Lemma 13.27.5. Let $\mathcal{A}$ be an abelian category with objects $A$, $B$. Any element in $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A)$ is $\delta (E)$ for some degree $i$ Yoneda extension of $B$ by $A$. Given two Yoneda extensions $E$, $E'$ of the same degree then $E$ is equivalent to $E'$ if and only if $\delta (E) = \delta (E')$.

**Proof.**
Let $\xi : B[0] \to A[i]$ be an element of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A)$. We may write $\xi = f s^{-1}$ for some quasi-isomorphism $s : L^\bullet \to B[0]$ and map $f : L^\bullet \to A[i]$. After replacing $L^\bullet $ by $\tau _{\leq 0}L^\bullet $ we may assume that $L^ i = 0$ for $i > 0$. Picture

Then setting $Z_{i - 1} = (L^{- i + 1} \oplus A)/L^{-i}$ and $Z_ j = L^{-j}$ for $j = i - 2, \ldots , 0$ we see that we obtain a degree $i$ extension $E$ of $B$ by $A$ whose class $\delta (E)$ equals $\xi $.

It is immediate from the definitions that equivalent Yoneda extensions have the same class. Suppose that $E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0$ and $E' : 0 \to A \to Z'_{i - 1} \to Z'_{i - 2} \to \ldots \to Z'_0 \to B \to 0$ are Yoneda extensions with the same class. By construction of $D(\mathcal{A})$ as the localization of $K(\mathcal{A})$ at the set of quasi-isomorphisms, this means there exists a complex $L^\bullet $ and quasi-isomorphisms

and

such that $s \circ t = s' \circ t'$ and $f \circ t = f' \circ t'$, see Categories, Section 4.26. Let $E''$ be the degree $i$ extension of $B$ by $A$ constructed from the pair $L^\bullet \to B[0]$ and $L^\bullet \to A[i]$ in the first paragraph of the proof. Then the reader sees readily that there exists “morphisms” of degree $i$ Yoneda extensions $E'' \to E$ and $E'' \to E'$ as in the definition of equivalent Yoneda extensions (details omitted). This finishes the proof. $\square$

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