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The notes I am reading say that groups in the title are isomorphic. Could someone explain to me why it is the case?

Here by $BO_n$ I mean the infinite Grassmann manifold of $n$-dimensional subspaces. It seems that it should somehow follow from the general theorem that $$H^*(B O_r ; \mathbb{Z}_2) \cong \mathbb{Z}_2\left[w_1, \ldots, w_r\right],$$ where $w_1,\ldots, w_r$ are the Stiefel-Whitney classes, however I don't really get it.

Thanks.

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    $\begingroup$ There is a fiber sequence $S^n\rightarrow BO(n)\rightarrow BO(n+1)$, hence the inclusion $BO(n)\rightarrow BO(n+1)$ is $n$-connected, which almost implies the result without doing any specific computation, but I'm not sure how to get injectivity in degree $n$ without a computation. $\endgroup$
    – Thorgott
    Apr 13 at 13:28

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As you state we have $H^*(BO_r; \mathbb{F}_2) \cong \mathbb{F}_2[w_1, \ldots, w_r]$, but you neglect to mention that $|w_i| = i$; with this the result is immediate since $\operatorname{rank}((\mathbb{F}_2[w_1, \ldots, w_r])_i) = \operatorname{rank}((\mathbb{F}_2[w_1, \ldots, w_i])_i)$ for all $i \leq r$.

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    $\begingroup$ Thank you! I was not sure about the grading on $\mathbb{F}_2\left[w_1, \ldots, w_r\right].$ By $(\mathbb{F}_2[w_1, \ldots, w_r])_i$ you mean the $i$th graded component of $\mathbb{F}_2[w_1, \ldots, w_r],$ right? In that case, isn't it true that $(\mathbb{F}_2[w_1, \ldots, w_r])_i=(\mathbb{F}_2[w_1, \ldots, w_i])_i$ for $i \leq r$? $\endgroup$
    – Haldot
    Apr 13 at 14:16
  • $\begingroup$ @Haldot You're welcome! Yes, I mean the $i$th graded component. For the last point, also yes. Not sure why I only wrote $\operatorname{rank}$ there, but either way the claim follows :) $\endgroup$ Apr 13 at 14:43

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