Let $A$ be a $K$-algebra, and $0 \to M' \to M \to M'' \to 0 $ be an exact sequence of $A$-modules that splits. Show that $$ 0 \to \operatorname{Ext}^n_A(X,M') \to \operatorname{Ext}^n_A(X,M) \to \operatorname{Ext}^n_A(X,M'') \to 0$$ and $$ 0 \to \operatorname{Ext}^n_A(M',X) \to \operatorname{Ext}^n_A(M,X) \to \operatorname{Ext}^n_A(M'',X) \to 0$$

are exact sequences of $K$-modules that split, for all $A$-modules $X$ and all $n\geq0$.

I'm unsure how to approach this problem. Do I have to unwind the definition of the corresponding induced maps to prove that the sequences are exact? I have tried using the long exact sequence induced by the short one, but I got nowhere.

  • $\begingroup$ hint: $M=M'\oplus M''$ and direct sum commutes with Hom and Ext. $\endgroup$
    – Simonsays
    Commented Jun 27, 2022 at 17:33
  • $\begingroup$ every functor preserves split epimorphisms and split monomorphisms. $\endgroup$ Commented Jun 27, 2022 at 18:56
  • $\begingroup$ But why the induced short sequence in $\operatorname{Ext}$ is an exact one? $\endgroup$ Commented Jun 27, 2022 at 19:21

1 Answer 1


You have two options.

  1. Every functor preserves split monomorphisms and split epimorphisms. This means that if you take the LES induced from the SES the corresponding maps will remain split mono and split epi. This shows that the boundary morphisms will be zero, so the LES is just a collection of the desired SES for each $n$.

  2. Any additive functor preserves split exact sequences. This is because a split exact sequence is up to isomorphism a sequence of the form $$0 \rightarrow X \xrightarrow{\operatorname{in}_X} X\oplus Y \xrightarrow{\operatorname{pr}_Y} Y \rightarrow 0$$ which is precisely the kind of data we assume the functor to preserve. So you might also simply use the fact that the Ext functors are additive.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .