# Exact sequences and normal series

Trying to understand exact sequences better, I was searching for a simple context in which they naturally arise. (The exact sequences that occur in algebraic topology occur in a very complicated context.)

Somewhere, possibly on this site, I found the following example: Start with a formally infinite normal series of groups, $$\dots \subset G_{n-1} \subset G_n \subset G_{n+1} \subset \cdots\,.$$ A finite sequence can be brought to this form by making the inclusions equalities in all but a finite number of cases.

From this series one can construct an exact sequence, $$\dots \xrightarrow{\partial_{n-2}} A_{n-1} \xrightarrow{\partial_{n-1}} A_{n} \xrightarrow{\partial_{n}} A_{n+1} \xrightarrow{\partial_{n+1}} \cdots,$$ with $$A_i = G_{i+2} / G_i$$ and $$\partial_{i}$$ the function $$g G_i \mapsto g G_{i+1}$$.

Now I can think that an exact sequence is "really" a normal series, which is great for the intuition. However, I cannot invert the construction. If an exact sequence is given, how can I find the $$G_i$$, and how unique is such a reconstruction? (A common normal subgroup of all $$G_i$$ certainly gets lost in the exact sequence, but is this everything that can happen?)

• I think this intuition might be misleading in general. An exact sequence gives you short exact sequences "locally" (by looking at image and kernel), which you can view as extensions of abelian groups. While it might be possible in the world of abelian groups to glue these extensions together (noncanonically) to a sequence as you're asking for, in general there are obstructions for that. For example, the corresponding elements in $Ext^1$ need to multiply to $0$, which gives counterexamples already for something like modules over an exterior algebra. Jul 25, 2021 at 15:25