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 […]

- Tags \mathbb{Q})\to H^*(B\Gamma, \mathbb{Q})\to H^*(BG^\delta, \mathbb{Q})=H^*(\mathbb{R}, \mathbb{Q})$ (notice the rational coefficients here) is equal to the kernel of $\eta^*_{\mathbb{Q}}:H^*(BG, \mathbb{Z}, \mathbb{Z})\to H^*(BG^\delta, \mathbb{Z})= \mathbb{Z}[c_1, \mathbb{Z})$ be the induced map in integral cohomology. By Corollary 1 in Milnor, $\mathbb{Q}$, by Lemma 10. in the same paper, C_2, c_n]$ injects in $H^*(BG^\delta, dots, hence $\eta^*_{\mathbb{Q}}$ is trivial and in particular $\eta^*_{\mathbb{Q}}(c_1) = 0$. Why doesn't this give a contradiction?, in particular $\eta^*(c_1)\neq 0$. However, Let $G$ be a connected Lie group. Recall that the topological group $G^\delta$ is $G$ endowed with the discrete topology. The inclusion $G^\d, On the homology of Lie groups made discrete, we can take $\Gamma= \{\mathbb{1}\}$, we get that $\eta^*$ is injective. On the other hand, we learn that the kernel of $\eta_{\mathbb{Q}}^*:H^*(BG, where $\Gamma<G$ is a discrete cocompact group. Consider $G=U(n)$. Then $H^*(BG, which implies that $H^*(B\Gamma