Hi, given a function $f:X \rightarrow Y$, not necessarily invertible, is there a conventional name for the function $$g_f := f^{-1} \circ f:X \rightarrow \mathcal{P}(X),$$ where $\mathcal{P}(X)$ denotes the power set of $X$?
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$\begingroup$ I was gonna call it the reflection under f, but maybe there are fields of mathematics where it is commonly used. I am not aware of them though. Just a physicist... $\endgroup$– tortortorApr 7, 2013 at 0:11
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3$\begingroup$ If you think of $f$ as modding out an equivalence relation on $X$, then this takes $x$ to its equivalence class. Other than that, I've never seen a term for this. $\endgroup$– Eric WofseyApr 7, 2013 at 0:35
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4$\begingroup$ I don't think there's a standard name, but perhaps you could use saturation. $\endgroup$– Benjamin DickmanApr 7, 2013 at 1:20
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1$\begingroup$ In line with what @Eric says I think $[x]_f$ would be a recognizable notation for $g_f(x).$ I suppose $[\cdot]_f$ could be used to single out global function. $\endgroup$– Aaron MeyerowitzApr 7, 2013 at 17:36
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2$\begingroup$ You could call it a fiber of $f$. The fiber that contains the point $x$. $\endgroup$– Tom GoodwillieApr 7, 2013 at 19:58
1 Answer
I do not know any name for it. However, note that the concept you are definining does not really depend on the function $f$, only on the equivalence relation induced by $f$ (which is sometimes called the kernel of $f$, unless you work with groups).
A set $S$ is called "saturated" with respect to an equivalence relation $\theta$ iff $S$ is a union of equivalence classes, or equivalently, if $S=f^{-1}(f(S))$ (where $f$ is some map inducing $\theta$).
Hence "saturation" (as Benjamin Dickman suggested in a comment) would be a natural choice.
But if the audience members are not mathematicians, "saturation" might mean something entirely different to them. I would suggest "$f$-neighborhood"; this term can be easily visualized, I think.
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$\begingroup$ I see, now I really like the saturation idea. Also, I see why the concept might not have (or deserve) its own name given that $g_f$ does not depend on $f$ itself. Thanks! I'll wait a day or two and then accept your answer as probably nothing else will show up. $\endgroup$ Apr 8, 2013 at 16:25