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Let $A$ be a finite dimensional algebra and $X,Y$ indecomposable modules and $n \geq 2$. We know that $Ext_A^n(X,Y) \cong Ext_A^1(\Omega^{n-1}(X),Y)$.

Now such an isomorphism should be given by sending a short exact sequence in $Ext_A^1(\Omega^{n-1}(X),Y)$ given by $0 \rightarrow Y \rightarrow Z \rightarrow \Omega^{n-1}(X) \rightarrow 0$ to the long exact sequence given by splicing together the short exact sequence $0 \rightarrow Y \rightarrow Z \rightarrow \Omega^{n-1}(X) \rightarrow 0$ with the start of a minimal projective resolution of $X$: $0 \rightarrow \Omega^{n-1}(X) \rightarrow P_{n-2} \rightarrow \cdots \rightarrow P_0 \rightarrow X \rightarrow 0$.

So an explicit isomorphism should be $\phi:Ext_A^1(\Omega^{n-1}(X),Y) \rightarrow Ext_A^n(X,Y)$ with $\phi( 0 \rightarrow Y \rightarrow Z \rightarrow \Omega^{n-1}(X) \rightarrow 0)= 0 \rightarrow Y \rightarrow Z \rightarrow P_{n-2} \rightarrow \cdots \rightarrow P_0 \rightarrow X \rightarrow 0.$

Is there a reference to quote for this?

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1 Answer 1

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These types of results, on the relationships between $Ext$, iterated extensions, the Yoneda product, and the long exact sequence in $Ext$ are discussed at length in Chapter III of Mac Lane's Homology. In particular, in that chapter he shows in section III.4 how to define Ext using iterated extensions, how to compose in section III.5, the relations to injective resolutions in III.8, and demonstrates the long exact sequence in III.9.

In particular, his definition of the connecting homomorphisms $Ext^n(A',C) \to Ext^{n+1}(A'',C)$ from a short exact sequence $0 \to A' \to A \to A'' \to 0$ is via splicing exact sequences, and since the isomorphism $Ext^1(\Omega^{n-1} X, Y) \cong Ext^n(X,Y)$ is via an iterated connecting homomorphism, this recovers your result.

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