Timeline for Example of a non-closed cocomplete symmetric monoidal category
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2013 at 20:28 | comment | added | Todd Trimble | @Buschi Sergio: that type of thing occurred to me too, although the analysis is probably more involved than what I wrote above. You might look at the paper by Day and Lack: arxiv.org/abs/math/0610439. Around page 13 they consider positive conditions for when categories of small presheaves are closed (they are looking a bit more generally at the enriched case). | |
| Jan 5, 2013 at 19:49 | comment | added | Buschi Sergio | Let $\mathcal{C}$ a category. Let $\mathcal{C}^>$ the category of presheaf on $\mathcal{C}$ and call $P\in \mathcal{C}^>$ a p-presheaf if it is isomorphic to a finite product of representable. Consider the full subcategory $\mathcal{A}\subset \mathcal{C}^>$ of the $P\in \mathcal{C}^>$ such that exist a epimorphism $e: \sum_{i\in I} P_i \to P$ from a (small) coproduct of p-presheaf to $P$. The inclusion $\mathcal{A}\subset \mathcal{C}^>$ create colimits and finite product, then $\mathcal{A}$ is cartesian monoidal and cocomplete (by product too). Is $\mathcal{A}$ closed? | |
| Jan 5, 2013 at 16:38 | vote | accept | Martin Brandenburg | ||
| Jan 5, 2013 at 16:38 | comment | added | Martin Brandenburg | Alright, now it makes sense. Thank you again! | |
| Jan 5, 2013 at 16:31 | comment | added | Todd Trimble | ... you can define objects and morphisms of a category isomorphic to the $V_+$ as I described it as elements of $V$, and define rules for composition, etc., etc. Just think of this as defining a partial order $V'$ in some way so that $V'$ has a terminal object $\top$ and the sub-partial order given by the complement of $\top$ is isomorphic to $V$ under the subset inclusion relation; you don't have to think of the partial order on $V'$ as literal subset inclusion itself (which is where I think the confusion arose). | |
| Jan 5, 2013 at 16:26 | comment | added | Todd Trimble | Well, one just works with isomorphs instead, using various ugly hacks. So instead of defining objects to be elements $x$ of $V$, define them to be e.g. ordered pairs $(\emptyset, x)$ where $x$ ranges over elements of $V$, and define $1$ to be something dumb like {$\emptyset$}. And define morphisms to be elements of the disjoint union of two classes where the first class consists of ordered triples $(1, x, y)$ such that $x \subseteq y$, and where the second class consists of, I don't know, elements of $(2, x)$ (which are supposed to stand for arrows $x \leq 1$). Under some such kludgy coding... | |
| Jan 5, 2013 at 16:12 | comment | added | Martin Brandenburg | A) Ok. B) My question was exactly why we can define this category in $V$. Doesn't this contradict again Cantor? | |
| Jan 5, 2013 at 15:38 | comment | added | Todd Trimble | Maybe I should have written $1 \wedge X$ instead of $1 \cap X$ in my last comment. The point is that $V_+$ as defined has finite meets, which play the role of cartesian products. | |
| Jan 5, 2013 at 15:33 | comment | added | Todd Trimble | $Y$ can't be any set; if $Y$ contains $X$ then we can show $Y^X$ is the top element $1$ (since clearly $1 \cap X \subseteq Y$, and $1$ is the maximal element). But you're right that we don't need $Y \subset X$ precisely; I think all we need is that $X$ contains some elements not belonging to $Y$. As for question (B), what's the problem? I'm simply defining a category whose objects are elements of $V$ plus an extra object $1$; we have a (unique) morphism $x \to y$ if either $x \subseteq y$ in $V$ or if $y = 1$. (It's easy enough to code up all the data in $V$, if you insist on this.) | |
| Jan 5, 2013 at 13:21 | comment | added | Martin Brandenburg | Thank you for this example, which is in fact quite amusing and shows how pathological non-accessible categories can be. Two questions: A) We don't need $Y \subset X$ for your argument, right? $Y$ could be any set, even $X$ or empty. B) Isn't there a problem with the construction of $V_+$? Namely, by a $V$-category $C$, I mean a subset $\mathrm{Mor}(C)$ of $V$ together with certain operations and properties. But why should $\mathrm{Mor}(V_+)$ be a subset of $V$? If this is not the case, we could choose a bigger universe $V'$ and consider $V^+$ as a $V'$-category, but it won't be cocomplete. | |
| Jan 5, 2013 at 4:50 | history | undeleted | Todd Trimble | ||
| Jan 5, 2013 at 4:50 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 18 characters in body; added 62 characters in body
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| Jan 5, 2013 at 4:38 | history | deleted | Todd Trimble | ||
| Jan 5, 2013 at 4:36 | history | answered | Todd Trimble | CC BY-SA 3.0 |