Lemma 13.27.3. Let $\mathcal{A}$ be an abelian category.

1. Let $X$, $Y$ be objects of $D(\mathcal{A})$. Given $a, b \in \mathbf{Z}$ such that $H^ i(X) = 0$ for $i > a$ and $H^ j(Y) = 0$ for $j < b$, we have $\mathop{\mathrm{Ext}}\nolimits ^ n_\mathcal {A}(X, Y) = 0$ for $n < b - a$ and

$\mathop{\mathrm{Ext}}\nolimits ^{b - a}_\mathcal {A}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(H^ a(X), H^ b(Y))$
2. Let $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. For $i < 0$ we have $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A) = 0$. We have $\mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(B, A)$.

Proof. Choose complexes $X^\bullet$ and $Y^\bullet$ representing $X$ and $Y$. Since $Y^\bullet \to \tau _{\geq b}Y^\bullet$ is a quasi-isomorphism, we may assume that $Y^ j = 0$ for $j < b$. Let $L^\bullet \to X^\bullet$ be any quasi-isomorphism. Then $\tau _{\leq a}L^\bullet \to X^\bullet$ is a quasi-isomorphism. Hence a morphism $X \to Y[n]$ in $D(\mathcal{A})$ can be represented as $fs^{-1}$ where $s : L^\bullet \to X^\bullet$ is a quasi-isomorphism, $f : L^\bullet \to Y^\bullet [n]$ a morphism, and $L^ i = 0$ for $i < a$. Note that $f$ maps $L^ i$ to $Y^{i + n}$. Thus $f = 0$ if $n < b - a$ because always either $L^ i$ or $Y^{i + n}$ is zero. If $n = b - a$, then $f$ corresponds exactly to a morphism $H^ a(X) \to H^ b(Y)$. Part (2) is a special case of (1). $\square$

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