Let $S^1\times\cdots \times S^1( n\text{ times })=\prod_n U(1)\to U(n)$ be the inclusion. This induces a map between classifying spaces

$$ \prod_nBS^1\to BU(n).$$ i.e., $$ (\mathbb{C}P^\infty)^{\times n}\to BU(n).$$

The induced map in cohomology $$ f:H^*(BU(n))=\mathbb{Z}[y_1\cdots,y_n]\to H^*((\mathbb{C}P^\infty)^{\times n})=\otimes_n H^*(\mathbb{C}P^\infty)=\mathbb{Z}[x_1]\otimes \mathbb{Z}[x_2]\otimes\cdots\otimes \mathbb{Z}[x_n], $$ $|x_i|=2$, $|y_i|=2i$ for all $i$.

Is $f$ a monomorphism? Why?

  • $\begingroup$ You have to check that all the functors preserve finite limits. I think it's true, but I'm not sure... $\endgroup$ – user40276 Dec 1 '14 at 6:26

Yes; in fact $f$ is an isomorphism onto $H^{\bullet}(BU(1)^n)^{S_n}$. This is an aspect of the splitting principle for complex vector bundles.

  • $\begingroup$ I still do not understand... $\endgroup$ – Shiquan Dec 2 '14 at 1:05
  • 1
    $\begingroup$ @RSQ: I'd like to be able to help you, but just saying "I don't understand" doesn't give me any feedback about how I can improve my explanation. What don't you understand? $\endgroup$ – Qiaochu Yuan Dec 2 '14 at 1:26

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