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In the derived category (i.e. the localisation of the homotopy category of chain complexes at all quasi-isomorphisms) $D(\mathcal{A})$ we know that morphisms can be thought of as roofs, or spans:

enter image description here

This is confirmed in, e.g. Dimca's Sheaves in Topology.

However, when we localise a category (at a right-multiplicative set, say) the morphisms are instead cospans (or coroofs, I guess?):

enter image description here

This is confirmed in, e.g. Schapira's Categories and homological algebra: an introduction to derived categories.

Question: why do we have this incongruence? Is it because Schapira considers a right-multiplicative set instead of a left one? Or is there something going on instead with the fact that Dimca uses cohomologically-graded cochain complexes (i.e. the differential increases degree)? Or is there something much deeper here, i.e. an actual fundamental difference.

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When a category $\mathcal{C}$ admits a calculus of right fractions with respect to $\mathcal{S}$, the localization $\mathcal{C} [\mathcal{S}^{-1}]$ may be described in terms of equivalence classes of spans. When it admits a calculus of left fractions, the localization may be described via equivalence classes of co-spans. When it admits both, the resulting constructions coincide. For details, see https://ncatlab.org/nlab/show/calculus+of+fractions

The homotopy category $K (\mathcal{A})$ with respect to quasi-isomorphisms admits both a calculus of right and left fractions.

Homological and cohomological situation is essentially the same, as one only changes the numbering.

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  • $\begingroup$ ok, that makes sense, thanks! so is there some sort of natural correspondence between the spans and the cospans? i.e. can we associate a span to every cospan or something similar? is there a geometric way of thinking about this? (these are all bonus questions, so don't feel you have to respond if you don't want!) $\endgroup$ – Tim Dec 7 '16 at 19:18
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    $\begingroup$ To turn a left fraction into a right fraction, one uses the "Ore condition", the one that involves a commutative square. This in fact gives a unique equivalence class. $\endgroup$ – user144221 Dec 7 '16 at 19:31

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