I'm pretty confused about the precise relation of the integral and the real cohomology of the classifying space $BG$ of a compact Lie group $G$. The natural map $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ certainly kills all torsion, but can it have non-torsion elements in the kernel? For instance, if $G=SU\_n$, then there is no torsion in $H^n(BG;\mathbb{Z})$ (see [Hatcher, Algebraic Topology, Theorem 4D.4]), so this map would be an injection for all $n$.

All expositions that I know (for instance [E. Brown, The cohomology of $BSO\_n$ and $BO\_n$ with integer coefficients]) somehow suggest that $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective, but this is certainly not the case for arbitrary spaces $X$ with torsion free $H^n(X;\mathbb{Z})$. The Corresponding statement for homology is true, since the singular chain complex is free, and so the universal coefficient theorem yiels $$ H\_n(X;\mathbb{Z})\otimes \mathbb{R} \xrightarrow{\cong} H\_n(X;\mathbb{R}). $$ However, this argument does not carry over to cohomology (or at least I don't see it).

What makes me suspicious is that if $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective for all $n$, then the long exact sequence induced from $\mathbb{Z}\to \mathbb{R} \to S^1$ would split, a thing I don't feel comfortable with.