Suppose you have two exact sequences $0 \to A \to B \to C \to 0$ and $0 \to E \to F \to G \to 0$ and a morphism $\{a_i\}$ between them.

Then the snake lemma gives us an exact sequence $$0 \to \text{ker}(a_1) \to \text{ker}(a_2) \to \text{ker}(a_3) \to \text{coker}(a_1) \to \text{coker}(a_2) \to \text{coker}(a_3) \to 0 $$

Now, one could factor $a_i = m_i \circ p_i $ where $p_i$ is a regular epi and $m_i$ is a mono.

This produces a sequence $ A/\text{ker}(a_1) \to B/\text{ker}(a_2) \to C/\text{ker}(a_3) $

This sequence is not (in general) exact.

In fact if it was the sequence $0 \to \text{ker}(p_1) \to \text{ker}(p_2) \to \text{ker}(p_3) \to 0$ should be exact, because $p_i$ are surjective, but this sequence coincides with $0 \to \text{ker}(a_1) \to \text{ker}(a_2) \to \text{ker}(a_3) \to 0$

Anyway, is there any connection between the exactness of this sequence and the snake lemma? Something like, its homology is precisely what is missing in the sequence $0 \to \text{ker}(a_1) \to \text{ker}(a_2) \to \text{ker}(a_3)$ to make it exact.

  • $\begingroup$ I can't find it right now, but I'm pretty sure I've seen a very similar question here. The answer was that this sequence is exact if and only if the connecting map $\ker(a_3)\to \operatorname{coker}(a_1)$ is zero. $\endgroup$ – Arnaud D. Jan 31 '18 at 15:22
  • $\begingroup$ Found it : math.stackexchange.com/questions/460941/… It's only about vector spaces, but I don't think it makes much of a difference... $\endgroup$ – Arnaud D. Jan 31 '18 at 15:28
  • $\begingroup$ You are right but the answer is essentially what I already wrote, I am interested in understanding if I can recover the $\delta$ map from that sequence. $\endgroup$ – Ivan Di Liberti Jan 31 '18 at 15:32
  • $\begingroup$ Ah, I see. Sorry for the misunderstanding. $\endgroup$ – Arnaud D. Jan 31 '18 at 15:36
  • $\begingroup$ No problem, probably my exposition was confused. $\endgroup$ – Ivan Di Liberti Jan 31 '18 at 15:36

The sequence

$$ 0 \to A / \ker(a_1) \to B / \ker(a_2) \to C / \ker(a_3) \to 0 $$

is exact on the left and on the right. The homology in the middle is precisely the image of the map $\ker(a_3) \to \mathrm{coker}(a_1)$, or equivalently the cokernel of the map $\ker(a_2) \to \ker(a_3)$.

Knowing what the right answer is, I imagine there is a straightforward diagram chase.

Alternatively, if you're at all familiar with the spectral sequences you can construct out of a bicomplex, then you can prove this by computing the total homology of the bicomplex

$$ \require{AMScd} \begin{CD} A @>>> B @>>> C \\ @VVV @VVV @VVV \\ A / \ker(a_1) @>>> B / \ker(a_2) @>>> C / \ker(a_3) \end{CD} $$

in two different ways.

Yet another proof is to consider the vertical short exact sequence of horizontal chain complexes:

$$ \require{AMScd} \begin{CD} 0 @>>> 0 @>>> 0 @>>> 0 @>>> 0 \\ @VVV @VVV @VVV @VVV @VVV \\ 0 @>>> \ker(a_1) @>>> \ker(a_2) @>>> \ker(a_3) @>>> 0 \\ @VVV @VVV @VVV @VVV @VVV \\ 0 @>>> A @>>> B @>>> C @>>> 0 \\ @VVV @VVV @VVV @VVV @VVV \\ 0 @>>> A / \ker(a_1) @>>> B / \ker(a_2) @>>> C / \ker(a_3) @>>> 0 \\ @VVV @VVV @VVV @VVV @VVV \\ 0 @>>> 0 @>>> 0 @>>> 0 @>>> 0 \end{CD} $$

An exact sequence of chain complexes gives a long exact sequence in homology. Since the middle row is exact, this proves the homology of the bottom row is canonically isomorphic to the homology of the top row, but shifted left one place.


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