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I'd like to understand how to find the homotopy groups of a fibration $F\rightarrow E \rightarrow B$ using a long exact sequence \begin{equation} \pi_n(F)\rightarrow \pi_n(E)\rightarrow \pi_n(B)\rightarrow \pi_{n-1}(F)\,. \end{equation} For concreteness, I want to focus on the example of first and second homotopy groups of real Grassmannians $\operatorname{Gr}(n,m) = O(n+m)/O(n)\times O(m)$, which were discussed in this post. As I understand, the relevant exact sequence is \begin{equation} \pi_2[O(n+m)] \rightarrow \pi_2[\operatorname{Gr}(n,m)] \rightarrow \pi_1[O(n)\times O(m)] \rightarrow \pi_1[O(n+m)] \rightarrow \pi_1[\operatorname{Gr}(n,m)] \rightarrow \pi_0[O(n)\times O(m)] \end{equation} For $N\geq 3$, there's $\pi_2[O(N)] = 0$, $\pi_1[O(N)] = Z_2$, $\pi_0[O(N)] = Z_2$ 2, so for $n,m\geq 3$ the sequence becomes \begin{equation} 0 \rightarrow \pi_2[\operatorname{Gr}(n,m)] \rightarrow Z_2\times Z_2 \rightarrow Z_2 \rightarrow \pi_1[\operatorname{Gr}(n,m)] \rightarrow Z_2\times Z_2\,. \end{equation} Question 1: From the first half of the sequence $0 \rightarrow \pi_2[\operatorname{Gr}(n,m)] \rightarrow Z_2\times Z_2$ follows that $\pi_2[\operatorname{Gr}(n,m)] = Z_2\times Z_2$ (right?), but this is not the correct result, which is $\pi_2[\operatorname{Gr}(n,m)] = Z_2$ 1. Where's the mistake and how to arrive at the correct result using the sequence?

Question 2: Can anything about $\pi_1[\operatorname{Gr}(n,m)]$ (which happens to be $Z_2$) be deduced from this sequence?

Michał Jan
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    The exactness of the sequence $0 \to \pi_2[\operatorname{Gr}(n, m)] \to \mathbb{Z}_2\times\mathbb{Z}_2$ only tells you that the map $\pi_2[\operatorname{Gr}(n, m)] \to \mathbb{Z}_2\times\mathbb{Z}_2$ is injective. In particular, you cannot conclude that it is an isomorphism (as you've noted, it isn't). To determine $\pi_2[\operatorname{Gr}(n,m)]$ you need to understand the next map $\mathbb{Z}_2\times\mathbb{Z}_2 \to \mathbb{Z}_2$. – Michael Albanese Oct 20 '23 at 15:37

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