Proving a map in a short exact sequence of homotopy groups of a fibration is a section I am working through the following question which has already been asked before, in order to show that $\pi_n(B,b)\cong \pi_n (E,e) \oplus \pi_{n-1} (F,e)$, if $p:(E,e)→(B,b)$ is a fibration for which the inclusion of $F=p^{-1}(b)$ is pointed null homotopic.
As suggested by Balarka, I was able to construct a section $r:\pi_{n-1}(F, e) \to \pi_n(B, b)$, by extending the null-homotopy to a ball, and factoring it through the quotient map. However, how do I prove that this is infact a section, i.e. the composition with the map $s:\pi_{n}(B, b) \to \pi_{n-1}(F,e)$ is actually the identity, $r \circ s = id.$
The explicit form of the map $s$ seems very tricky to me, and it seems difficiult to get an handle on how the composition will look. 
 A: Although you've already constructed Balarka Sen's section $s:\pi_{n-1}(F,e)\to\pi_n(B,b)$, I'll run through the construction so that my notation doesn't come out of nowhere. 
Let $f:S^{n-1}\to F$ represent an element of $\pi_{n-1}(F,e)$. Since $i$ is null-homotopic, the composition
    $$S^{n-1}\xrightarrow{\ f \ }F\xrightarrow{\ i \ }E.$$
is as well. Hence, there exists an extension to the cone $CS^{n-1}=D^n$, whose restriction to $S^{n-1}$ is $i\circ f$, which is really just $f$ itself:
    $$\overline{i\circ f}:(D^n,S^{n-1})\to (E,F),\quad \overline{i\circ f}|_{S^{n-1}} = f.$$
Composing this with $p:E\to B$ gives a map $p\circ\overline{i\circ f}:D^n\to B$. Note that for all $x\in S^{n-1}$,
    $$(p\circ\overline{i\circ f})(x) = (p\circ f)(x) = b.$$
Thus, by the universal property of quotients, $p\circ\overline{i\circ f}$ factors through a map $\hat{f}:S^n\to B$ making the diagram:
$$\require{AMScd} \begin{CD}
    D^n @>{\pi}>> D^n/\partial D^n=S^n\\ @V{p\circ\overline{i\circ f}}VV @VV{\hat{f}}V\\
    B @>>{id_B}> B
    \end{CD}$$
commute. Hence, we make the assignment $s[f] = [\hat{f}]$. To see why this is a section, recall that the boundary map $\partial$ in the long exact sequence
$$\cdots\xrightarrow{\ \ \ }\pi_n(F,e)\xrightarrow{\ i_* \ }\pi_n(E,e)\xrightarrow{\ p_* \ }\pi_{n}(B,b)\xrightarrow{\ \partial \ }\pi_{n-1}(F,e)\xrightarrow{\ i_* \ }\pi_{n-1}(E,e)\xrightarrow{\ \ \ }\cdots$$
is defined as the composition
    $$\partial:\pi_n(B,b)\cong \pi_n(B,\{b\},b)\xrightarrow{\ p_*^{-1} \ }\pi_n(E,F,e)\xrightarrow{\ \delta \ }\pi_{n-1}(F,e)$$
where $\delta[g] = [g|_{S^{n-1}}]$. Now, we would like to show that $\partial\circ s=id_{\pi_{n-1}(F,e)}$. Note that $p_*$ sends $[\overline{i\circ f}]\in\pi_n(E,F,e)$ to $[\hat{f}]\in\pi_n(B,b)$. Hence, 
    $$(\partial\circ s)[f] = \partial[\hat{f}] = (\delta\circ p_*^{-1})[\hat{f}] = \delta[\overline{i\circ f}] = [\overline{i\circ f}|_{S^{n-1}}] = [f].$$
