Baer Sum notation requires clearence. I am working on Baer sum and I have my book by Rotman, Introduction to Homology, and also MacLanes book Homology and they use notation I am puzzled on.
I have understood baer sum of extensions through a different manner but I want to understand their notation as I understand it is used to extend it from just short exact sequences.
Let 
$$E: B\overset{i}{\hookrightarrow} E\overset{\pi}{\twoheadrightarrow}A$$
and
$$F: B\overset{j}{\hookrightarrow} F\overset{p}{\twoheadrightarrow}A$$
be two extensions of $A$ by $B$. First step is the direct sum
$$E\oplus F: B\oplus B\overset{i\oplus j}{\hookrightarrow} E\oplus F\overset{\pi\oplus p}{\twoheadrightarrow}A\oplus A$$
Nothing difficult so far, another short exact sequence that makes no mysteries to me. They define the diagonal morphism $\bigtriangleup_B:B\to B\oplus B$ such that $\bigtriangleup_B(b)=b\oplus b$ and the codiagonal one with $\bigtriangledown_B:B\oplus B\to B$, such that $\bigtriangledown_B(a\oplus b)=a+b$. From this they demonstrate that for $f,g:A\to B$ we have $f+g=\bigtriangledown_B\circ(f\oplus g)\circ \bigtriangleup_A$ Which again doens't bother me. It is easy to show from the given definitions. My issues arises in how to interpretate this
$$\bigtriangledown_A\circ(E\oplus F)\circ \bigtriangleup_B$$
where it is the entire extension. I do not understand how I am to view the chain, modules and morphisms in this instance and notation, especially the middle part. I do know that the baer sum is equal to $X/N$ with $X=\{a\oplus b\in E\oplus F:\pi(a)=p(b)\}$ and $N=\{i(n)\oplus -j(n):n\in B\}$.
How to connect these two definitions currently eludes me. The main issue is that I do not know how to interpretate that notation and as hence cannot make a connection. I'd appreciate if someone could show me step by step in the composition how the given chain, from direct sum as it is fairly easy to understand, becomes and then how it ends up and why.
Best regards and thanks for any assistence.
 A: I guess the author meant the following : let $\operatorname{Ext}^1(A,B)$ be the set of extension of $A$ by $B$. Then $\operatorname{Ext}^1(A,B)$ is a covariant functor in $B$ and a contravariant functor in $A$.
So, if $f:A'\rightarrow A$ and $E\in\operatorname{Ext}^1(A,B)$, you can take its image in $\operatorname{Ext}^1(A,B)$, this image will be denoted by $E\circ f\in\operatorname{Ext}^1(A',B)$. Namely, $E\circ f$ is the pull-back $E\circ f=A'\times_A E$ :
$$\require{AMScd}
\begin{CD}
0@>>>B@>>>E\circ f@>>>A'@>>>0\\
@.@|@VVV@VfVV@.\\
0@>>>B@>>>E@>>>A@>>>0
\end{CD}$$
And similarly, if $g:B\rightarrow B'$ is any morphism, write $g\circ E\in\operatorname{Ext}^1(A,B')$ for the corresponding extension. It is dually the pushout :
$$\require{AMScd}
\begin{CD}
0@>>>B@>>>E@>>>A@>>>0\\
@.@VgVV@VVV@|@.\\
0@>>>B'@>>>g\circ E@>>>A@>>>0
\end{CD}$$
Now you have an extension $E\oplus F\in\operatorname{Ext}^1(A\oplus A,B\oplus B)$. Just take $\nabla_B\circ(E\oplus F)\circ\Delta_A$. In other words, the Baer sum of $E$ and $F$ is exactly your quotient $X/N$.
