bio | website | mimuw.edu.pl/~mrp |
---|---|---|
location | Poland | |
age | ||
visits | member for | 4 years, 4 months |
seen | Jun 25 at 10:00 | |
stats | profile views | 2,768 |
I eat vegetarians.
If you repeat ``p or not p'' often enough, it becomes the truth.
Jun 25 |
comment |
Van Kampen colimits
Thank you for putting this into answer. |
Jun 25 |
accepted | Van Kampen colimits |
Jun 24 |
comment |
Van Kampen colimits
@JoonasIlmavirta, I am afraid some people are overenthusiastic about the moderation toolbox. Yes, I wanted to remove the question from both "unanswered" and "open" question lists. In fact, I would have deleted the question itself, if it were not for Marc Hoyois, who offered me his time. I think it is fair to keep the track of the conversation. On the other hand, it is clear that my question based on some terminological confusion and therefore, strictly speaking, it cannot be answered --- because there can be no answer to a no-question. |
Jun 23 |
comment |
Van Kampen colimits
If you post your observations as an answer than I will be happy to accept it. |
Jun 23 |
comment |
Van Kampen colimits
@ZhenLin, MarcHoyois --- that's right. Somehow I thought that if you strictify a pseudofunctor, then the concept of limits strictifies automatically. But it seems that it strictifies only in one direction. Hm... I just have unlearned something... |
Jun 22 |
awarded | Cleanup |
Jun 22 |
revised |
Van Kampen colimits
rolled back to a previous revision |
Jun 22 |
comment |
Van Kampen colimits
@MarcHoyois, no, wait, I'm having hard moments today --- we are talking about discrete colimits, which are the same as in Cat (since the inclusion Set ---> Cat is left adjoint). So, what is your counterexample? BTW, we do not have to talk about 2-limits in Cat, because every limit in Cat is automatically a 2-limit. |
Jun 22 |
revised |
Van Kampen colimits
Error spotted by Marc Hoyois in the question |
Jun 22 |
comment |
Van Kampen colimits
@MarcHoyois, right --- because colimits in Set are not the same as colimits in Cat. Silly me :-) |
Jun 22 |
asked | Van Kampen colimits |
Jun 9 |
revised |
A cosmos where coproduct injections are not monic
Fixed some grammar and typos |
Jun 9 |
comment |
A cosmos where coproduct injections are not monic
Dimitri, if you have a moment, please, write up your example in full detail --- just for the sake of having a reference to the construction (I suspect it may be a good instructional example). |
Jun 7 |
revised |
A cosmos where coproduct injections are not monic
Added even more explicit example |
Jun 7 |
answered | A cosmos where coproduct injections are not monic |
Jun 5 |
comment |
A cosmos where coproduct injections are not monic
Hi Mike, just two quick notes (till I find more time...): a) I believe it suffices to assume distributivity (rather than extensivity)to prove that coproduct's injections are mono, b) do not categories of Chu spaces serve as obvious examples, or am I missing something?... |
Mar 7 |
awarded | Yearling |
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...) |