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?