# cohomology monomorphism between grassmannians and product of projective spaces

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?

• You have to check that all the functors preserve finite limits. I think it's true, but I'm not sure... – 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.