Timeline for Does $\mathbf{Cat}$ have the Cantor–Schröder–Bernstein property?
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Aug 19, 2023 at 6:10 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 18, 2023 at 18:47 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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S Aug 18, 2023 at 11:20 | vote | accept | Tian Vlašić | ||
S Aug 18, 2023 at 11:20 | vote | accept | Tian Vlašić | ||
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Aug 17, 2023 at 21:37 | history | edited | Tian Vlašić |
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Aug 17, 2023 at 21:19 | vote | accept | Tian Vlašić | ||
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Aug 17, 2023 at 21:17 | history | edited | LSpice | CC BY-SA 4.0 |
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Aug 17, 2023 at 19:10 | vote | accept | Tian Vlašić | ||
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Aug 17, 2023 at 11:05 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 22:28 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 22:06 | answer | added | Tian Vlašić | timeline score: 3 | |
Aug 15, 2023 at 20:16 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 20:02 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 19:56 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 19:44 | comment | added | Tian Vlašić | Sorry, I meant bimorphism which is an epic embedding functor. | |
Aug 15, 2023 at 19:43 | history | edited | Tian Vlašić | CC BY-SA 4.0 |
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Aug 15, 2023 at 19:38 | answer | added | Joel David Hamkins | timeline score: 22 | |
Aug 15, 2023 at 19:35 | comment | added | Joel David Hamkins | The latter order has a minimal element, which gives an initial object in the category, but the former does not. (but I don't know anything about biomorphic, sorry...) | |
Aug 15, 2023 at 19:33 | comment | added | Tian Vlašić | Ok, so as I thought. I see how this might produce counterexamples. Embeddings should not be too hard to construct. However, how can we prove there is no biomorphic functor $\mathcal{F}: (\mathbb{Q}, \geq ) \rightarrow (\mathbb{Q}^{\geq 0}, \geq )$ or, for that matter, maybe $\mathcal{F}: (\mathbb{Q}^{\geq 0}, \geq ) \rightarrow (\mathbb{Q}, \geq )$? | |
Aug 15, 2023 at 19:29 | comment | added | Joel David Hamkins | The objects are the elements of the order, and whenever $x< y$, then there is precisely one morphism from $x$ to $y$, plus the identity morphisms. | |
Aug 15, 2023 at 19:26 | comment | added | Tian Vlašić | @JoelDavidHamkins Precisely how do you view ordered sets as categories? What are the morphisms? | |
Aug 15, 2023 at 19:19 | comment | added | Joel David Hamkins | Can't one take the standard violations of CSB for orders, say, and translate this to categories? For example, the orders $\langle\mathbb{Q},<\rangle$ and $\langle\mathbb{Q}^{\geq 0},<\rangle$ each order-embed into each other, but are not isomorphic. Viewing these orders as categories would seem to produce a similar counterexample of your desired form. | |
S Aug 15, 2023 at 19:10 | review | First questions | |||
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S Aug 15, 2023 at 19:10 | history | asked | Tian Vlašić | CC BY-SA 4.0 |