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$$\dots\xrightarrow{p_*}\pi_{n+1}(B)\xrightarrow{\partial}\pi_n(F)\xrightarrow{\text{inc}_*}\pi_n(E)\xrightarrow{p_*}\pi_n(B)\xrightarrow{\partial}\pi_{n-1}(F)\xrightarrow{\text{inc}_*}\dots$$ From the above how can one deduce a short exact sequence $$0\to coker[\pi_{n+1}(B)\xrightarrow{\partial}\pi_n(F)]\xrightarrow{i}\pi_n(E)\xrightarrow{j}\ker[\pi_n(B)\xrightarrow{\partial}\pi_{n-1}(F)]\to 0$$

Do we need to use the snake lemma?

Thanks for any help.

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    $\begingroup$ It's a general fact about long exact sequence, there's nothing special about this particular one. You should try proving it yourself before reading the solution, it's a very instructive game of playing with the definitions. $\endgroup$ – Najib Idrissi Mar 24 '16 at 15:22
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    $\begingroup$ No snake lemma needed. $\endgroup$ – Cheerful Parsnip Mar 24 '16 at 15:26
  • $\begingroup$ Do $i$ and $j$ represent the inclusions? Also It seems that there is a "jump" in the short exact sequence, I thought it should be of the form $0\to \text{coker}\to \pi_n(E)\to\text{Image}$ rather than $0\to \text{coker}\to \pi_n(E)\to\text{Ker}$ $\endgroup$ – yoyostein Mar 25 '16 at 4:41