Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero
Suppose is a homomorphism of groups such that the restriction is the trivial map. Then, is also the trivial map.
Given: is the Baer-Specker group, is the restricted direct product of the direct factors. is a homomorphism such that the restriction is the trivial map.
To prove: for all .
|Step no.||Assertion/construction||Facts used||Given data||Previous steps used||Construction|
|1||Suppose is a prime number. Then, for any element of of the form , with ,||is a homomorphism and||[SHOW MORE]|
|2||Pick two distinct primes . Every element of can be expressed as the sum of elements and where all the and values are integers.||This follows from the Chinese remainder theorem applied in each coordinate.|
|3||for any||is a homomorphism||Steps (1), (2)||Express as per Step (2), then use Step (1) to show that . Thus, .|