How does one split up a long exact sequence into short exact sequences?

Say you have some longs exact sequences of modules $$ 0\longrightarrow M_1\stackrel{\phi_1}{\longrightarrow}M_2\stackrel{\phi_2}{\longrightarrow}M_3\stackrel{\phi_3}{\longrightarrow}M_4\stackrel{\phi_4}{\longrightarrow}\cdots $$ I've read it's possible to split this into short exact sequences. What exactly does that mean? Would it be written as short exact sequences, one appended to another like $$ 0\longrightarrow N_1\longrightarrow M_1\longrightarrow N'_1\longrightarrow 0\longrightarrow N_2\longrightarrow M_2\longrightarrow N'_2\longrightarrow 0 \longrightarrow\cdots? $$ If so, how does this work? Merci.


2 Answers 2


You can think of the long exact sequence $$0\longrightarrow M_1\stackrel{\phi_1}{\longrightarrow}M_2\stackrel{\phi_2}{\longrightarrow}M_3\stackrel{\phi_3}{\longrightarrow}M_4\stackrel{\phi_4}{\longrightarrow}\cdots$$ as a collection of short exact sequences $$0\longrightarrow M_1\stackrel{\phi_1}{\longrightarrow}M_2\stackrel{\phi_2}{\longrightarrow}\mathrm{Image}(\phi_2)\longrightarrow 0$$ $$0\longrightarrow\mathrm{Coker}(\phi_2)\stackrel{\phi_3}{\longrightarrow} M_4\stackrel{\phi_4}{\longrightarrow}\mathrm{Image}(\phi_4)\longrightarrow 0$$ $$\vdots$$ where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $\phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.

  • $\begingroup$ Thanks Alex, I understand now. $\endgroup$
    – GGGG
    Jan 23, 2012 at 19:52
  • $\begingroup$ Is this some sort of reverse to the Snake lemma? $\endgroup$
    – gary
    May 27, 2018 at 20:08
  • $\begingroup$ I am a beginner in this topic. I mean, if I want to retrieve the original sequence from the collection produced, I just have to suppress the 0s? Is there any issue with this? Why do we put the zeros in the splitting? $\endgroup$ Oct 21, 2019 at 17:17
  • $\begingroup$ There is a dual s.e.s. to the one you posted. It is $$ 0\to\operatorname{Coim}\phi_{i-1}\xrightarrow{\phi_{i-1}}M_i\xrightarrow{\phi_i}\operatorname{Ker}\phi_{i+1}\to 0. $$ $\endgroup$ Jul 1 at 17:41

Just to expand on @AlexBecker's answer above.

Let's say we have a long exact sequence of $A$-modules $$\dots \to M_{i-1} \xrightarrow{f_i} M_i \xrightarrow{f_{i+1}} M_{i+1} \dots$$

If we let $N_i = \operatorname{Im}(f_i) = \operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 \to N_i \xrightarrow{\iota} M_i \xrightarrow{\pi} M_i / N_i \to 0 \ \ \ \ \ \ \ \ \ (*)$$ where $\iota : N_i \to M_i$ is the inclusion map and $\pi : M_i \to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.

Now since $N_{i+1} = \operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/\operatorname{ker}(f_{i+1}) \cong \operatorname{Im}(f_{i+1}) \iff M_i/N_i \cong N_{i+1}$. If we let $f : M_i/N_i \to N_{i+1}$ be this isomorphism then letting $\pi' = f \circ \pi$ we obtain a short exact sequence

$$0 \to N_i \xrightarrow{\iota} M_i \xrightarrow{\pi'} N_{i+1} \to 0 \ \ \ \ \ \ \ \ \ (**)$$

which we can then rewrite by definition of the $N_i$'s as

$$0 \to \operatorname{Im}(f_i) \xrightarrow{\iota} M_i \xrightarrow{\pi'} \operatorname{Im}(f_{i+1}) \to 0 \ \ \ \ \ \ \ \ \ (**)$$

Now here's a Lemma that we will use below

Lemma: If we have $A$-modules $M, M', M''$, then $0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0$ is exact $\iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $\operatorname{Coker}(f)$ onto $M''$

Replacing $\pi'$ in the exact sequence $(**)$ by $\pi''$ which is the isomorphism induced by $\pi'$ of $\operatorname{Coker}(\iota)$ onto $N_{i+1}$ we obtain the following short exact sequence

$$0 \to \operatorname{ker}(f_{i+1}) \xrightarrow{\iota} M_i \xrightarrow{\pi''} \operatorname{Coker}(\iota) \to 0 \ \ \ \ \ \ \ \ \ (***)$$

So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.

  • $\begingroup$ Just to clarify: isn't $\pi'$ just simply $f_{i+1}$? $\endgroup$ Nov 1, 2020 at 11:54

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