46
$\begingroup$

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.

$\endgroup$
47
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thanks Alex, I understand now. $\endgroup$ – GGGG Jan 23 '12 at 19:52
  • $\begingroup$ Is this some sort of reverse to the Snake lemma? $\endgroup$ – gary May 27 '18 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? $\endgroup$ – Francesco Bilotta Oct 21 '19 at 17:17
20
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.