Let $$\dots A_i\stackrel {f_i}\to B_i \stackrel {g_i}\to C_i \stackrel {h_i}\to A_{i+1}\to \dots$$ be a long exact sequence of Abelian groups.

Is it true that if there are maps $k_i:D_i\to E_i$ such that $\ker k_i\simeq \ker g_i$ and $\operatorname{coker} k_i \simeq \operatorname{coker}g_i$, that then a long exact sequence $$\dots A_i\to B_i \stackrel {k_i}\to C_i \to A_{i+1}\to \dots$$ exists?

If yes, what are the maps?

  • $\begingroup$ What are you studying? What text is this drawn from, if any? If not, how did the question arise? What kind of approaches (to similar problems) are you familiar with? What kind of answer are you looking for? Basic approach, hint, explanation, something else? Is this question something you think should be able to answer? Why or why not? $\endgroup$ – Shaun Mar 13 at 20:55
  • $\begingroup$ Unless $B_i = D_i$ and $E_i = C_i$, the answer is obviously no, since $k_i$ cannot have both $B_i$ and $D_i$ as domain, or both $E_i$ and $C_i$ as codomain. But if these groups are the same, then the kernel and cokernel conditions are exactly what is required to replace $g_i$ with $k_i$ in the original sequence and have it remain exact. $\endgroup$ – Paul Sinclair Mar 14 at 0:35

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