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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 $\text{Ext}^n$.

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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 $$ M \longrightarrow I_N[n] $$ 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

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  • $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Commented Feb 3, 2022 at 14:13
  • $\begingroup$ Sorry for not getting it yet, but how do the injective extensions correspond to ordinary ones? $\endgroup$
    – user30211
    Commented Feb 3, 2022 at 17:46
  • $\begingroup$ @ZhenLin Of course, I was answering the case in the question. As a matter of fact, my argument work whenever the base category is Grothendieck. In general it is not clear to me that Ext and Yoneda-Ext agree $\endgroup$
    – Leo Alonso
    Commented Feb 3, 2022 at 17:57
  • $\begingroup$ @KindBubble It is because injective resolutions are universal up to homotopy. $\endgroup$
    – Leo Alonso
    Commented Feb 3, 2022 at 17:57
  • $\begingroup$ @LeoAlonso what is the equivalence relation on extensions? Is it weak equivalence? $\endgroup$
    – user30211
    Commented Feb 3, 2022 at 18:18

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