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Is a homomorphism out of a free abelian group determined by its value at the basis elements?

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Yes, as any element in the free group has unique expression up to order as a $\,\Bbb Z-$combination of those free basis elements. – DonAntonio Feb 19 '13 at 3:54

Yes, this is what free abelian groups are designed to do. In more details, if $A$ is a free abelian group on the set $S$ then, for all abelian groups $A'$ there is a natural bijection between the set of functions $f:S\to A'$ and the set of group homomorphisms $\psi: A\to A'$.

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This is a dumb question, but for instance if $f$ is such a group homomorphism, then is $f(ne_i) = nf(e_i)$ or is it $f(n)f(e_i)$? – user62917 Feb 19 '13 at 3:59
The former. $f$ is not even defined at $n$, since $n$ is an integer, and not (in general) an element of the group. – MJD Feb 19 '13 at 4:00

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