The Stacks project

Proof. By definition we can find a distinguished triangle

in $D(R)$ such that $E$ is represented by a bounded complex of finite free $R$-modules and such that $H^ i(L) = 0$ for $i \geq m$. Then $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(L, N) = 0$ for any $R$-module $N$ and $n \leq -m$, see Derived Categories, Lemma 13.27.3. By the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ associated to the distinguished triangle we see that $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, N) \to \mathop{\mathrm{Ext}}\nolimits ^ n_ R(E, N)$ is an isomorphism for $n < -m$ and injective for $n = -m$. Thus it suffices to prove that $M \mapsto \mathop{\mathrm{Ext}}\nolimits _ R^ n(E, M)$ commutes with filtered colimits when $E$ can be represented by a bounded complex of finite free $R$-modules $E^\bullet $. The modules $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(E, M)$ are computed by the complex $\mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M)$, see Derived Categories, Lemma 13.19.8. The functor $M \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(E^ p, M)$ commutes with filtered colimits as $E^ p$ is finite free. Thus $\mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M_ i)$ as complexes. Since filtered colimits are exact (Algebra, Lemma 10.8.8) we conclude. $\square$