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We have that $\mathrm{Ext}^n(M, N) = \mathrm{Hom}(M, N[n])$, in other words, an element of $\mathrm{Ext}^n(M, N)$ is a map $$ I_M \longrightarrow N[n] $$ where $M \to I_M$ is an injective resolution, in other words, induces a sequence $$ 0 \to M \to I_1 \to \cdots I_n \to N \to 0 $$ which corresponds to the classical $n$-extensions. The last map is a surjection because the derived category map goes trough the homology of the complex.