The Stacks project

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). $\square$