#10963: Axioms and more functorial constructions
-------------------------------------+-------------------------------------
Reporter: nthiery | Owner: stumpc5
Type: enhancement | Status: needs_info
Priority: major | Milestone: sage-6.2
Component: categories | Resolution:
Keywords: days54 | Merged in:
Authors: Nicolas M. Thiéry | Reviewers: Simon King, Frédéric
Report Upstream: N/A | Chapoton
Branch: | Work issues: merge with #15801
public/ticket/10963-doc- | once things stabilize
distributive | Commit:
Dependencies: #11224, #8327, | f2ceccfb4fbd4cea8daf58c3294af98133ba1b23
#10193, #12895, #14516, #14722, | Stopgaps:
#13589, #14471, #15069, #15094, |
#11688, #13394, #15150, #15506, |
#15757, #15759, #15919 |
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Comment (by darij):
Replying to [comment:589 nthiery]:
> Replying to [comment:572 darij]:
> > I'll try to think of an example with quotients (that does not involve
schemes...). For direct products, for example the direct product of two
groups in the category of groups is their free product (
http://en.wikipedia.org/wiki/Free_product ), while the direct product of
two abelian groups in the category of abelian groups is their cartesian
product. If you have two abelian groups A and B, their direct product as
groups is not the same as their direct product as monoids, not even as a
monoid!
>
> Yes! That's precisely why there is basically nothing about "direct
products" in
> the category infrastructure: the fact that it's not covariant (aka
> uniform across categories) means there is nothing we can share
> automatically. This is unlike cartesian products or quotients!
>
> Just to state the obvious about the covariance for quotients (or
> equivalently homomorphic images): take B a subcategory of A, and Y an
> homomorphic image in B; since B-morphisms are A-morphisms, Y is also
> an homomorphic image in A).
OK, now I see what you mean by "covariant" :) The problem is that the
quotient is a certain *universal* homomorphic image, and a quotient in B
(despite still being a homomorphic image in A) will not always have this
universality property in A.
I'm still at a loss for a good counterexample; these are rare, but
fundamentally there is no difference between quotients and direct products
here: neither is more universal than the other. There are some easy
counterexamples if your notion of a "quotient" includes coequalizers (not
just cokernels, which are the narrowest reading of the word "quotient")
and your notion of a "subcategory" is the wider sense in which you are
using this word (i.e., rings form a subcategory of additive groups).
{{{
Indeed, take two ring homomorphisms from ZZ[x] to
ZZ[x,y], the first one sending x to x, the second
one sending x to y. Their coequalizer (roughly
speaking, "ZZ[x,y] modulo identifying images under
the first homomorphism with corresponding images
under the second") is
- the additive group ZZ[x,y] modulo the additive
subgroup <x-y, x^2-y^2, x^3-y^3, ...>, if you
are working in the category of additive groups;
- the ring ZZ[x,y] modulo the ideal generated by
x-y, x^2-y^2, x^3-y^3, ..., if you are working
in the category of rings.
The former is bigger than the latter, because e.g.
the difference x^2y^3-x^3y^2 is in the ideal but
not in the additive subgroup.
}}}
> Fair enough. Let it be Christmas! Done.
Thanks a lot!
> Oh, I see your point now. Ok, that can be easily
> improved. Done. Please check!
Cleared up, thank you.
> Ah, interesting, you are the first one to raise this point :-) So far,
> I was taking for granted that it was natural to consider the
> additional structure as being "encoded" in the morphisms. Ok, we
> probably need to clarify this, even though we are just following the
> terminology that is in use in other systems like Axiom, ...
I fear this will come up in practice a lot, e.g., when you consider Lie
algebras which have a grading and a filtration, and the filtration is
*not* the one you would get from the grading. (These appear everywhere.)
So I'm not very happy to have "graded Xes" as a subcategory of "Xes"...
Anyway I've added some doc now. Is it on the right track?
> Yes indeed! One of the very next steps is precisely to improve the
> support for morphisms (see #10668). And without the full subcategory
> business, there is not much homsets stuff to inherit and share between
> categories.
Well... any operation on the homset of a big category is also an operation
on the homset of any subcategory, if "subcategory" is understood in the
mathematical sense. Maybe by saying "full subcategory" you really mean
"subcategory in the mathematical sense", i.e. subcategory without
additional data?
> The point is that often, in practice, what's useful in similar
> situations is actually the full subcategory, even if mathematically
> speaking the natural category is not the full one. For example, for
> ModulesWithBasis (modules with a distinguished basis), we definitely
> want to use the distinguished bases to compute with morphisms. But
> restricting to morphisms that preserve the distinguished bases would
> be really boring.
Good point -- this is probably the right statement to make about full
subcategories.
> Done. A full note actually, inspired by the discussion above. Please
check.
Well explained. (No, I'm not saying this is something that needs to be
fixed; only documented. I don't see much of an advantage in having
operation-polymorphic axioms.)
Best regards,\\
Darij
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Ticket URL: <http://trac.sagemath.org/ticket/10963#comment:603>
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