Suppose I have the following exact sequence:

$$\begin{matrix} A_0& \xrightarrow{\quad\quad} & B_0 \xrightarrow{\quad\alpha\quad} & C_0\\ \uparrow&&&\downarrow \\ C_1 & \xleftarrow{\quad\beta\quad}& B_1 \xleftarrow{\quad\quad}& A_1 \end{matrix}$$

Then I obtain two short exact sequence using $ \alpha$ and $ \beta$ ( Kernels and cokernels ) with $A_0$ and $A_1$ in the middle.

Any help as to how I get these sequences?


  • $\begingroup$ I have edited your question; tell me if that's correct. $\endgroup$ – user122283 Jun 18 '14 at 15:30
  • $\begingroup$ No :( It's a sequence of 6 terms, $\beta$ goes from $B_1$ to $C_1$ and there's a map from $ C_0$ to $A_1$. $\endgroup$ – CeCe Jun 18 '14 at 15:37
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    $\begingroup$ Maybe my edit is better :) $\endgroup$ – Vladimir Jun 18 '14 at 18:25

$$ 0\longrightarrow \operatorname{Coker} \beta\longrightarrow A_0\longrightarrow \operatorname{Ker\alpha}\longrightarrow0 $$ because, by the exactness of the original sequence, $\operatorname{Im}(A_0\to B_0)=\operatorname{Ker}\alpha$ and $\operatorname{Ker}(A_0\to B_0)=C_1/\operatorname{Im}\beta\equiv\operatorname{Coker}\beta$.

The second sequence is similar.

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  • $\begingroup$ Similar as in $$ 0\longrightarrow \operatorname{Coker\alpha} \longrightarrow A_1\longrightarrow \operatorname{Ker\beta}\longrightarrow0 $$?? $\endgroup$ – CeCe Jun 18 '14 at 18:40
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    $\begingroup$ Yes sure. Just rotate your original sequence (counter)clockwise by $180^\circ$ and you will see that all you need is to change the subscripts $0\to1$, $1\to0$. $\endgroup$ – Vladimir Jun 18 '14 at 18:46

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