# Yoneda Ext theorem and extensions

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 $$\text{Ext}^n$$.

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
• Out of curiosity: I think the theorem for $\textrm{Ext}^1$ can be proved without assuming enough injectives. The OP asks about modules over a ring but we could equally ask about any abelian category. Commented Feb 3, 2022 at 14:13