Timeline for A cosmos where coproduct injections are not monic
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8, 2015 at 17:16 | vote | accept | Mike Shulman | ||
| Jun 8, 2015 at 17:08 | comment | added | Mike Shulman | @BuschiSergio Yes, I do. CjSLat has monic injections, since it has zero morphisms. Are you suggesting that PFrm doesn't? | |
| Jun 7, 2015 at 12:05 | comment | added | Buschi Sergio | Do you know the categories CJSLat and PFRm in "Sketches of an Elephant.." by P. Johnstone Vol.II p. 475 p.1.1.5 ? | |
| Jun 7, 2015 at 9:23 | answer | added | Michal R. Przybylek | timeline score: 8 | |
| Jun 6, 2015 at 22:56 | answer | added | Dimitri Chikhladze | timeline score: 4 | |
| Jun 6, 2015 at 18:59 | comment | added | Mike Shulman | I don't know anything about Chu spaces. If they are an example, please explain in an answer! | |
| Jun 5, 2015 at 23:07 | comment | added | Michal R. Przybylek | 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?... | |
| Jun 5, 2015 at 3:40 | comment | added | Mike Shulman | @DimitriChikhladze Good question! (-: What are some examples of commutative algebraic theories without a zero operation (since then there would be zero morphisms)? | |
| Jun 5, 2015 at 3:36 | comment | added | Mike Shulman | @EricWofsey Nice! I can't think offhand of an example of a complete and cocomplete category such that coproduct injections are neither monic in $C$ nor in $C^{\mathrm{op}}$, can you? | |
| Jun 4, 2015 at 18:20 | comment | added | Dimitri Chikhladze | There is a closed monoidal structure on a category of algebras over a commutative algebraic theory. Do these categories always have that property? | |
| Jun 4, 2015 at 0:15 | comment | added | Eric Wofsey | Since coproduct injections are monic in $\mathcal{C}^{op}$, $[A\coprod B,A]=[A,A]\times[B,A]\to [A,A]$ is epic, so the two maps $[A,A]\to [P,A]$ are equal. But this implies the original maps $P\to A$ were equal, since they are adjoint to the composition of the maps $[A,A]\to[P,A]$ with the map $1\to[A,A]$ that is adjoint to the identity on $A$ (where $1$ is the monoidal unit). | |
| Jun 4, 2015 at 0:13 | comment | added | Eric Wofsey | More generally, here's something that rules out a lot of potential counterexamples: if coproduct injections are monic in $\mathcal{C}^{op}$ but not in $\mathcal{C}$, then $\mathcal{C}$ admits no closed monoidal structure. To show this, suppose a coproduct injection $A\to A\coprod B$ coequalizes two parallel arrows $P\to A$, and apply the functor $[-,A]$ to everything, where $[-,-]$ is the internal hom of the closed monoidal structure. | |
| Jun 3, 2015 at 23:40 | comment | added | Karol Szumiło | For what it's worth, $\mathrm{Set}^{\mathrm{op}}$ has no closed monoidal structure. Just by playing with the adjunction you can see that the unit of such a structure has to be a finite set and, if it has $m$ elements, then the number of functions $X \to Y$ would have to be a power of $m$ for any two finite sets $X$ and $Y$. | |
| Jun 3, 2015 at 22:44 | comment | added | Mike Shulman | @EricWofsey No, I don't. (You're right, $\mathrm{Set}^{\mathrm{op}}$ is also a fine example, e.g. $\emptyset \times X \to X$ is not epic if $X$ is nonempty.) So if you could give a closed monoidal structure on either of them, that would answer the question. | |
| Jun 3, 2015 at 22:21 | comment | added | Eric Wofsey | Do you know that commutative rings (or, for that matter, $Set^{op}$) do not admit any closed monoidal structure? | |
| Jun 3, 2015 at 22:12 | history | asked | Mike Shulman | CC BY-SA 3.0 |