Yoneda Ext theorem and extensions

Question

Consider the category of chain complexes over a ring $R$.

We can show that $\text{Ext}^1(M, N)$ classifies extensions using the triangulated category structure: the homotopy kernel of a map $N \rightarrow M[1]$, $K$, gives rise to the designated triangle $M \rightarrow K \rightarrow N$. This gives a simple picture of Yoneda's extension theorem for $\text{Ext}^1(M, N)$, which also seems to rely on the "threeness" of a triangulated category.

My question is, is there an extension of this view to $\text{Ext}^n(M, N)$?

The proof of the situation for $\text{Ext}^1$ is quite simple using triangulated categories and it would be nice if the situation could be handled the same way for $\text{Ext}^n$.

Answer 1

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 $$ M \longrightarrow I_N[n] $$ where $N \to I_N$ is an injective resolution. From this we obtain a distinguished triangle, $$ M \longrightarrow I_N[n] \longrightarrow C \overset{+}\longrightarrow $$ The homotopical invariant gives an extension of complexes $$ 0 \longrightarrow I_N[n] \longrightarrow C \longrightarrow M \longrightarrow 0 \ $$ which, plugin-in the beginning of the resolution of $N$, yields the classical $n$-extension $$ 0 \longrightarrow N \longrightarrow I^0_N \longrightarrow I^1_N \cdots \longrightarrow I^{n-1}_N \longrightarrow M \longrightarrow 0 $$ as wanted