What is the classifying space, $K(G,1)$ for $\mathbb Z[1/2]$? I'm wondering about the classifying space for the diadic fractions $\mathbb Z[1/2]$? I have no idea how to begin answering the question, so my apologies for showing a lack of effort.
More generally, given an abelian group $G$, one can view it as a $\mathbb Z$-module, and so we can form its localization at some element. Is there any way to relate $K(G,1)$ with $K(S^{-1}G,1)$?
 A: The Baumslag-Solitar group $G=\mathbf{Z}[1/2]\rtimes\mathbf{Z}$ acts properly (and freely) on a contractible space, namely the product $P$ of the hyperbolic plane and of a trivalent tree $T$, which can be chosen to be either the Bruhat-Tits tree of $\mathrm{SL}_2(\mathbf{Q}_2)$, or the Bass-Serre tree when $\mathbf{Z}[1/2]\rtimes\mathbf{Z}$ is view as an ascending HNN-extension of $\mathbf{Z}$. Taking the quotient by the $\mathbf{Z}[1/2]$-action, on gets a classifying space for the latter (which in addition is locally compact).
[It's unnecessary here, but one can notice that there is a $G$-invariant subset $Q$ in $P$, of the form $\{(x,y):b(x)+b'(y)=0\}$ for suitable Busemann functions $b,b'$ on the hyperbolic plane and on the tree, and $Q$ itself is homeomorphic to $T\times\mathbf{R}$; then one can take the quotient of $Q$ by the $\mathbf{Z}[1/2]$-action to get a smaller classifying space (closed in the previous one).]
A: You can construct such a classifying space as a mapping telescope.  Let $f:S^1\to S^1$ be a degree $2$ map and let $T$ be the telescope of the sequence $S^1\to S^1\to S^1\to\dots$ where the maps are all $f$.  The homotopy groups of $T$ will then be the colimits of the induced sequence of maps on the homotopy groups of $S^1$.  This means $\pi_n(T)=0$ for $n\neq 1$, and $\pi_1(T)$ is exactly the localization $\mathbb{Z}[1/2]$, so $T$ is a $K(\mathbb{Z}[1/2],1)$.
More generally, if $G$ is an abelian group and $X$ is a $K(G,1)$ space, for each $n\in\mathbb{Z}$ there is a map $n:X\to X$ which induces multiplication by $n$ on $\pi_1$.  The mapping telescope of iterating this map is then a $K(G[1/n],1)$.  If you want to invert an entire multiplicatively closed set $S$, you can similarly take a mapping telescope of maps $n:X\to X$ for different values of $n\in\mathbb{Z}$, as long as the set of primes that are factors of infinitely many of the $n$ is the same as the set of primes that divide elements of $S$.
A: This is analogous to Eric Wofsey's answer, but approaches the problem by homology first.
One can also solve this by showing first that the mapping telescope $\mathcal{S}$ given by
$$S^1 {\to} S^1 \to S^1\to\dots$$
with all maps degree $p$ for a fixed prime is the Moore space for $\mathbb Z[\frac{1}{p}]$, and since this is abelian, the Hurewicz morphism is in fact an isomorphism on to  $\pi_1(\mathcal{S})$.
To see that it is a Moore Space, we assume that the maps are cellular, giving the mapping telescope the structure of unions of CW-complexes (and hence one itself.)
With this, we have that the space satisfies the conditions of the following theorem:
Theorem (Hatcher 3.33): If $X$ is the union of a directed et of subspaces $X_{\alpha}$ with the property that each compact set in $X$ is contained in $X_{\alpha}$ then the natural map $\varinjlim H_i(S_\alpha) \to H_i(\mathcal{S})$ is an isomorphism.
This is not difficult to prove, since compactness forces each cycle in $\mathcal{S}$ to be contained in some $X_\alpha$, making the assignment injective, and surjectivity is similar.
Observing that the direct limit of induced maps $H_1(S) \to H_1(S)$ is precisely $\mathbb Z[\frac{1}{p}]$, we obtain the result.
