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Let $F\hookrightarrow E\xrightarrow pB$ be a fibre bundle. Consider the long exact sequence in homotogy groups :

$$\cdots\rightarrow \pi_n(F)\xrightarrow{i_*}\pi_n(E)\xrightarrow{p_*}\pi_n(B)\xrightarrow{\partial_*}\pi_{n-1}(F)\rightarrow\cdots$$

Suppose $p:E\to B$ has a section $s:B\to E$ , $p\circ s=id_B$ then $s_*$ gives a right inverse for $p_*$.

Can we say that $\pi_n(E)=\pi_n(F)\oplus\pi_n(B)$?

Clearly if $\pi_n(F)=\pi_{n-1}(F)=0$ then this is true. But what about otherwise?

Thank you.

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    $\begingroup$ If $p$ admits a section then all the maps $p_*$ are onto, so the maps $\partial_*$ are identically zero and hence the $\iota_*$ are injective. This gives short exact sequences for $n>1$ that split by virtue of the splitting lemma. $\endgroup$ – Pedro Tamaroff Jul 9 '17 at 7:15
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I am posting the comment by Pedro Tamaroff that answers my question -

If $p$ admits a section then all the maps $p_∗$ are onto, so the maps $∂_∗$ are identically zero and hence the $i_∗$ are injective. This gives short exact sequences for $n>1$ that split by virtue of the splitting lemma.

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