Let $A$ be a Krull-Schimdt category. Suppose $A$ is a full subcategory of an abelian category $B$. Let $C$ be the category of complexes with terms in $A$. Call $S$ the class of all the exact sequences with terms in $C$ such that they are split exact in every degree. A monomorphisms is called an $S$-monomorphism if it can be completed to an exact sequence in $S$, the same for epimorphism. And you can define $S$-injectives using $S$-monomorphisms and in the same way you can define $S$-projectives. I want to prove that in the category $C$, the $S$-injectives and the $S$-projectives coincide.

Take $X\in A$ and define $P^i(X)^\bullet$ the complex in $C$ with $P^i(X)^i=P^i(X)^{i+1}=X$ and zeros in the other degrees, and the obvious differentials. These objects are $S$-projectives. Define $I^i(X)^\bullet$ in the following way: $I^i(X)^i=I^i(X)^{i-1}=X$ and zeros in the other degrees and the differential in the obvious way, these objects are $S$-injectives. I proved that $C$ has enough $S$-injectives (and $S$-projectives) because if you have $X^\bullet=(X^i,d^i)$ then take $I(X^\bullet)^\bullet=\bigoplus_iI^i(X^i)^\bullet$. My first question is:

why the $S$-injectives and $S$-projectives coincide?

My second question is:

let $f$ be a morphism in $C$. Then $f$ factors through an $S$-injective if and only if $f$ is homotopic to zero.

how can I prove it?

  • $\begingroup$ Regarding your first question, do you ask for a proof why the $S$-projectives and $S$-injectives have the claimed form or don't you see why $P^i(X)=I^{i+1}(X)$? Also I think you mean that the $P^i(X)$ for $X$ indecomposable should be the indecomposable $S$-projective objects. $\endgroup$ – Julian Kuelshammer Nov 5 '12 at 20:28
  • $\begingroup$ I see that $P^i(X)=I^{i+1}(X)$ but why if I take a general $S$-projective then it is $S$-injective? (and viceversa) $\endgroup$ – Nick Nov 6 '12 at 5:18

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