TITLE: About the cohomology of $BG^\delta$. Making a Lie group discrete
QUESTION [9 upvotes]: Let $G$ be a connected Lie group. Recall that the topological group $G^\delta$ is $G$ endowed with the discrete topology. The inclusion $G^\delta \to G$ induces a map between the classifying spaces $\eta: BG^\delta\to BG$. 
Question 1
Let $\eta^*:H^*(BG,\mathbb{Z})\to H^*(BG^\delta,\mathbb{Z})$ be the induced map in integral cohomology.
By Corollary 1 in Milnor, On the homology of Lie groups made discrete,
we get that $\eta^*$ is injective.
On the other hand, by Lemma 10 in the same paper, we learn that  the kernel of 
$\eta_{\mathbb{Q}}^*:H^*(BG,\mathbb{Q})\to H^*(BG^\delta,\mathbb{Q})$ (notice the rational coefficients here)
is equal to the kernel of $\eta^*_{\mathbb{Q}}:H^*(BG,\mathbb{Q})\to H^*(B\Gamma,\mathbb{Q})$, where $\Gamma<G$ is a discrete cocompact group.
Consider  $G=U(n)$. Then $H^*(BG,\mathbb{Z})= \mathbb{Z}[c_1,c_2,\dots,c_n]$ injects in $H^*(BG^\delta, \mathbb{Z})$, in particular $\eta^*(c_1)\neq 0$.
However,  we can take $\Gamma= \{\mathbb{1}\}$, which implies that  $H^*(B\Gamma,\mathbb{Q})=H^*(\mathbb{R},\mathbb{Q})$, hence $\eta^*_{\mathbb{Q}}$ is trivial and  in particular $\eta^*_{\mathbb{Q}}(c_1) = 0$.

Why doesn't this give a contradiction?

REPLY [13 votes]: I will only attempt to answer your first question. The reason there is no contradiction is that it is not true for arbitrary spaces that $H^{\ast}(X;\mathbb Q) = H^{\ast}(X;\mathbb Z) \otimes \mathbb Q$.
For instance, take $X = B\mathbb Q$. Then $H^2(X;\mathbb Z) = \text{Ext}(\mathbb Q,\mathbb Z)$ is a $\mathbb Q$ vector space of uncountable dimension. In particular, we can find a map $B\mathbb Q \to K(\mathbb Z,2)$ that is injective on second cohomology with $\mathbb Z$ coefficients. But clearly the map induced on rational cohomology is zero.