The question Topologizing free abelian groups on Mathoverflow asks about the existence and uniqueness of a topology for the free abelian group on $X$ such that the natural inclusion is a homeomorphism onto its image and addition and inverse map are continuous.
There are four different answers and some interesting discussions along the comments. The final answer is affirmative if and only if the topological space is completely regular. In the Abelian case (as in others) it can be regarded as $\mathbb Z[X]$ endowed with the weakest topology with respect to which all homomorphic extensions of continuous maps from $X$ to topological groups are continuous (Sipacheva, Free Boolean Topological Groups).
The topology on these spaces is important because their homotopy groups compute the homology with integer coefficients of the topological spaces they're the span of (Dold-Thom theorem).
In this sense, McCord provided a systematic construction of abelian groups whose homotopy groups computed the homology of the space in question with coefficients in ''good'' groups (I think there are some technical conditions). For a (pointed) space $(X,x_0)$ and a topological abelian group $G$, $B(G,X)$ was the space of functions $f:X\to G$ with finite support with the extra convention that $f(x_0)=e$ for all $f$.
For each $n\geq 0$ $B_n(G,X)$ was the subspace of $B(G,X)$ consisting of functions whose support had at most cardinality $n$, with the quotient topology with respect to the map
$$ q_n:(X\times G)^n \to B_n(G,X); $$
$q((g_1,x_1),\dots,(g_n,x_n))$ is the function $f:X\to G$ such that $f(x_i)=g_i$ and $f(x)=e$ for $x\in X\setminus \{x_1,\dots,x_n\}$ (with the convention that $q(g,x_0)=e$ for all $g\in G$. Then
$$ B(G,X)= \bigcup_n B_n(G,X). $$
For $G=\mathbb Z$ with the discrete topology, $B(\mathbb Z,X)$ is the group completion of the topological monoid $SP^\infty X$, the infinite symmetric product of $X$.
Is $B(\mathbb Z,X)$ isomorphic to $\mathbb Z[X]$ as topological abelian groups?
In the paper (Sipacheva,Free Boolean Topological Groups), the author defines free Boolean topological groups are defined similarly: the abstract free Boolean group generated by the set $X$ endowed with the weakest topology with respect to which all homomorphic extensions of continuous maps from $X$ to topological Boolean groups are continuous. Is this equivalent to $B(\mathbb Z_2,X)$?
$\bullet$ M. C. McCord, Classifying spaces and infinite symmetric products.
EDIT
As @Tyrone points out in a comment, McCord considers compactly generated spaces. In the accepted answer to the original question, the space $X$ must be normal for the free abelian topological group $\mathbb Z[X]$ to exist. You can assume that all the spaces here are locally finite CW-complexes.