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](https://link.springer.com/article/10.1007/BF02564625),
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?
> 
> 
> 



About the cohomology of $BG^\delta$. Making a Lie group discrete
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.


