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In $\textit{Atiyah and Bott: The Yang-Mills equations over riemann surfaces}$, there is a sentence about the topology of BU(n) (i.e. the classifying space of U(n)):

Atiyah-Bott

Here $K(\mathbb{Z},n)$ means the Eilenberg-MacLane space with the $n^{th}$ homotopy group $\mathbb{Z}$. Does anybody know what's the meaning of the symbol ${\simeq \atop \mathbb{Q}}$? Does anyone know a reference (or a proof) for the statement? Thanks a lot!

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    $\begingroup$ My guess is that there is a natural map, say $f$ in one direction such that $\pi_n(f)$ is an isomorphism after tensoring with the rationals. In otherwords, the two spaces become equivalent once you localize the category of topological spaces at 0. I could be wrong though $\endgroup$ Sep 1, 2014 at 0:53
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    $\begingroup$ Standard: rational homotopy equivalence, meaning the two spaces become homotopy equivalent after rationalizing (when the Z's become Q's on the right). A suitable map from left to right is given by the Chern classes. $\endgroup$
    – Peter May
    Sep 1, 2014 at 1:06
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    $\begingroup$ Pages 163ff of Griffiths-Morgan. pages.uoregon.edu/njp/GM81.pdf $\endgroup$
    – nsrt
    Sep 1, 2014 at 9:26

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Griffiths and Morgan, Rational Homotopy Theory and Differential Forms (see nsrt's comment above):

$\mathrm{BU}(n)$ is the Grassmannian of $n$-planes in $\mathbb{C}^\infty$. Recall that $$\mathrm{H}^\ast(\mathrm{BU}(n),\mathbb{Z})\cong\mathbb{Z}[c_1,\cdots,c_n]$$ Where $c_i\in\mathrm{H}^{2i}(\mathrm{BU}(n),\mathbb{Z})$ is the $i$th Chern class of the universal vector bundle. This gives a map $$\mathrm{BU}(n)\xrightarrow{f}\prod^n_{i=0}K(\mathbb{Z},2i)$$ Which induces an isomorphism on $\mathbb{Q}$-cohomology.

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