Zero element of extensions of $M$ by $N$ The set of equivalence classes $E(M,N)$ of extensions of $M$ by $N$ with Baer sum of extensions is an abelian group. I would like to know why the trivial extension ($0 \to N \to M\oplus N \to M \to 0$) is identity element of $E(M,N)$.
Suppose that \begin{align*}
0 \to N \to E \to M \to 0
\end{align*} 
is an extension of $M$ by $N$. By forming the pullback $X=\{(e, (g(e),n))\mid  e\in E, n \in N\}$ over $A$:
\begin{align*}
\begin{array}{c}
&  & & & & X & \to & M \oplus N & \\
&  & & & & \downarrow & & \downarrow &  \\
& 0 & \to &  N & \overset{f}{\to} &  E & \overset{g}{\to} &  M  &  \to  & 0
\end{array}
\end{align*} 
Let $F = X/ \{(f(n),(0,-n)) \mid n \in N \}$. I am stuck how to prove $F \cong E$. 

Ignore my comment, I'm not sure what I was thinking. Anyways, I will assume that you have already checked you get a well-defined exact sequence from $F$; I will denote the map $N\to F$ by $f_1$ and $F\to M$ by $g_1$. Define $\varphi:F\to E$ by $\varphi(e,g(e),n)=f(n)+e$. You can check that this is a homomorphism, and that this is well-defined on $F$ because it is zero on the ideal you quotiented by to get $F$. Then, check that the following diagram is commutative:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
0 & \ra{} & N & \ra{f_1} & F & \ra{g_1} & M & \ra{} & 0 & \\
&& \da{\text{id}} & & \da{\varphi} & & \da{\text{id}} & &  \\
0 & \ra{} & N & \ra{f} & E & \ra{g} & M & \ra{} & 0  \\
\end{array}
$$
which by the Five Lemma implies $\varphi$ is an isomorphism.
