# How do you split a long exact sequence into short exact sequences?

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.

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.

• Thanks Alex, I understand now. – GGGG Jan 23 '12 at 19:52
• Is this some sort of reverse to the Snake lemma? – gary May 27 '18 at 20:08
• 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? – Francesco Bilotta Oct 21 '19 at 17:17

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.

• Just to clarify: isn't $\pi'$ just simply $f_{i+1}$? – chickenNinja123 Nov 1 '20 at 11:54