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1. • CommentRowNumber1.
   • CommentAuthorTim_Porter
   • CommentTimeMar 7th 2010
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   Author: Tim_Porter
   Format: MarkdownAt present in this entry, the arrows to be inverted are called 'weak equivalences'. In [[proper homotopy theory]], there is a calculus of fractions where the arrows to be inverted are cofinal inclusions. There is thus a slight terminological clash, as the cofinal inclusions are in no way 'weak equivalences'. It is also the case that in say Gabriel's localisation theory the morphisms to be inverted are not really weak equivalences in any non-contrived way. Suggestions as to how to change things a smidgin to get around this.

   At present in this entry, the arrows to be inverted are called 'weak equivalences'. In proper homotopy theory, there is a calculus of fractions where the arrows to be inverted are cofinal inclusions. There is thus a slight terminological clash, as the cofinal inclusions are in no way 'weak equivalences'. It is also the case that in say Gabriel's localisation theory the morphisms to be inverted are not really weak equivalences in any non-contrived way. Suggestions as to how to change things a smidgin to get around this.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 9th 2010
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   Author: DavidRoberts
   Format: MarkdownYou could import terminology and call them quasi-isomorphisms, or something similar, like pseudoisomorphisms.

   You could import terminology and call them quasi-isomorphisms, or something similar, like pseudoisomorphisms.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJan 14th 2011
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   Author: Mike Shulman
   Format: MarkdownItexA month ago I noticed that the third condition in the definition of [[calculus of fractions]] was wrong (the source and target of a morphism made no sense for what it was being composed with), so I made the smallest change that made it at least meaningful. Today I finally got around to looking up the actual definition, and I *think* that the whole thing was reversed from how it should be to match the second condition, so I fixed it. Could anyone more familiar with the notion verify that it is now correct, or that it is now wrong?

   A month ago I noticed that the third condition in the definition of calculus of fractions was wrong (the source and target of a morphism made no sense for what it was being composed with), so I made the smallest change that made it at least meaningful. Today I finally got around to looking up the actual definition, and I think that the whole thing was reversed from how it should be to match the second condition, so I fixed it. Could anyone more familiar with the notion verify that it is now correct, or that it is now wrong?

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 14th 2011
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   Author: DavidRoberts
   Format: MarkdownItexIt's now correct. Thanks for picking that up.

   It’s now correct. Thanks for picking that up.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeJan 15th 2011
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   Author: Mike Shulman
   Format: MarkdownItexThanks!

   Thanks!

6. • CommentRowNumber6.
   • CommentAuthorZhen Lin
   • CommentTimeOct 1st 2013
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   Author: Zhen Lin
   Format: MarkdownItexThe article says that the left/right convention is the opposite of the one used by Gabriel and Zisman, but as far as I can tell the convention is actually the _same_. Rather, it is that GZ focuses on calculi of left fractions, because the class of anodyne extensions in the Hurewicz-style homotopy category of $sSet$ admits a calculus of left fractions.

   The article says that the left/right convention is the opposite of the one used by Gabriel and Zisman, but as far as I can tell the convention is actually the same. Rather, it is that GZ focuses on calculi of left fractions, because the class of anodyne extensions in the Hurewicz-style homotopy category of $sSet$ admits a calculus of left fractions.

7. • CommentRowNumber7.
   • CommentAuthorKarol Szumiło
   • CommentTimeOct 1st 2013
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   Author: Karol Szumiło
   Format: MarkdownItexYou are right. Left fractions are cospans an right fractions are spans, so the conventions indeed agree. I have removed the misleading remark. I have also corrected an example that said that in the quotient of a category of fibrant objects by the homotopy relation acyclic fibrations satisfy the left calculus of fractions. It is actually the right calculus of fractions. Moreover, I have changed "acyclic fibrations" to "weak equivalences" since I believe that this is a more natural way of thinking about this example, especially that in the quotient the distinction between fibrations and non-fibrations disappears.

   You are right. Left fractions are cospans an right fractions are spans, so the conventions indeed agree. I have removed the misleading remark. I have also corrected an example that said that in the quotient of a category of fibrant objects by the homotopy relation acyclic fibrations satisfy the left calculus of fractions. It is actually the right calculus of fractions. Moreover, I have changed “acyclic fibrations” to “weak equivalences” since I believe that this is a more natural way of thinking about this example, especially that in the quotient the distinction between fibrations and non-fibrations disappears.

8. • CommentRowNumber8.
   • CommentAuthorzskoda
   • CommentTimeOct 1st 2013
   • (edited Oct 1st 2013)
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   Author: zskoda
   Format: MarkdownItex> Left fractions are cospans an right fractions are spans, so the conventions indeed agree. Depending what you call left: assuming either the Leibniz or counterLeibniz convention for the composition in a category (to match with left and right fractions for monoids and rings). In ring theory $s^{-1}t$ is a left fraction (no hesitation in all literature). If you assume that the composition is the operation (say, for an additive category with one object) then (in Leibniz convention, which is more standard nowdays) $s^{-1} \circ t$ is corresponding to the __cospan__ $(t,s)$, and the right fraction $t\circ s^{-1}$ corresponds to the __span__ $(s,t)$. Of course, among those who use concatenation notation for composition (or ; symbol, like in the category theory for computer science), the counterLeibniz notation is common. The mnemo rule for what is right Ore and right cancelability is that one has universal and existential quantifiers. Those variables which come with _existential_ quantifiers are on the right (in Leibniz order, or in ring context), for both "right" conditions. Say in ring context, to avoid talking composability (domains and codomains), a multiplicative set $S\subset R$ is right Ore if $\forall r\in R\forall s\in S\exists r'\in R\exists s'\in S, s r'=r s'$ and $\forall s\in S\forall n\in R$ $s n = 0\implies \exists s'\in S, n s' = 0$.

   Left fractions are cospans an right fractions are spans, so the conventions indeed agree.

   Depending what you call left: assuming either the Leibniz or counterLeibniz convention for the composition in a category (to match with left and right fractions for monoids and rings). In ring theory $s^{-1}t$ is a left fraction (no hesitation in all literature). If you assume that the composition is the operation (say, for an additive category with one object) then (in Leibniz convention, which is more standard nowdays) $s^{-1} \circ t$ is corresponding to the cospan $(t,s)$, and the right fraction $t\circ s^{-1}$ corresponds to the span $(s,t)$. Of course, among those who use concatenation notation for composition (or ; symbol, like in the category theory for computer science), the counterLeibniz notation is common.

   The mnemo rule for what is right Ore and right cancelability is that one has universal and existential quantifiers. Those variables which come with existential quantifiers are on the right (in Leibniz order, or in ring context), for both “right” conditions. Say in ring context, to avoid talking composability (domains and codomains), a multiplicative set $S\subset R$ is right Ore if $\forall r\in R\forall s\in S\exists r'\in R\exists s'\in S, s r'=r s'$ and $\forall s\in S\forall n\in R$ $s n = 0\implies \exists s'\in S, n s' = 0$.