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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.