Timeline for Relations with "for each" composition and its properties (coming from profunctors with end composition)
Current License: CC BY-SA 4.0
13 events
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| May 3 at 16:35 | history | edited | Cole Comfort | CC BY-SA 4.0 |
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| May 3 at 16:35 | comment | added | Cole Comfort | My apologies, I didn't read Paul's comment properly. I didn't mean to cause any insult. I was just very sloppy. @Emily In Set-enriched profunctors, this does not give you a composition, but I am pretty sure it does for Bool-enriched ones. However, if you want to stay in Set-enriched profunctors, one of the Kan extensions makes Prof into a closed bicategory. | |
| May 3 at 16:27 | history | edited | Cole Comfort | CC BY-SA 4.0 |
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| May 3 at 15:55 | comment | added | Emily | (I guess another thing to try would be to show that $*$ makes $\mathsf{Prof}$ into a skew-bicategory: maybe while we don't have natural isomorphisms $\mathfrak{p}*\mathrm{Hom}_{\mathcal{C}}\cong\mathfrak{p}$ and $\mathrm{Hom}_{\mathcal{D}}*\mathfrak{p}\cong\mathfrak{p}$, I could see there being a non-invertible map from $\mathfrak{p}*\mathrm{Hom}_{\mathcal{C}}$ to $\mathfrak{p}$ or the other way around in a way that makes $\mathsf{Prof}$ into a skew-bicategory.) | |
| May 3 at 15:51 | comment | added | Emily | Incidentally, do you know if there's an analogue for profunctors of what we're calling the "apartness composition" for relations here? I figure the end formula $$\mathfrak{q}\mathbin{*}\mathfrak{p}=\int_{B\in\mathcal{D}}\mathfrak{p}^{B}_{A}\mathbin{\textstyle\coprod}\mathfrak{q}^{C}_{B}$$ won't give $\mathsf{Prof}$ a unital composition, but maybe there's a more refined analogue of the "apartness composition" for profunctors that works? | |
| May 3 at 15:50 | comment | added | Emily | Hey Cole, thank you so much for your answer and welcome to MO! I've defined the $\square$ composition with a product instead of a coproduct, and this is what leads it to being non-unital. That said, I can't overstress how much I appreciate your answer, as it gives a number of great resources on the apartness composition as well as on the general categorical stuff going on with it, which I was really eager to read. Again, thank you so much! | |
| May 3 at 15:44 | comment | added | Peter LeFanu Lumsdaine | I think this answer is mistaken — it applies to the “apartness composition” defined by $\forall b.\, a \sim_R b \vee b \sim_S c$ mentioned by Paul Taylor in comments, but the question itself is about the composition defined with $\wedge$ not $\vee$. As Paul says, that composition has no units except on subsingletons. If $I$ is a unit on $X$, then for any $R \subseteq X \times Y$, whenever $(x \sim_R y)$ we have $(\forall x',\ x \sim_I x' \wedge x' \sim_R y )$, so a fortiori $\forall x',\, x' \sim_R y$. Taking $R$ to be equality and $y = x$, we get for any $x$ that $\forall x',\, x' = x$. | |
| S May 3 at 13:31 | review | First answers | |||
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| S May 3 at 13:31 | history | edited | Cole Comfort | CC BY-SA 4.0 |
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| May 3 at 7:14 | history | edited | Cole Comfort | CC BY-SA 4.0 |
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| May 3 at 7:07 | review | Late answers | |||
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| S May 3 at 6:51 | review | First answers | |||
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| S May 3 at 6:51 | history | answered | Cole Comfort | CC BY-SA 4.0 |