bio | website | |
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visits | member for | 4 years, 11 months |
seen | Sep 27 at 11:23 | |
stats | profile views | 788 |
Sep 17 |
comment |
Categorical definition of the ideal product within the category of rings
No. As Martin pointed out in his question, the pushout/tensor product corresponds to the sum of ideals. |
Aug 20 |
accepted | In the category of sets epimorphisms are surjective - Constructive Proof? |
Aug 19 |
comment |
In the category of sets epimorphisms are surjective - Constructive Proof?
Thanks for the many answers; I liked this one best: I'm actually not working in Set but rather in Enriched Categories so this should be a proof that might translate to my setting. |
Aug 18 |
asked | In the category of sets epimorphisms are surjective - Constructive Proof? |
Jul 2 |
awarded | Curious |
Apr 4 |
comment |
A question on the Grothendieck construction
My intuition is that your intuition is right - depending on the grothendieck construction used you'd end up with lax/colax variants of a 2-coend. I'll have to think about this a bit: There's actually two grothendieck constructions for such distributors: mathoverflow.net/questions/128393/… |
Feb 28 |
awarded | Good Question |
Feb 4 |
comment |
What spaces can be obtained from $\mathbb{R}^{n}$ by taking quotient spaces and subspaces?
Concerning your friends question: Take a look at Bill Lawvere's answer in mathoverflow.net/questions/127841/… "My paper about Volterra's functionals [...] discusses the unsuitability for functional analysis (as well as for homotopy theory) of the attempt to characterize continuity or cohesion using open sets or other contravariant structure." |
Jan 24 |
comment |
Generalisation of the Grothendieck construction for presheaves as a lax pullback
Ah sry; misread the question then. |
Jan 23 |
revised |
Generalisation of the Grothendieck construction for presheaves as a lax pullback
added 99 characters in body |
Jan 23 |
answered | Generalisation of the Grothendieck construction for presheaves as a lax pullback |
Nov 26 |
comment |
Recovering measure from the family of sigma-subalgebras
This seems to be a homework question: However, i guess this is possible and straightforward if $\mathcal F$ is generated (as a $\sigma$-agebra) by the union $\bigcup\mathcal F_\alpha$ and furthermore the family $\mathcal F_\alpha$ is compatible in the following sense: If $A\in\mathcal F_\alpha$ and $B\in\mathcal F_\beta$ then is $A\cap B$ is in both $\mathcal F_\alpha$ and $\mathcal F_\beta$. |
Nov 26 |
answered | About (co)limits of accessible categories |
Nov 25 |
awarded | Citizen Patrol |
Nov 21 |
comment |
Kan extensions and the yoneda embedding.
Interesting: $f_!(\{x\})$ is allways the complement of the image $f(X)$ except when $x$ is the only element in the respective fiber of $f$. So in general $f_!$ should maybe say something about the triviality of fibers. Let's see where this leads for general quasiorders aka 2-enriched categories. Here $\{x\}$ needs to be replaced with the respective down-set. |
Nov 14 |
comment |
Ends and Coends - Analogues for higher arity - Horn Filling
I thought about it but could not come up with a 3-ary replacement for the hom. |
Nov 14 |
revised |
Ends and Coends - Analogues for higher arity - Horn Filling
Fixed the latex formating |
Nov 14 |
revised |
Ends and Coends - Analogues for higher arity - Horn Filling
Modified: Title |
Nov 14 |
asked | Ends and Coends - Analogues for higher arity - Horn Filling |
Oct 28 |
awarded | Yearling |