Timeline for The name for the quotient property
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 22, 2014 at 3:26 | comment | added | Vadim | @RamirodelaVega Yes, this is what I saw, so it seems to be the best (or, at least, the only) name for the property. If you put it as an answer, and no one suggests something else, I will accept it as the answer. Thanks! | |
| Oct 22, 2014 at 3:24 | comment | added | Vadim | @AlexDegtyarev In many cases it would just be convenient to call a function that is known to satisfy this property, but not necessary continuous, call it something without rewriting the property every time. | |
| Oct 22, 2014 at 3:20 | comment | added | Vadim | @FrancescoPolizzi Consider $f:\mathbb{R}\rightarrow S^1$ given by $(\cos 2\pi x,\sin 2\pi x)$, this is a surjective, continuous and open map, so quotient, but it is not closed. Now restrict the same function to domain $[0,1]$, it is now surjective, continuous and closed map, so still quotient, but not open. An example from Munkres, the projection from $\mathbb{R}^2$ to $x$-axis restricted to domain $\mathbb{R}\times\{0\}\cup[0,+\infty)\times\mathbb{R}$ is a non-open non-closed quotient. | |
| Oct 22, 2014 at 3:18 | comment | added | Vadim | @FrancescoPolizzi In this context openness (or closeness) is stronger in the sense that if a function is surjective continuous + open (or close) then it is a quotient map, but not vice versa. | |
| Oct 22, 2014 at 0:59 | comment | added | Ramiro de la Vega | Sometimes it is said that $f$ is open with respect to saturated sets (i.e. sets of the form $f^{-1}(A)$). | |
| Oct 21, 2014 at 21:11 | comment | added | Alex Degtyarev | I think it's not quite the same as openness: you do not require that the image of any open set should be open. Why do you need a name? It doesn't seem to make much sense without the continuity, and with continuity, you already know the name :) | |
| Oct 21, 2014 at 21:05 | comment | added | Francesco Polizzi | What is (if any) the relation of (???) with the usual concept of open map? I mean, continuity + (???) imply that $f$ is open? | |
| Oct 21, 2014 at 20:37 | review | First posts | |||
| Oct 21, 2014 at 20:39 | |||||
| Oct 21, 2014 at 20:36 | history | asked | Vadim | CC BY-SA 3.0 |