For any set $S$ one can consider the free abelian group $\mathbb{Z}[S]$ generated by this set. Now suppose, there is a topology on $S$ given. Is it possible to find a topology on $\mathbb{Z}[S]$ in such a way, that:
i) The map $S\rightarrow \mathbb{Z}[S]$ is a homeomorphism onto the image.
ii) The addition and the inverse map are continuous
And if it is possible, is this topology unique?
I don't know if such a topology is unique, but it exists if and only if $S$ is completely regular. This includes locally compact hausdorff spaces and CW complexes.
With Freyd's Adjoint Functor Theorem, it can be shown that the forgetful functor from abelian top. groups to top. spaces has a left adjoint. This is essentially the same proof as in the discrete case. Explicitely, $\mathbb{Z}[S]$, the free abelian top. group over the top. space $S$, is the usual free abelian group endowed with the weak topology for all homomorphisms $\mathbb{Z}[S] \to A$, such that $S \to \mathbb{Z}[S] \to A$ is continuous. Here, $A$ is an arbitrary abelian top. group. In order to show that this topology exists, we may assume that $A$ is, as a group, a quotient of $\mathbb{Z}[S]$, so that these $A$ form a set. But the description of the topology does not change and even without Freyd's Theorem it is easy to see that $\mathbb{Z}[S]$ thus becomes an abelian top. group satisfying the desired universal property.
Now I claim that the three assertions
$S \to \mathbb{Z}[S]$ is a homeomorphism onto its image.
$S$ is a subspace of an abelian top. group.
$S$ is completely regular.
are actually equivalent!
1) implies 2), that's clear. Now assume 2), thus $S \subseteq A$ for some top. abelian group. Extend the inclusion $S \to A$ to a continuous homomorphism $\mathbb{Z}[S] \to A$. Every open subset of $S$ can be extended to an open subset of $A$. Pull it back to $\mathbb{Z}[S]$. This is an open subset of $\mathbb{Z}[S]$ which restricts to the given oben subset of $S$. This proves 1).
2) implies 3), this follows from the fact that every topological group is completely regular and subspaces of completely regular spaces are obviously completely regular.
Finally assume 3), i.e. $S$ carries the initial topology with respect to all continuous functions $S \to \mathbb{R}$. Endow $\mathbb{R}^{(S)}$ with the initial topology with respect to all homomorphisms $\mathbb{R}^{(S)} \to \mathbb{R}$, such that the restriction $S \to \mathbb{R}$ is continuous. Then $\mathbb{R}^{(S)}$ is an abelian topological group and $S \to \mathbb{R}^{(S)}$ is an embedding, thus 2).
I also believe that (but cannot prove)
In another comment, it was suggested to endow $\mathbb{Z}[S]$ with the final topology with respect to $S \to \mathbb{Z}[S]$. But this does not even yield a translation invariant topology: If $S=\{a,b\}$ with the only nontrivial open subset $\{a\}$, then $\{a\}$ is open in $\mathbb{Z}[S]$, but $\{b\}$ is not.
But maybe, if $S$ is a completely regular space, the topology of $\mathbb{Z}[S]$ used above is the final topology?
Free topologican groups, abelian or not, have been studied since the 40s by Markov and Graev (the original references are somewhat obscure). You need your space $S$ to be completely regular for the construction to work. There is a survey by O. V. Sipacheva (THE TOPOLOGY OF FREE TOPOLOGICAL GROUPS, Journal of Mathematical Sciences, Vol. 131, No. 4, 2005) which is maybe too thorough if you just want to get a feeling of this concept, but the first two or three pages contain interesting and easy-to-grasp information on this class of topological groups. The construction itself is not that difficult, and the topology is indeed unique, but they are sometimes difficult to delve into, especially the non-Abelian ones, and a very valuable source of counterexamples.
[edit] I have read the question more carefully and it seems to amount to the following: For a (well behaved) topological space $S$, can one endow the free group $\mathbb Z[S]$ with a group topology such that $S$ becomes a topological subspace of this group? If so, is this construction unique? The answer to the first question is yes, as I have just said. Moreover, $S$ becomes a closed subspace of $\mathbb Z[S].$ I'm not sure about the uniqueness, though. The universal property which characterizes the free topological abelian group is, as long as I remember, the following: it is the only group topology on the free abelian group for which the inclusion mapping $S\to \mathbb Z[S]$ becomes a topological embedding and such that for every continuous mapping $f:S\to G,$ where $G$ is a topological Hausdorff abelian group, the unique homomorphism which extends $f$ to $\mathbb Z[S]$ becomes continuous. This is at first sight a stronger property than the one contained in your question.
There is a unique topology on $\mathbb{Z}[S]$ for which the map $S\rightarrow\mathbb{Z}[S]$ is continuous such that $\mathbb{Z}[S]$ has the above-mentioned universal property. It is the final topology with respect to the injective map $S\rightarrow\mathbb{Z}[S]$. Any map, continuous or not, $S\rightarrow A$, $A$ an abelian group induces a unique homomorphism of groups $\mathbb{Z}[S]\rightarrow A$ such that $S\rightarrow\mathbb{Z}[S]\rightarrow A=S\rightarrow A$. In particular, if $A$ is an abelian topological group, and $S\rightarrow A$ continuous, then this implies that $\mathbb{Z}[S]\rightarrow A$ is continuous. As I noted in my comment, however, it is not at all clear to me that the group operations on $\mathbb{Z}[S]$ will be continuous with this topology.
The answer is yes. This follows from the fact that the group ${\mathbb Z}[S]$ should satisfy a universal property:
Let $A$ be any topological Abelian group and let $f: S \to A$ be a continuous function. Then $f$ can be extended uniquely to a continuous homomorphism ${\mathbb Z}[S] \to A.$
Suppose $G$ and $H$ are two copies of ${\mathbb Z}[S]$ with possibly different topologies. $G$ and $H$ both contain copies of $S.$ Applying the above unviersal property in two directions allows us to see that the identity from $G$ to $H$ is a homeomorphism. More precisely, we get a map $G \to H$ and another $H \to G.$ Composing them one way gives us a map $G \to G$ which must be the identity by the universal property. Composing them the other way gives as a map $H \to H$ which again must be the identity. So the two maps are inverses and each is a homeomorphism.
[Edit: adding the existence portion of the argument]
Consider $(S \cup S^{-1} \cup \{1 \})^{\mathbb Z},$ where $S^{-1}$ is just another ``copy" of $S.$ An element $s \in S$ shall correspond to $s^{-1}$ in $S^{-1}.$ This is a topological space, though I can't remember what topology to put on it. Let $X$ be the subset consisting of all elements with only finitely many coordinates different from $1.$
Now we quotient $X$ in a technically complicated but conceptually simple way. If $x \in X$ has $x_i = s$ and $x_{i+1} = s^{-1}$ for some $s \in S.$ Then we identify $x$ with $y$ where $y=x$ except at places $i,$ and $i+1$ where $y_i = 1 = y_{i+1}.$ We do the same if the $s,s^{-1}$ appear in the other order. Also, if $\sigma$ is a permutation of ${\mathbb Z}$ then we identify $x$ with the $y$ where $y_i = x_{\sigma(i)}.$
To define a group operation, we take two infinite "words", slide them so that they have disjoint "support" and concatenate. (Writing down the details here is harder than letting the reader guess what I mean.)
