Timeline for Why does mathematics seem to have a polarity bias?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 18 at 3:26 | comment | added | KConrad | I wrote about missing the chance to use the term "variant functor" 10 years ago in a comment on the page mathoverflow.net/questions/175833/… | |
| Jan 9 at 19:12 | comment | added | David White | @DevSinha I have "coproduct vs product" in D, which is exactly "not more common." However, the example of coalgebras over a coring is one where every coalgebra is also an algebra, but not vice versa, so I put it in B. I guess your comment is about C, where I've got that products seem more intuitive. It's true that addition corresponds to coproduct, but I don't think students think about it that way till much later. | |
| Jan 9 at 19:00 | comment | added | Dev Sinha | As Will points out below, coproducts of sets are not less common. I've thought about the algebra/ coalgebra question and think of the answer here as algorithmic. To specify an algebra means giving a rule to produce something from two things. To specify a coalgebra means creating a "lookup" function - e.g. find all occurrences of 36 in the multiplication table. I'm no computer scientist, but I conjecture that if this can be made well-posed the lookup question will be more intensive. | |
| Jan 9 at 18:55 | comment | added | David White | I edited again to address Mike's most recent comment. | |
| Jan 9 at 18:55 | history | edited | David White | CC BY-SA 4.0 |
Edited again to address points raised by Mike Shulman.
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| Jan 9 at 5:17 | comment | added | Mike Shulman | I still think the comment about monoidal structures is misleading. I wouldn't say that you get a monoidal structure "for free" from a closed structure, since you can have a closed category that is not monoidal. Moreover, this is a bad example of (B), since the symmetry between monoidal structures and closed structures is fairly complete; either suffices to characterize the other, but neither suffices to construct the other. And I am at least doubtful of your claim about "most category theorists". | |
| Jan 9 at 4:50 | comment | added | Todd Wilcox | I hope it's ok if this answer gets more votes than any of the coanswers. | |
| Jan 8 at 13:54 | comment | added | David White | @MikeShulman I edited to address your comments. I wouldn't want to mislead any future readers, so added clarity about what I was trying to say. | |
| Jan 8 at 13:53 | history | edited | David White | CC BY-SA 4.0 |
Edited to address issues raised by Mike Shulman.
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| S Jan 8 at 10:32 | history | suggested | joriki | CC BY-SA 4.0 |
delete duplication
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| Jan 8 at 10:05 | review | Suggested edits | |||
| S Jan 8 at 10:32 | |||||
| Jan 8 at 8:20 | comment | added | Mike Shulman | Also a "covariant functor" is not the dual of a "variant functor" -- it is actually the non-co concept, with a "co-covariant functor" being popularly known as a "contravariant functor". | |
| Jan 8 at 8:19 | comment | added | Mike Shulman | I don't understand your comment about monoidal structures and internal homs. Either one can be characterized uniquely as an adjoint to the other, but may not exist: you can have a non-closed monoidal category or a non-monoidal closed category. | |
| S Jan 8 at 2:19 | history | suggested | CommunityBot | CC BY-SA 4.0 |
list formatting for accessibility
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| Jan 8 at 1:52 | review | Suggested edits | |||
| S Jan 8 at 2:19 | |||||
| Jan 8 at 1:12 | comment | added | Alison Miller | I'd say that the extent to which homology/cohomology is an example of B depends substantially on your approach. (And also there are cases like de Rham theory where there is a cohomology theory but no homology theory.) | |
| Jan 7 at 20:45 | comment | added | David White | @provocateur Yes. I don't think I wrote that there were "more examples" of the thing than the co-thing in A, but indeed the goal was to think about "more natural" or "more interesting." Sorry if the answer gave the wrong impression. However, for (B) there really could be "more examples" since to be a coalgebra implies being an algebra, plus more, for example. | |
| Jan 7 at 20:11 | comment | added | LSpice | I think that the comment about creating vs. destroying that you reference is @SeanSanford's. | |
| Jan 7 at 20:11 | comment | added | provocateur | For A), isn't it true though that each example can be translated into a co-example? The problem seems to be that we find the original examples more interesting and natural than the co-examples, not that there are more of one than the other. | |
| Jan 7 at 18:17 | history | answered | David White | CC BY-SA 4.0 |