Timeline for Can we build a continuous function from "fibers"/preimages defined over a topological base?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2014 at 7:11 | answer | added | Dominic van der Zypen | timeline score: 1 | |
| Sep 9, 2014 at 11:29 | comment | added | Dominic van der Zypen | It is not always possible, if we manage to construct a map $F:B^* \to A^*$ with the properties $(1)-(3)$ above but which doesn't respect arbitrary intersections (only finite ones). Then there can't be a map $f$ with the desired properties, because $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$ respects arbitrary intersections. | |
| Sep 7, 2014 at 9:00 | comment | added | Max Suica | D'oh. If $Y$ has the indiscrete topology there is only one admissible $F^*$. But in that case, any function from $X$'s carrier set to $Y$ is going to be continuous. I guess I'd like to pivot the question and also ask whether there are conditions that let us uniquely define $f$. I'm interested in particular in a case where $X$ is normal and $Y = \mathbb{R},$ or where $X$ or $Y$ (or both) are sober. | |
| Sep 7, 2014 at 8:50 | history | edited | Max Suica | CC BY-SA 3.0 |
Added necessary conditions to definition of F^*
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| Sep 7, 2014 at 8:37 | history | asked | Max Suica | CC BY-SA 3.0 |