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Given a set $S$, the free abelian group is defined as the set of formal sums of elements of $S$. Is this restricted to formal sums of a finite subset of elements of $S$? For example, is $S = \mathbb{R}$, then is $\sum_{z\in \mathbb{Z}}a_z z$ a legitimate element of the free abelian group on $\mathbb{R}$? What does the free abelian group on $\mathbb{R}$ even look like? What about the free abelian group on other infinite sets like $\mathbb{R}^2$, $\mathbb{Z}$, and the set of functions $f: \mathbb{R}\to\mathbb{R}$?

Anon
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  • Think of free meaning the object you invent with the fewest amounts of rules. So we want an abelian group, the elements of which are elements include $\mathbb{R}$, and there are certain elements we need to have, namely the finite sums $\sum_{j=1}^n k_j x_j$ with each $k_j\in \mathbb{Z}$. But we don't need to add the infinite sums, and therefore, these should not be elements of the free group. I don't know if there's a good way of seeing why this should be the correct perspective without using the adjunction with the forgetful functor. – WoolierThanThou Sep 10 '19 at 15:54
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    If $X$ is any set, infinite or not, the free abelian group on $X$ is (isomorphic to) the group of all functions $f:X\to \mathbb{Z}$ for which the set ${x\in X:f(x)\neq 0}$ is finite, with pointwise addition of functions. Each element $x$ of $X$ corresponds to the function taking $x$ to $1$ and all $y\neq x$ to $0$. – Luiz Cordeiro Sep 10 '19 at 15:55
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    Adding to @WoolierThanThou, I think the reason why we care about the free abelian groups is really because of the universal property they have (given by the adjunction), so it is actually the correct motivation. The description of it as "the set of finite formal sums[...]" is just a realization of the group. – Luiz Cordeiro Sep 10 '19 at 15:59

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