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Roots descend through a surjective homomorphism
PDF p. 13: Divisibility of the Picard group
Statement
Let q: A → B be a surjective homomorphism of abelian groups and let n be a positive integer. If multiplication by n is surjective on A, then multiplication by n is surjective on B.
Facts & Assumptions
Given: q, A, B and n as stated.
Goal: Find an n-th additive root of every element of B.
Proof
For b in B, surjectivity of q gives a in A with q(a) = b. Surjectivity of multiplication by n on A gives x in A with nx = a.
Since q is a homomorphism, n q(x) = q(nx) = q(a) = b. Thus q(x) is the required root, and multiplication by n on B is surjective.