Timeline for unordered configuration space of pointed space
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2015 at 21:38 | history | edited | Johannes Hahn | CC BY-SA 3.0 |
Fixed the definition of F(X,k)
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| May 14, 2015 at 12:26 | comment | added | Craig Westerland | Hey @GabrielC.Drummond-Cole, excellent point. | |
| May 14, 2015 at 11:58 | comment | added | bananastack | ah, thanks, I hadn't noticed it was configuration space and not the symmetric power... | |
| May 14, 2015 at 11:32 | comment | added | Gabriel C. Drummond-Cole | @CraigWesterland of course, the answer is not yes even at the set-theoretical level for $X$ finite. | |
| May 14, 2015 at 6:51 | vote | accept | Shiquan Ren | ||
| May 14, 2015 at 6:50 | history | edited | Shiquan Ren | CC BY-SA 3.0 |
added 70 characters in body
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| May 14, 2015 at 6:47 | comment | added | Włodzimierz Holsztyński | Could you introduce the notation, especially $\ F(X,k)\ $ ? | |
| May 14, 2015 at 6:18 | answer | added | Dylan Thurston | timeline score: 6 | |
| May 14, 2015 at 6:16 | history | edited | Shiquan Ren | CC BY-SA 3.0 |
added 116 characters in body
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| May 14, 2015 at 4:58 | comment | added | Craig Westerland | Because one of the $x_i$ could be $*$, rendering the resulting $(k+1)$-tuple not a configuration of distinct points. I think that the answer is undoubtedly yes: there is an inclusion (by some set-theoretic nonsense), but no, it is almost surely not a reasonable map (in particular, it's not obvious that there is a continuous map). Note however that there is a continuous inclusion when $X$ admits an injective self map $f:X \to X$ which misses a point (say $*$); then $[x_1, ..., x_k] \mapsto [f(x_1), ..., f(x_k), *]$ works. | |
| May 14, 2015 at 4:48 | comment | added | bananastack | sorry, it's a bit late, why is it not well-defined? | |
| May 14, 2015 at 4:30 | history | asked | Shiquan Ren | CC BY-SA 3.0 |