Expliciting description of the connecting homomorphism between Yoneda Ext groups I have learned from Stacks 06XP that there is a natural interpretation of higher $\mathrm{Ext}$ groups as Yoneda extensions. Where a $i$-th Yoneda extension of $B$ by $A$ is an equivalence class of exact sequences of the form
$$
E : 0 \to A \to Z_{i-1} \to Z_{i-2} \to \ldots Z_0 \to B \to 0
$$
Given a short exact sequence $0 \to A \to B \to C \to 0$, and consider the long exact sequence when we take $\mathrm{Hom}(-, E)$.  I would like to ask if there is an explicit description of the connecting homomorphism $\delta : \mathrm{Ext}^i(A, E) \to \mathrm{Ext}^{i+1}(C, E)$ in terms of Yoneda extensions?
In particular, I would like to know if there is a description of extensions $\mathrm{Ext}^1(A, E)$ that lie in the kernel of $\mathrm{Ext}^2(C, E) \to \mathrm{Ext}^2(B, E)$.
I know there is a description of the first connecting homomorphism $\mathrm{Hom}(A, E) \to \mathrm{Ext}^1(C, E)$: given morphism $\beta \in \mathrm{Hom}(A, E)$, we can consider the pushout diagram:
$\require{AMScd}$
\begin{CD}
0 @>>> A @>>> B @>>> C @>>> 0\\
@. @V \beta VV @VVV @|\\
0 @>>> E @>>> X @>>> C @>>> 0
\end{CD}
then the connecting homomorphism sends $\beta$ to the class $[X] \in \mathrm{Ext}^1(C, E)$.
Thanks in advance!
 A: You can just splice the short exact sequence at the end of the long extension.  That is, given a long extension $$0 \to E \to Z_{i-1} \to \cdots \to Z_0 \to A \to 0,$$ the connecting homomorphism associated to the short exact sequence $0 \to A \to B \to C \to 0$ takes this to the long(er) extension $$0 \to E \to Z_{i-1} \to \cdots \to Z_0 \xrightarrow{(*)} B \to C \to 0,$$ where the only new map is $(*)$, which is defined as the composite $Z_0 \to A \to B$.
From this description, we see that the extensions in the kernel of $\DeclareMathOperator{\Ext}{Ext} \Ext^2(C, E) \to \Ext^2(B, E)$ are those of the form $$0 \to E \to Z \to B \to C \to 0.$$  To see that these extensions are indeed in the kernel, pull it back along $B \to C$ to get $$0 \to E \to {?} \to B \times_C B \to B \to 0.$$  Since $B \times_C B \cong B \oplus \ker(B \to C) \cong B \oplus \operatorname{im}(Z \to B)$, we see that this extension must be $$0 \to E \to Z \to B \oplus \operatorname{im}(Z \to B) \to B \to 0,$$ which is the direct sum of $$0 \to E \to Z \to \operatorname{im}(Z \to B) \to 0 \to 0$$ with $$0 \to 0 \to 0 \to B \xrightarrow{1} B \to 0.$$  The former is equivalent to $0 \to E \xrightarrow{1} E \to 0 \to 0$, so the pullback is equivalent to $$0 \to E \xrightarrow{1} E \xrightarrow{0} B \xrightarrow{1} B \to 0,$$ which is the zero in the $\Ext$ group.
