I've been struggling with the following exerciese:
Let $f\in\mathrm{Hom}_{\mathbb{Z}}(\prod_{i\geq 0}\mathbb{Z}, > \mathbb{Z})$, where $\prod_{i\geq 0}\mathbb{Z}$ denotes the $\mathbb{Z}$-module of integral sequences. For $n\geq 0$, set $\mathbb{e}_{n} := (0, \ldots, 0, \underbrace{1}_{n^{\mathrm{th}}\mathrm{-entry}}, 0, \ldots)$. Show that $f(\mathbb{e}_{n}) \neq 0$ only for finitely many $n\in\mathbb{N}$, and that if $f(\mathbb{e}_{n}) = 0$ for all $n\in\mathbb{N}$, then $f = 0$.
That those statements are true becomes obvious in light of the fact that $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i\geq 0}\mathbb{Z}, \mathbb{Z})$ is a free module. However, I'm looking for an easy and direct proof, which would work without showing freeness of $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i\geq 0}\mathbb{Z}, \mathbb{Z})$.