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.