Timeline for Do monoid homomorphisms from $X^X$ to a group factor through $\text{Sym}(X)$? [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| May 1, 2018 at 3:04 | review | Reopen votes | |||
| May 1, 2018 at 8:22 | |||||
| Apr 23, 2018 at 13:57 | comment | added | Benjamin Steinberg | BTW, of you ask for uniqueness of $f'$ the answer is no because for the trivial homomorphism $X^X$ to $Sym(X)$ you have two extensions: the identity and the trivial homomorphism. | |
| Apr 23, 2018 at 13:54 | comment | added | Benjamin Steinberg | You are asking the backward universal property. If M is a monoid, then its group of units is the universal group with a morphism INTO $M$. So any homomorphism from a group $G$ into $M$ factors through $Sym(X)$. If a monoid contains a right or left zero, all its group images are trivial. | |
| Apr 23, 2018 at 11:45 | review | Reopen votes | |||
| Apr 23, 2018 at 12:33 | |||||
| Apr 23, 2018 at 8:05 | history | closed |
Will Sawin Dan Petersen abx YCor Gro-Tsen |
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| Apr 23, 2018 at 7:13 | review | Close votes | |||
| Apr 23, 2018 at 8:09 | |||||
| Apr 23, 2018 at 6:59 | vote | accept | Dominic van der Zypen | ||
| Apr 23, 2018 at 6:58 | answer | added | Will Sawin | timeline score: 6 | |
| Apr 23, 2018 at 6:46 | comment | added | S. carmeli | oh, I see, sorry. | |
| Apr 23, 2018 at 6:41 | comment | added | Dominic van der Zypen | $X^X$ is not a group in your example (I require $G$ above to be a group). | |
| Apr 23, 2018 at 6:40 | comment | added | S. carmeli | No. For example the identity map $X^X \to X^X$ don't factor through $Sym(X)$ for a finite set $X$ because $Sym(X)$ has less elements than $X^X$. | |
| Apr 23, 2018 at 6:35 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |