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In an additive category, given a morphism $f : A\to B$ let $\operatorname{im} f = \ker(\operatorname{coker}(f))$ and $\operatorname{coim} f = \operatorname{coker}(\ker(f))$, if they exist. If they do, then there is a unique morphism $\tilde f$ such that $f = \operatorname{im} f \circ \tilde f \circ \operatorname{coim} f$.

An additive category is "semi-Abelian", if every morphism a has a kernel and cokernel and the $\tilde f$ is mono and epi for all $f$.

I put 'semi-Abelian' in quotation marks because this definition conflicts with the on the nLab, but this definition is given in "Derived Functors in Functional Analysis" by J. Wengenroth, which was originally introduced by Palamodov in "Homological Methods in the Theory of Locally Convex Functions" (see J. Wengenroths comment).

Are "semi-Abelian" categories regular? If not do they have regular epi-mono factorizations?

"Regular" should be interpreted in the weakest sense possible in terms of limits, i.e. "pullbacks along strong epis exists".


This might be "obvious", but it isn't to me. A̶l̶s̶o̶,̶ ̶I̶'̶m̶ ̶a̶l̶s̶o̶ ̶i̶n̶t̶e̶r̶e̶s̶t̶e̶d̶ ̶i̶n̶ ̶t̶h̶e̶ ̶q̶u̶e̶s̶t̶i̶o̶n̶ ̶w̶i̶t̶h̶ ̶"̶r̶e̶g̶u̶l̶a̶r̶"̶ ̶r̶e̶p̶l̶a̶c̶e̶d̶ ̶b̶y̶ ̶"̶p̶r̶o̶t̶o̶m̶o̶d̶u̶l̶a̶r̶"̶,̶ ̶b̶u̶t̶ ̶I̶'̶m̶ ̶g̶o̶i̶n̶g̶ ̶t̶o̶ ̶p̶o̶s̶t̶p̶o̶n̶e̶ ̶t̶h̶a̶t̶ ̶q̶u̶e̶s̶t̶i̶o̶n̶ ̶f̶o̶r̶ ̶n̶o̶w̶.̶

EDIT: Nevermind this latter question: in a left exact pointed category, we have "additive = protomodular + all subobjects are normal" (Bourceux-Bourn Thm. 3.2.16). At least this covers all the cases where the category is left exact.

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    $\begingroup$ I took this notion from Palamodov's Homological Methods in the Theory of Locally Convex Functions [ Uspehi Mat. Nauk 26 (1971), no. 1(157), 3–65]. In a footnote on page 8 Palamodov writes that this terminology differs slightly from that used by Raikov (Semi-Abelian categories, Soviet Math. Dokl. 10 (1969), 1242-1245). $\endgroup$ Commented Jun 26, 2017 at 6:56
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    $\begingroup$ The proper title of the question should thus be "... in the sense of Palamodov". $\endgroup$ Commented Jun 26, 2017 at 7:01
  • $\begingroup$ @JochenWengenroth Thank you for clarifying this issue. $\endgroup$ Commented Jun 26, 2017 at 7:45

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