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I am reading topology of Lie groups by Mimura and Toda and got to the part where they are beginning to compute $H^*(O(n))$, page 120.

If we let $r_m :S^m \to O(m+1)$ be the map that sends $v$ to the the reflection by the hyperplane perpendicular to $v$ or, $r(v)(v')=v'-2 v$. Then we can form the map $$ s: (D^{m+1}_+,S^m) \xrightarrow{r_{m+1}} (O(m+2),O(m+1)) \xrightarrow{\text{proj}}(O(m+2)/O(m), S^m) $$ Now the authors claim that this induces an isomorphism in cohomology. $$s^*:H^{m+1}(O(m+2)/O(m), S^m) \xrightarrow{\cong} H^{m+1}(D^{m+1}_+,S^m) $$ Could someone illuminate this isomorphism?

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    Either your title or the last equation you wrote in your question is wrong. (and $m$ is $n$, no?)2012-01-31
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    There, fixed it.2012-02-01
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    I think it's a little diagram chase using the exact sequences for a pair (and the fact that $S^{m+1} \cong O(m+2)/O(m+1)$ via the given map).2012-02-01
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    It is due to the fibration $S^m \to O(m+2)/O(m) \to S^{m+1}$.2012-02-02

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