Characterizing inverse elements in extension groups under the Baer sum operation For objects $A$ and $B$ in an abelian category, there is an isomorphism between $\operatorname{Ext}^1(A,B)$ defined as a derived functor, and the group of equivalence classes of short exact sequences of the form $B \hookrightarrow \_\_ \twoheadrightarrow A$. In the latter characterization of the group $\operatorname{Ext}^1(A,B)$, how can we easily see what the inverse elements are? So given $B \hookrightarrow X \twoheadrightarrow A$,
how do we characterize the element $Y$ in the category such that the Baer sum of $B \hookrightarrow X \twoheadrightarrow A$ and $B \hookrightarrow Y \twoheadrightarrow A$ splits? 

To provide the details of this characterization of $\operatorname{Ext}^1(A,B)$, we say that two sequences $B \hookrightarrow X \twoheadrightarrow A$, and $B \hookrightarrow Y \twoheadrightarrow A$ are equivalent if there is some map $X \to Y$ that makes the diagram
$$
\require{AMScd}
  \begin{CD}
     0 @>>> B @>>> X @>>> A @>>> 0\\
     @. @V\mathrm{id}VV @VVV @VV{\mathrm{id}}V @.\\
     0 @>>> B @>>> Y @>>> A @>>> 0\\
  \end{CD}
$$
commute. Then we define the group structure with the following operation (see Wiebel's Homological Algebra). For our two sequences $B \hookrightarrow X \twoheadrightarrow A$, and $B \hookrightarrow Y \twoheadrightarrow A$, let $X \times_A Y$ denote the pullback of $X$ and $Y$ over $A$. With the maps $B \hookrightarrow X$ and $B \hookrightarrow Y$ we can think of an image of $B \times B$ living in this pullback. Let $\Delta$ denote the skew diagonal, the set of elements like $(-b,b)$, in the image of $B \times B$ in $X \times_A Y$. Then the Baer sum of those two extensions is the sequence $B \hookrightarrow (X \times_A Y)/\Delta \twoheadrightarrow A$. This operation gives us a group structure, where the class of split exact sequences acts as the identity.
 A: I’ll assume for ease of explanation that we are working in a module category, although the conclusion is true in any abelian category.
The inverse of the element of $\text{Ext}^1(B,A)$ represented by
$$0\longrightarrow A\stackrel{\alpha}{\longrightarrow}X
\stackrel{\beta}{\longrightarrow}B\longrightarrow0$$
is represented by
$$0\longrightarrow A\stackrel{-\alpha}{\longrightarrow}X
\stackrel{\beta}{\longrightarrow}B\longrightarrow0.$$
The Baer sum of this with the original sequence has middle term $(X\times_BX)/\nabla$, where $\nabla$ is the unskew diagonal consisting of elements of the form $\left(\alpha(a),\alpha(a)\right)$ where $a\in A$, and the map $(X\times_BX)/\nabla\to B$ is given by $[(x_1,x_2)]\mapsto\beta(x_1)=\beta(x_2)$.
This map is split by the map $B\to(X\times_BX)/\nabla$ given by $b\mapsto\left[(\bar{b},\bar{b})\right]$, where for each $b$, $\bar{b}$ is any element of $X$ with $b=\beta\left(\bar{b}\right)$. Note that this is independent (modulo $\nabla$) of the choices of $\bar{b}$ and is a module homomorphism.
