Hom on sequences of integers is determined by values on finite sequences I have found this problem intended for school students (math olympiad) and couldn't solve it myself.
Consider the short exact sequence of abelian groups
$0 \to\bigoplus \limits_{i \in \mathbb{N}} \mathbb{Z}_i \to \prod\limits_{k \in \mathbb{N}} \mathbb{Z}_k \to X \to 0$
The inclusion is obvious (finite sequences of integers to all sequences).
The problem is to show that $Hom(X, \mathbb{Z}) = 0$. 
I thought it can be done by simple categorical cosiderations  but apparently it can't. We have 
$0 \to Hom(X, \mathbb{Z}) \to Hom( \prod\limits_{k \in \mathbb{N}} \mathbb{Z}_k, \mathbb{Z})  \to Hom(\bigoplus \limits_{i \in \mathbb{N}} \mathbb{Z}_i, \mathbb{Z}) \cong  \prod\limits_{i \in \mathbb{N}} \mathbb{Z}_i$
but the middle one doesn't give anything nice because $Hom(_,X)$ doesn't respect products. 
So I am out of ideas.

I know the proof now but I haven't solved it myself.
Note that for a prime $p$ a   sequence of the form $(p^na_n)$ is divisible in $X$ by $p^k$ for all $k \in \mathbb{N}$ because all but finitely many of  $p^na_n$ are. Therefore for any  $f \in Hom(X, \mathbb{Z}), f(p^na_n) $ is divisible by $p^k$ for all $k$ so $f(p^na_n) = 0$.
Now take $x_n \in X$, it can be represented as $(2^na_n + 3^nb_n)$ because gcd$(2^k, 3^k) = 1 $. Hence $f(x_n) = f(2^na_n) + f(3^nb_n) = 0$ as desired. 
