bio | website | mimuw.edu.pl/~mrp |
---|---|---|
location | Poland | |
age | ||
visits | member for | 3 years, 4 months |
seen | 4 hours ago | |
stats | profile views | 2,181 |
I eat vegetarians.
If you repeat ``p or not p'' often enough, it becomes the truth.
Jul 18 |
comment |
opposite category
In fact I have the feeling that all of these generalizations work perfectly in the contexts, where we actually know how they should work, and rarely work outside of these well-known contexts. |
Jul 18 |
comment |
opposite category
@DimitriChikhladze, you have to substitute the category of internal discrete two-sided fibrations by the category of internal bimodules. But to be honest --- I don't think that this approach (or the approach of Mark Weber stated in my answer) is the right one. |
Jul 17 |
revised |
opposite category
Additional explanation |
Jul 17 |
answered | opposite category |
Jul 2 |
awarded | Curious |
May 6 |
answered | Non-continuous higher differentiability |
May 2 |
revised |
What is the co-form of Grothendieck construction?
some minor corrections |
May 2 |
revised |
What is the co-form of Grothendieck construction?
Connection with the other answer. |
May 2 |
comment |
What is the co-form of Grothendieck construction?
I didn't see your answer when I was writing mine --- I'll edit my answer to make a connection with yours. |
Apr 29 |
comment |
Generalizing indexed coproduct from $\mathrm{Set}$ to other monoidal categories
Because your grupoid $\mathbb{B}$ consists of all objects from $\mathbf{FinSet}$, and every functor preserves isomorphisms, this (strong) monoidal structure restricts to a monoidal structure on $\mathbb{B}$. Now, what kind of generalization are you looking for? Obviously, you may replace $\mathbf{FinSet}$ by any category with any (strong) monoidal structure, and everything will remain true. |
Apr 29 |
comment |
Generalizing indexed coproduct from $\mathrm{Set}$ to other monoidal categories
Brent, I have no idea of what you are trying to achieve, but I can tell you what you are doing. First, forget about $\mathbf{Set}$, because you are talking about finite sets. So, you have observed that the category of finite sets $\mathbf{FinSet}$ has finite products. Finite products in $\mathbf{FinSet}$ are generated by the (strong) monidal structure $\langle 1, \times \rangle$ on $\mathbf{FinSet}$. (cont) |
Apr 25 |
answered | What is the co-form of Grothendieck construction? |
Apr 5 |
revised |
An isomorphism of categories
Some minor corrections |
Apr 3 |
comment |
A question on the Grothendieck construction
Your quick candidate is almost the right one --- since you take a colimit in Cat, you cannot obtain a non-trivial 2-category. The right way is to postcompose $I^{op}\to Cat$ with the embedding $Cat \to 2Cat$ and take the lax colimit. This construction is sometimes called a "2-Grothendieck construction" (notice, that in case of 2-categories, there are four different Grothendieck constructions). |
Apr 3 |
answered | An isomorphism of categories |
Mar 31 |
comment |
An isomorphism of categories
@ZhenLin, thanks for the explanation :-) |
Mar 31 |
comment |
An isomorphism of categories
Second, how is $j$ defined? Does it send everything from $\mathrm{FinSet}$ to the terminal $1$ (what does $*$ mean?)? If so, then $j$ is terminal in $[\mathrm{FinSet}, C]$ and $j\circ f$ is terminal in $[I^{op}, C]$. Thus $[I^{op}, C]/{(j\circ f)} = [I^{op}, C]$ --- I do not understand how, in the general case, could $[I^{op}, C]$ be isomorphic to $[F^{op}, C]$. For example, if $I = 1$ then $F$ is just an arbitrary finite set. |
Mar 31 |
comment |
An isomorphism of categories
@ZhenLin, MaMing, I have no idea what you are talking about. First, there are typing errors in the question: a) an adjoint to $[F^{op},C] \rightarrow [I^{op},C]$ has type $[I^{op},C] \rightarrow [F^{op},C]$, b) you cannot apply $j$ to $f$, so perhaps you mean $j\circ f$, right?, c) it is not clear which category of elements you have in mind (i.e. you cannot compose $I^{op} \rightarrow C$ with $F \rightarrow I$; do you compose with the dual of the projection?). (cont) |
Mar 20 |
comment |
The most unexpected and/or the least natural category theory theorem?
Todd, some of my students are sometimes better than myself (so I wonder if it is not high time to find another job :-) But, what I was really trying to say is: if we think that something is unnatural (in category theory), then we have to rethink it and study harder. There are a lot of things in category theory that seems to be unnatural/unexpected to me, and the only reason why I refrain from posting about them is to not get comments like the first one under your post :-) |
Mar 20 |
comment |
The most unexpected and/or the least natural category theory theorem?
@PeterArndt, no --- you are given positive connectives and quantifications, so it suffices to define finite disjunctions in the internal logic. But this is easy when you can quantify over all propositions. |