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In other words I want to show that the induced map $\pi_k BSO(n-1) \to \pi_k BSO(n)$ is 0 for $k\leq n-1$. The fiber of this fibration is $S^{n-1}$. I am having trouble with the case $k=n-1$.

I am realizing the classifying spaces $BSO(n-1)$ as the stiefel manifold $S^\infty_n \times_{SO(n) \times 1} pt$ and the sphere bundle of $BSO(n)$ as $S^\infty_n \times_{SO(n) \times 1} S^{n-1}$ so that these are homeomorphic, and we do indeed have the fibration mentioned above.

We have the long exact sequence $\pi_{n-1} S^{n-1} \xrightarrow{i^*} \pi_{n-1} BSO(n-1) \xrightarrow{s^*}\pi_{n-1} BSO(n) \to \pi_{n-2}S^{n-1}$.

I need to show that $s^*$ is zero. One way of doing this is showing that $i^*$ is an isomorphism.

Since $BSO(1)$ is contractible, we get that $BSO(2)$ is simply connected, and hence that $BSO(n)$ is simply connected. This is the most I have been able to get inductively.

My motivation is that this is a step in page 537 of tom dieck's topology book, where he shows that the thom space $MSO(k)$ is $k-1$ connected.

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If $S^n$ is an $n$-sphere, $\pi_k(S^n)=1$ for $k<n$. Let $p_n:ESO(n)\rightarrow BSO(n)$ be the universal bundle. The Serre exact sequence gives: $..\pi_k(S^n)\rightarrow \pi_k(ESO(n))\rightarrow \pi_k(BSO(n))\rightarrow \pi_{k-1}(S^n)..$ For $k<n$, you have $\pi_k(ESO(n))=\pi_{k-1}(S^n)=1$. This implies that $\pi_k(BSO(n))=1$. Thus the map $\pi_k(BSO(n-1))\rightarrow \pi_k(BSO(n))$ is zero.

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    $\begingroup$ The fibration $ESO(n) \to BSO(n)$ you described has fiber $SO(n)$, not $S^n$. To compute the homotopy groups of $BSO(n)$ this way I will need the homotopy groups of $SO(n)$ through the fibration $SO(n-1) \hookrightarrow SO(n) \to S^{n-1}$. This is not easy past $n=2$. $\endgroup$ – user062295 Apr 7 '16 at 20:03

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