has

E_{2}^{0, 0} = E_{2}^{0, 1} = E_{2}^{0, 2} = E_{2}^{2, 0} = F_{2},

and

E_{2}^{1, 0} = E_{2}^{1, 1} = 0

. A periodic free resolution of the cyclic group gives

H_{1} (F; Z) = Z /4

and

H_{2} (F; Z) = 0

. Hence

H^{1} (F; F_{2}) ≅ Hom (Z /4, F_{2})

H^{2} (F; F_{2}) ≅ Ext (Z /4, F_{2})

; both are one-dimensional.

The edge map

H^{2} (B; F_{2}) \to H^{2} (E; F_{2})

is induced by multiplication by

4

and is therefore zero modulo two. Consequently

d_{2} : E_{2}^{0, 1} ⟶ E_{2}^{2, 0}

is an isomorphism. The differential out of

E_{2}^{0, 2}

is zero, because the total space has a one-dimensional

H^{2}

which must survive in filtration zero. Thus for every

r \geq 3

and

p + q \leq 2

E_{r}^{0, 0} = E_{r}^{0, 2} = F_{2}

and all other groups in that range vanish.

H^{0} (F; F_{2}) = H^{1} (F; F_{2}) = H^{2} (F; F_{2}) = F_{2} .

The surviving filtration-zero class is the restriction of the degree-two class of

E

, so

im H^{2} (f; F_{2}) = H^{2} (F; F_{2}) .

Solved by gpt-5.6-sol high.

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