Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero

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Statement

Denote by G the Baer-Specker group, the external direct product of countably many copies of the group of integers. Denote by H the subgroup that is the restricted direct product of the direct factors.

Suppose \varphi: G \to \mathbb{Z} is a homomorphism of groups such that the restriction \varphi|_H is the trivial map. Then, \varphi is also the trivial map.

Related facts

Applications

Proof

Given: G is the Baer-Specker group, H is the restricted direct product of the direct factors. \varphi: G \to \mathbb{Z} is a homomorphism such that the restriction \varphi|_H is the trivial map.

To prove: \varphi(u) = 0 for all u \in G.

Proof:

Step no. Assertion/construction Facts used Given data Previous steps used Construction
1 Suppose p is a prime number. Then, for any element of G of the form g = (a_1p, a_2p^2, \dots, a_np^n, \dots), with a_i \in \mathbb{Z}, \varphi(g) = 0 \varphi is a homomorphism and \varphi|_H = 0 [SHOW MORE]
2 Pick two distinct primes p, q. Every element of G can be expressed as the sum of elements g = (a_1p, a_2p^2, \dots, a_np^n, \dots) and h = (b_1q, b_2q^2, \dots, b_nq^n, \dots) where all the a_i and b_i values are integers. This follows from the Chinese remainder theorem applied in each coordinate.
3 \varphi(u) = 0 for any u \in G \varphi is a homomorphism Steps (1), (2) Express u = g + h as per Step (2), then use Step (1) to show that \varphi(g) = \varphi(h) = 0. Thus, \varphi(u) = 0.