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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?

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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.

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  • $\begingroup$ Probably it's obvious but why the map on rational cohomology is zero? I tried to compute $H^2(B\mathbb{Q},\mathbb{Q}) = H^2(\mathbb{Q},\mathbb{Q}) = Ext^2_{\mathbb{Z}[\mathbb{Q}]}(\mathbb{Z},\mathbb{Q})$ but I can't see why it should be trivial. $\endgroup$ – Warlock of Firetop Mountain Mar 27 at 18:04
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    $\begingroup$ I would compute it differently, $H^2(B\mathbb Q;\mathbb Q) = \text{Hom}(H_2(B\mathbb Q;\mathbb Q),\mathbb Q)$, and $H_2(B\mathbb Q;\mathbb Q) = \text{Tor}_1(\mathbb Q,\mathbb Q) = 0$ since $\mathbb Q$ is flat. $\endgroup$ – Jens Reinhold Mar 27 at 19:17

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