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

## 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.

## 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, $\varphi(u) = 0$.