# 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.

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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.

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Thanks Alex, I understand now. –  GGGG Jan 23 '12 at 19:52