Timeline for are quotients by equivalence relations "better" than surjections?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 19, 2019 at 22:56 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
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| May 19, 2019 at 22:20 | vote | accept | Kevin Buzzard | ||
| Dec 9, 2018 at 22:18 | comment | added | Kevin Buzzard | Me too. I asked explicitly at mathoverflow.net/questions/317241/… | |
| Dec 8, 2018 at 18:21 | comment | added | Andrej Bauer | You're right, my example doesn't work. But I'll bet that there are silly examples. | |
| Dec 8, 2018 at 8:09 | comment | added | Kevin Buzzard | It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow". | |
| Dec 8, 2018 at 8:05 | comment | added | Kevin Buzzard | Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Y\to Z$ in $P(Z)$ with $Y=\overline{Y_0}$, I choose a subset $X_0$ of $Y$, I get a map $X\to Y$ in $P(Y)$, but I don't think that $X\to Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right? | |
| Dec 8, 2018 at 4:03 | comment | added | Santana Afton | @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$? | |
| Dec 8, 2018 at 1:05 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
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| Dec 8, 2018 at 1:04 | comment | added | Andrej Bauer | For any set $S$ let $\bar{S} = \{\{x\} \mid x \in S\}$ and let $b_S : \bar{S} \to S$ be given by $b_S(\{x\}) = x$. Given $S \subseteq X$ let $i_S : S \to X$ be the canonical inclusion. Define $P(X) = \{ i_S \circ b_S : \bar{S} \to X \mid S \subseteq X\}$. Then I think we get representatives of injections that are closed under composition (because they're not composable). | |
| Dec 8, 2018 at 0:53 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
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| Dec 8, 2018 at 0:11 | comment | added | Kevin Buzzard | "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely. | |
| Dec 7, 2018 at 23:15 | history | answered | Andrej Bauer | CC BY-SA 4.0 |