Timeline for Does $\mathbf{Cat}$ have the Cantor–Schröder–Bernstein property?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2023 at 11:20 | vote | accept | Tian Vlašić | ||
| Aug 17, 2023 at 21:19 | vote | accept | Tian Vlašić | ||
| Aug 18, 2023 at 11:20 | |||||
| Aug 16, 2023 at 8:21 | comment | added | Peter LeFanu Lumsdaine | @TianVlašić: Given a map $f \newcommand{\Q}{\mathbb{Q}} : \Q_{\geq 0} \to \Q$ (a functor between them as categories, or equivalently, a monotone map of orders), clearly the image of $f$ must lie within $\Q_{\geq f(0)} \subsetneq \Q$. We expect this to mean that $f$ can’t be epi; and to justify this expectation, take $g, g'$ to be any maps/functors out of $\Q$ that agree on all arguments $x \geq f(0)$ but differ on some $x < f(0)$, e.g. $g = \mathrm{id} : \Q \to \Q$, and $g'(x) = x$ for $x \geq f(0)$ but $g'(x) = x-1$ for $x < f(0)$. Then $gf = g'f$ but $g \neq g'$, so $f$ is not epi. | |
| Aug 15, 2023 at 22:43 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 17 characters in body
|
| Aug 15, 2023 at 22:09 | comment | added | jdc | @user2275150: Presumably that the restrictions of $g, g'$ to the set $\{q \in \mathbb Q:q \geq f(0)\}$ agree (but the values on rationals less than $f(0)$ are not the same). | |
| Aug 15, 2023 at 20:09 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 239 characters in body
|
| Aug 15, 2023 at 20:02 | comment | added | Peter LeFanu Lumsdaine | [cont’d] If you already know how to recognise epis/monics the category of preorders (or posets, or linear orders) then you can skip the details of my previous comment, by noting that the functors from these categories to $\mathrm{Cat}$ are faithful functors, so they reflect monos/epis, and hence any bimorphism between these orders as categories would be a bimorphism between them as orders. | |
| Aug 15, 2023 at 20:01 | comment | added | Joel David Hamkins | Thanks , Peter. | |
| Aug 15, 2023 at 19:59 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 24 characters in body
|
| Aug 15, 2023 at 19:58 | comment | added | Peter LeFanu Lumsdaine | @TianVlašić: In fact, Joel’s counterexample answers both parts of your question: these two orders are not connected by a bimorphism in $\mathrm{Cat}\newcommand{\Q}{\mathbb{Q}}$. The easiest way to see this is probably just by hand. First, for any $f : \Q_{\geq 0} \to \Q$, take two maps $g,g':\Q \to \Q$ which agree above $f(0)$ but differ below it; these show $f$ cannot be epi. Conversely, any epi $e : \Q \to \Q_{\geq 0}$ must hit $0$, by a similar argument; then taking some preimage $x$ of $0$, and some $x' < x$, the maps $1 \to \Q$ picking out $x$ and $x'$ show $e$ is not monic. | |
| Aug 15, 2023 at 19:50 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 16 characters in body
|
| Aug 15, 2023 at 19:45 | comment | added | Joel David Hamkins | Sure, I did this. | |
| Aug 15, 2023 at 19:44 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 16 characters in body
|
| Aug 15, 2023 at 19:42 | comment | added | Tian Vlašić | Could you please remark that this answers the first part of my question? | |
| Aug 15, 2023 at 19:38 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |