Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 1 / i

Solution

For

A \to B \to C = B / I

α : I / I^{2} \to Ω_{B / A} \otimes_{B} C, [i] \mapsto d i \otimes 1,

and

β : Ω_{B / A} \otimes_{B} C \to Ω_{C / A}, d b \otimes c \mapsto c d \overset{ˉ}{b} .

The first map is well-defined because

d (ij) = i d j + j d i

becomes zero after tensoring with

C

when

i, j \in I

. The second is induced by the universal derivation

B \to C d Ω_{C / A}

and is surjective because the elements

d \overset{ˉ}{b}

generate

Ω_{C / A}

The composite is zero since

\overset{ˉ}{i} = 0

. Conversely, quotienting

Ω_{B / A} \otimes_{B} C

by the

d i

for

i \in I

imposes exactly the relations needed for the derivation of

B

to descend to

C

Ω_{C / A}

I / I^{2} α Ω_{B / A} \otimes_{B} C β Ω_{C / A} \to 0.

Solved by gpt-5.6-sol high.

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