A long exact sequence of free Abelian group is the direct sum of very short exact sequences.
The definition of short exact sequences doesn't seem to be very common from what I can see online:
An exact sequence is very short if it has at most 2 non-trivial terms.
I can see why this is true but can someone give me a formal way of proving this? Also, how different is this from writing a long exact sequence from short exact sequences?
Denote our long exact sequence by $(A_{\bullet}, d)$ and et $C_{i}$ denote the kernel of $d_{i} : A_{i} \rightarrow A_{i+1}$. By the exactness of the sequence we also have $C_{i} = Im(d_{i-1})$, so the sequence factors as in the diagram below. 
Now every free abelian group is by definition a free $\mathbb{Z}$-module, and free modules are projective, so any short exact sequence of free abelian groups splits, i.e. if $$ 0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0 $$ is exact then $A \cong A' \oplus A''$. Thus for each $i$ we have $A_{i} \cong C_{i} \oplus C_{i+1}$.
Let $C^{i}_{\bullet}$ denote the very short exact sequence $$\cdots \longrightarrow 0 \longrightarrow C_{i} \longrightarrow C_{i} \longrightarrow 0 \longrightarrow \cdots,$$ where the first $C_{i}$ is in the place of $A_{i}$ in the original sequence. Then $$ A_{\bullet} \cong \bigoplus_{i \in \mathbb{Z}} C^{i}_{\bullet},$$ as the object in the $i$th place is isomorphic to $C_{i} \oplus C_{i+1} \cong A_{i}$.
Let the chain complex be called $C_*$. Let $Z_*\subset C_*$ be the subgroup of cycles, which by exactness are exactly equal to the boundaries as well. Since $\partial_k\colon C_k\to C_{k-1}$ goes between two free abelian groups, the kernel $Z_k$ can be represented as a direct summand of $C_k$. (This requires an argument.) So write each $C_k=Z_k\oplus Y_k$. Then $\partial_k|_{Z_k}$ is trivial and $\partial_k|_{Y_k}\colon Y_k\overset{\cong}{\to} Z_{k-1}$ since boundaries map onto cycles, and the kernel of $\partial_k$ intersects $Y_k$ trivially. Therefore $$C_*=\bigoplus_{k}(0\to Y_k\to Z_{k-1}\to 0),$$
This fails for arbitrary abelian groups as the exact sequence $$0\to \mathbb Z \overset{\times 2}\to \mathbb Z\to \mathbb Z/2\mathbb Z\to 0$$ shows. It can't be written as a direct sum of smaller exact sequences.
