Timeline for How is the Julia set of $fg$ related to the Julia set of $gf$?
Current License: CC BY-SA 3.0
9 events
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| Aug 3, 2016 at 20:41 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| Aug 7, 2012 at 16:06 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Aug 6, 2012 at 17:58 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Aug 5, 2012 at 18:37 | comment | added | Tom Leinster | OK, thanks. So "full preimage" is a synonym for "preimage". | |
| Aug 5, 2012 at 8:27 | comment | added | Alexandre Eremenko | Full $f$-preimage of a point $w$ is the set of all $z$ such that $f(z)=w$. Full preimage of a set is the union of full preimages of its points. | |
| Aug 4, 2012 at 21:41 | comment | added | Tom Leinster | Sorry, Alexandre, I'm still not certain that I understand you. I simply don't know what the term "full preimage" means (and my attempts to look it up have come to nothing). So: what is the definition of full preimage? | |
| Aug 4, 2012 at 19:54 | comment | added | Alexandre Eremenko | "Preimage" should be understood as "full preimage" in what I wrote. I think the general facts that a) the family $g(f_n)$ is normal on the set where $f_n$ is normal, and b) the family $f_n(g)$ is normal on the full $g$-preimage of the set where $f_n$ is normal are immediate consequences of definition of normality. | |
| Aug 4, 2012 at 19:37 | comment | added | Tom Leinster | Thanks, Alexandre. Just to check: by "full g-preimage", you just mean "g-preimage" (i.e. preimage under g), right? So the word "full" is redundant? | |
| Aug 4, 2012 at 15:32 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |