Integral versus real (universal) characteristic classes

Question

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.

Answer 1

If $X$ is a space of finite type (meaning that the homology groups $H_i(X)$ are all finitely generated, a condition which applies in particular to $X=BG$ for $G$ a compact Lie group) then for each $n$ the map $H^n(X)\to H^n(X;\mathbb{R})$ is injective if and only if $H_{n-1}(X)$ is torsion free. Here and below integer coefficients are omitted from the notation.

To see this, note that the universal coefficient sequence $$ 0\to \operatorname{Ext}(H_{n-1}(X),A)\to H^n(X;A)\to \operatorname{Hom}(H_n(X),A)\to 0 $$ is natural with respect to homomorphisms $A\to A'$ of abelian coefficient groups. If $H_{n-1}(X)$ is torsion free, the ext groups $\operatorname{Ext}(H_{n-1}(X),\mathbb{Z})$ and $\operatorname{Ext}(H_{n-1}(X),\mathbb{R})$ both vanish, and the map $H^n(X)\to H^n(X;\mathbb{R})$ is injective if and only if $\operatorname{Hom}(H_n(X),\mathbb{Z})\to \operatorname{Hom}(H_n(X),\mathbb{R})$ is injective (which it clearly is).

Conversely, if $H_{n-1}(X)$ has torsion then so does $H^n(X)$, and this torsion is in the kernel of $H^n(X)\to H^n(X;\mathbb{R})$.

Answer 2

This is also equivalent to the corresponding statement for homology: Since $Hom(C,\mathbb{Z}) \otimes \mathbb{R} = Hom(C,\mathbb{R})$ for $C$ a free abelian group, the cochain complex with $\mathbb{R}$-coefficients is isomorphic to the cochain complex with $\mathbb{Z}$-coefficients tensored with $\mathbb{R}$. The same homological algebra as in the homology universal coefficient theorem gives you that $H^n(X) \rightarrow H^n(X; \mathbb{R})$ precisely kills torsion. (Which of course comes from $H_{n-1}$ if you invoke the universal coefficient sequence as in Marks answer.)