Characterisation of trivial $n$-extensions $\newcommand{\Ext}{\operatorname{Ext}}$Let $\mathcal C$ be an abelian category. For $M, N\in\mathcal C$, we know that the set $\Ext^1(M, N)$ of all equivalence classes of short exact sequences $0\to N\to E\to M\to 0$ is a group whose neutral element is the equivalence class of $0\to N\to N\oplus M\to M\to 0$.
How can we characterise trivial $n$-extesions? Let's look at $2$-extensions. The product of a trivial and an arbitrary $1$-extension is zero, so e.g. 
$$\xi\colon\qquad0\to L\to E\xrightarrow{\bigl(\begin{smallmatrix}p\\0\end{smallmatrix}\bigr)} N\oplus M\to M\to 0$$
and 
$$\xi'\colon\qquad0\to L\to L\oplus N\xrightarrow{(0\; i)} E'\to M\to 0.$$
These two extensions thus belong to the same equivalence class $0\in\Ext^2(M, N)$. By definition, $\xi\sim\xi'$ if and only if there is a commutative diagram $\xi\to\xi'$ such that the outer morphisms are identities.
How do I find that commutative diagram rendering $\xi$ and $\xi'$ equivalent?
 A: There is not such a morphism $\xi \to \xi'$, but there is a morphism $\xi' \to \xi$ which is the identity on outer objects. The morphism $\xi' \to \xi$ is
$$\require{AMScd}
\begin{CD}
0 @>>> L @>>> L \oplus N @>>> E' @>q>> M @>>> 0 \\
& @V=VV @Vp \oplus 0VV @V0 \oplus qVV @V=VV \\
0 @>>> L @>p>> E @>>> N\oplus M @>>> M @>>> 0. 
\end{CD}
$$
Note the inner maps are not isomorphisms. (Both maps in the inner square compose to zero.) In contrast, a morphism $\xi \to \xi'$ would give a diagram
$$\begin{CD}
0 @>>> L @>>> E @>>> N \oplus M @>>> M @>>> 0 \\
& @V=VV @Vs \oplus s'VV @Vt \oplus t'VV @V=VV \\
0 @>>> L @>>> L \oplus N @>>> E' @>>> M @>>> 0.
\end{CD}$$
Examining the first square gives $s' = 0$ and $s:E \to L$ splits $0 \to L \to E \to N \to 0$; then the second square composes to zero, so $t = 0$ and $t': M \to E'$ splits $0 \to N \to E' \to M \to 0$.
In general, the equivalence relation on $\mathrm{Ext}^n$ for $n \geq 1$ is the equivalence relation generated by morphisms $\xi \to \xi'$ which are the identity on the outer objects. It is a special feature of the case $n=1$ that such morphisms are always isomorphisms (the Five Lemma) and consequently the equivalence relation is exactly isomorphism of diagrams with equality on the outer objects.
