For example, given an additive category $\mathcal{C}$ with a (both sided) multiplicative system S, the localization $S^{-1}\mathcal{C}$ is also additive. Yet to prove that the addition of morphisms in $S^{-1}\mathcal{C}$ is independent of the choice of roofs is lengthy which repeatedly uses Ore Condition and Cancellation to find a rather long common roof. This rather mundane kind of argument also appears in other similar proofs and so I'd like to know if there is a clean way to utilize Ore Condition and Cancellation to give a systematic treatment (or lemma) once for all when dealing with similar situations?
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$\begingroup$ I don't know how to answer your question. When I got into derived categories I thought this "calculus of fractions" thing was important, but then I never seem to have used it. But what's the point of this comment? If you only care about additivity, it's enough to show that the natural map from a coproduct to a product is an isomorphism. Maybe that helps, maybe you knew that already. $\endgroup$– bananastackCommented Sep 13, 2014 at 15:58
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$\begingroup$ In the case if you are interested in $S$ that do not necessarily satisfies the Ore condition, you may have a look at our paper arxiv.org/abs/1304.6059 (the arxiv will show the corrected version of this text in a few days). $\endgroup$– Mikhail BondarkoCommented Sep 13, 2014 at 19:46
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$\begingroup$ @user125763. I partially agree with you that it seems to me that calculus of fractions is only used in the very beginning to establish some basic theorems (properties), e.g. the one I mentioned in the question, and thereafter we just use these theorems. However, for every beginner who wants to build up some confidence in the general theory, the initial routine check is non-avoidable. I am just wondering if there is a neat and unified way of doing that when using Ore Condition and Cancellation. That's why I asked if there is some lemma which can be availed of when coming across such situations. $\endgroup$– W.Z.Commented Sep 14, 2014 at 4:37
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$\begingroup$ By the way, I am not so sure the fact that the natural map between binary coproducts and products is an isomorphism in an additive category can used to give rise to the independence of the choice of roofs when defining addition in question. $\endgroup$– W.Z.Commented Sep 14, 2014 at 4:39
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$\begingroup$ @W.Z. I'm not sure it is "non-avoidable" as you write, but I see your point. Also, regarding your last comment, I am sure whatever you proof one finds is bound to be equivalent to any other. I just pointed that fact out as having a different perspective sometimes illuminates the matter. I didn't claim anything deep or meaningful (and that's partially also why I did not write it as an answer). Cheers! $\endgroup$– bananastackCommented Sep 14, 2014 at 4:44
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