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Let $f\colon \prod \mathbb{Z} \rightarrow \mathbb{Z}$ be a $\mathbb{Z}$-module homomorphism. I want to show that $f(e_i) = 0$ for almost all $i$ where $e_i$ are the standard unit vectors. I assume $f(e_i) \neq 0$ for infinitely many $i$. We can assume without restriction that $f(e_i) > 0$ for all $i \geq 1$. Now is the following a valid argument and if not why? It is $f(\sum e_i) = \sum (f(e_i)) = \infty$, a contradiction. Or do I have to use a trickier argument?

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    an homomorphism only commutes with finite sums. So $f(\sum e_i)$ exists, but it isn't related to $\sum f(e_i)$, and it doesn't imply that this sum should converge. I remember that this problem is quite tricky.2011-11-25
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    You should specify what your product is being taken over.2011-11-25

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