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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
Aug 15, 2023 at 19:12
S Aug 15, 2023 at 19:10 history asked Tian Vlašić CC BY-SA 4.0