Which Short Exact Sequences Can I Extract From A Doubly Infinite Exact Sequence?

Question

I know how if we have a short exact sequence of $R$ modules, $0 \rightarrow A_1 \rightarrow A_2 \rightarrow A_3 \rightarrow 0$ , we can deduce properties about the known modules from the unknown modules, such as that $A_3 \cong \frac{A_2}{A_1}$ and that $A_2 \cong A_3 \oplus A_1$.

What I am unsure about is that is we have a long exact sequence $\ldots ^{}\rightarrow A_{i+2}\rightarrow A_{i+1} \rightarrow A_i \rightarrow A_{i-1} \rightarrow A_{i-2} \rightarrow \ldots$, which short exact sequences are we allowed to construct from it?

I know that we can get $0 \rightarrow Im(A_{i+1}) \rightarrow A_i \rightarrow Im(A_i) \rightarrow 0$ as a short exact sequence. But if, say $A_{i-2}$ is the zero module then is the sequence $0 \rightarrow Im(A_{i+1}) \rightarrow A_i \rightarrow A_{i-1} \rightarrow 0$ necessarily exact?

Are there any other canonical short exact sequences which can be constructed, under certain conditions?

I am not familiar with a lot of category theoretical terms.

Answer 1

Your second deduction is untrue. You cannot, in general, conclude from a short sequence

$$ 0\longrightarrow A_1 \longrightarrow A_2 \longrightarrow A_3 \longrightarrow 0 $$

an isomorphism

$$ A_2 \cong A_3 \oplus A_1 $$

For instance, this is not the case for the short exact sequence of abelian groups

$$ 0 \longrightarrow \mathbb{Z} \stackrel{2}{\longrightarrow} \mathbb{Z} \stackrel{\pi}{\longrightarrow} \mathbb{Z}_2 \longrightarrow 0 \ , $$

where the first morphism is multiplication by $2$ and the second one sends every integer number to its class modulo 2. Certainly you don't have an isomorphism

$$ \mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}_2 \ , $$

have you?

In order to deduce your isomorphism, the short exact sequence must split. Which is always true, for instance, if $A_3$ is free.

Answer 2

$0 \rightarrow \mathrm{Im}(A_{i+1}) \rightarrow A_i \rightarrow A_{i-1} \rightarrow 0$ must be exact.

Because if $A_{i-2}=0$, then $\mathrm{Im}(A_{i-1})=0$, then $\mathrm{Im}(A_{i})=\mathrm{Ker}(A_{i-1})=A_{i-1}$.

This is just a special case of the short exact sequence you get.