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Suppose you have an abelian category $\bf A$, and $A\to B\to C$, $A'\to B'\to C'$ two exact sequences, in a diagram $$ \begin{array}{cccccccc} 0 &\to & A &\to& B &\to& C &\to & 0\\ &&\downarrow && \downarrow && \downarrow \\ 0 &\to & A' &\to& B' &\to& C' &\to & 0 \end{array} $$ (i.e., suppose you have a morphism of exact sequences $(f,g,h)$); let $(\cal E,M)$ be a factorization system on $\bf A$, and suppose $f,h$ lie in $\cal E$.

Under which conditions (on the category, or on the factorization systems) $g\in\cal E$? Same question for $f,h\in\cal M$ $\Rightarrow g\in\cal M$.

  1. I'm not assuming that $(\cal E,M)$ is proper (so there can be no relation between $(\cal E,M)$ and $(Epi, Mono)$);
  2. For the motivating example I have in mind with this question, assuming something more on the factorization system is more natural than assuming it for the category.
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1 Answer 1

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One condition on the factorization system that should work is that "factorizations preserve exact sequences", i.e. if you have a map between exact sequences and you $(E,M)$-factor it componentwise, then the intermediate objects also form an exact sequence. If this is true, then you can factor your given map of exact sequences and apply the ordinary 5-lemma, using the fact that $f\in E$ iff the $M$-part of its $(E,M)$-factorization is an isomorphism.

However, this assumption may not be much easier to check in an example than to check your desired conclusion directly! (-:

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  • $\begingroup$ Thank you! Is a class of OFS satisfying this "exactness" property already known? I am in particular working in the context of simple and semi-left-exact factorization systems (where there is a functorial, canonical procedure to produce the factorization). $\endgroup$
    – fosco
    Commented Mar 1, 2016 at 13:37
  • $\begingroup$ I don't know, I've never encountered it before. $\endgroup$ Commented Mar 1, 2016 at 16:17

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