To elaborate on the question from the title, $\mathbb{R}\_\delta$ is the additive group of real numbers (without any topology) and $K(\mathbb{R}\_\delta,n)$ is an Eilenberg-MacLane space. I would like to know what its integral (co-)homology is, but I'm also happy about any results in this direction.


Bill Thurston has stated in a comment ([Nontrivial finite group with trivial group homologies?](https://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies#comment130307_52552)) that the homology of $B\mathbb{R}\_\delta = K(\mathbb{R}\_\delta,1)$ has rank $2^\omega$ in each degree, but I don't know why this holds and I would also like to have results about $n>1$.


The background behind this question is that I'd like to show that $K(\mathbb{R}\_\delta,n)$ cannot be a retract of a (smooth) manifold, and I'm hoping that (co-)homology will help me to figure this out.



What is the (co-)homology of $K(\mathbb{R}_\delta,n)$?
Given any $\mathbb Q$-vector space $A$ (in your case, $A=\mathbb R\_\delta$), the integral homology of $K(A,n)$, which is the same as its rational homology, is given by the cofree graded-cocommutative coalgebra cogenerated by $A$ in degree $n$.
In particular, for $n$ an even number, we have
$$
H\_m(K(A,n),\mathbb Z)=H\_m(K(A,n),\mathbb Q)= \begin{cases}Sym^k A &\text{if $m=kn$}
\\0 &\text{otherwise}\end {cases}
$$


The rational cohomology of $K(A,n)$ is obtained by the universal coefficient theorem (UCT):
$$
H^m(K(A,n),\mathbb Q)=\hom(H\_m(K(A,n),\mathbb Q),\mathbb Q).
$$
The integral cohomology is more messy to describe: it involves a non-trivial $ext$-term in the UCT short exact sequence (see the wikipedia entry on the [UCT](https://en.wikipedia.org/wiki/Universal_coefficient_theorem#Universal_coefficient_theorem_for_cohomology) for more details).


