bio | website | mimuw.edu.pl/~mrp |
---|---|---|
location | Poland | |
age | ||
visits | member for | 3 years, 11 months |
seen | Dec 21 '14 at 23:47 | |
stats | profile views | 2,621 |
I eat vegetarians.
If you repeat ``p or not p'' often enough, it becomes the truth.
Dec 21 |
comment |
Cantor's theorem for presheaves?
@QiaochuYuan, you're right --- it's not obvious (i.e. what I have written is wrong), because there can be non-trivial compositions in $\mathbb{C}$. Yes, it is obvious if $\mathbb{C}$ is small (we can take the coproduct over the image of $F$, to get an object that by the assumption of essentially surjectivness would have a monomorphism from any object, what would lead to a contradiction). On the other hand, I don't think that there are foundational issues here --- what if we asked the same question about $\mathbf{FinSet}$ instead of $\mathbf{Set}$? |
Dec 21 |
comment |
Cantor's theorem for presheaves?
This means that there are at least $2^{|\mathbb{C}|}$ non-isomorphic objects in $\mathbf{Set}^{\mathbb{C}^{op}}$, so by classical Cantor's argument there can be no essentially surjective functor $\mathbb{C} \rightarrow \mathbf{Set}^{\mathbb{C}^{op}}$. |
Dec 21 |
comment |
Cantor's theorem for presheaves?
Todd, isn't this obvious? Let us consider sets $1$ and $2$. There are functions $! \colon 2 \rightarrow 1$ and $j \colon 1 \rightarrow 2$. Therefore any function from objects $|\mathbb{C}^{op}| \rightarrow \{1, 2\}$ uniquely extends to a functor $\mathbb{C}^{op} \rightarrow \mathbf{Set}$ lying over $!$ and $j$. Indeed, for the uniqueness, if $F, G \colon \mathbb{C}^{op} \rightarrow \mathbf{Set}$ are two such functors with $F \approx G$, then for every $A \in \mathbb{C}$ we have $F(A) = G(A)$. (cont...) |
Dec 15 |
comment |
Existence of internal toposes/inner models in a topos
Thanks, David, I'll take a look into the paper. But, anyway, I think it was Jean Benabou who first introduced the notion of an internal topos (BTW: I meant JEAN Benabou, of course. I can't believe I've written John. Shame on me...) |
Dec 15 |
revised |
Adjoint of simplicial left Kan extension
in the mentioned context the subscript 0 actually means "underlying set" |
Dec 15 |
comment |
Existence of internal toposes/inner models in a topos
David, what do you mean by "arithmetic universe" defined by Joyal? And what is your definition of an "internal topos"? Are your definitions of arithmetic universe and internal topos the same as given by John Benabou (and later described in "Problemes dans les topos")? |
Dec 14 |
answered | Adjoint of simplicial left Kan extension |
Dec 12 |
comment |
The (un)reasonable (non-)ubiquity of the Grothendieck construction
I am still confused. Are you interested in enriched functors $X \rightarrow \mathbb{V}$ or classical functors $C \rightarrow \mathbf{Cat}(\mathbb{V})$? |
Dec 12 |
comment |
Adjoint of simplicial left Kan extension
It is generally true that if $\mathcal{E}$ is cotensored and $R$ preserves cotensors, then the underlying adjunction $L \dashv R$ lifts to enriched adjunction (by classical Yoneda lemma). And your $R$ clearly preserves any limit that exists in $\mathcal{E}$. |
Dec 12 |
comment |
The (un)reasonable (non-)ubiquity of the Grothendieck construction
I do not understand how Definition 3.1 from the paper answers your question --- this definition deals with a completely trivial situation of a (lax) functor from a (classical) category to a (classical) 2-category. |
Dec 4 |
comment |
Map of adjunctions
@PiotrAchinger, thanks --- I haven't thought about the problem this way. Dimitri, good point, thanks. |
Dec 3 |
revised |
Map of adjunctions
Corrected typos |
Dec 3 |
revised |
Map of adjunctions
Added motivation |
Dec 3 |
asked | Map of adjunctions |
Nov 12 |
awarded | Nice Answer |
Oct 22 |
revised |
Completion of a category
Added missing "op" |
Oct 22 |
comment |
Completion of a category
@GlenMWilson, I've finally written the whole answer --- hope you'll find it interesting. |
Oct 22 |
revised |
Completion of a category
The second part of the disscussion |
Oct 22 |
awarded | Necromancer |
Oct 21 |
answered | Completion of a category |