Talk:Image of Preimage under Mapping

"Still ambiguous. $f: S \to T$ cannot be simply composed with $f^{-1}: \powerset T \to \powerset S$. You mean ${f\sqbrk \cdot} \circ {f^{-1}\sqbrk \cdot}$?"


 * Sorry, I have corrected the link in $f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$ to Definition:Composition of Relations, as of course $f^{-1}$ is not necessarily a mapping.


 * Does that make more sense now? --prime mover (talk) 10:44, 3 July 2022 (UTC)


 * I think both makes the sense but the notation is confusing.
 * You defined $f$ as $f : S \to T$ but you mean below $f \sqbrk \cdot : \powerset S \to \powerset T$, both mapping are denoted by $f$.


 * No I do not. $f$ is a mapping from $S$ to $T$. $f \sqbrk X$ (I puke on the ugly $f \sqbrk \cdot$ btw) means "the set of elements of the codomain corresponding to all the elements of $X$ by $x$", that is: $\set {y \in \Cdm f: x \in X}$. --prime mover (talk) 12:35, 3 July 2022 (UTC)


 * Unfortunately, the difference between $\map f \cdot$ and $f \sqbrk \cdot$ is not visible. Maybe helpful to write more concretely, for example:


 * It's standard notation. If $f$ is a mapping from $S$ to $T$, then $\map f x$ means one thing, while $f \sqbrk X$ means something totally different. I have provided links to exactly what each concept means. Given a mapping $f: S \to T$ then $\map f x$ is defined for each $x$ in $S$ to associate exactly one element of $T$ with that element $x$. $f \sqbrk X$ is the set of elements of $T$ for all $x \in X$. $f^{-1} \sqbrk Y$ is the set of elements $x$ of $S$ such that there exists $y \in Y$ such that $\map f x = y$. --prime mover (talk) 12:35, 3 July 2022 (UTC)


 * I'd be really surprised if you hadn't come across this notation before.


 * $f \circ f^{-1}$ denotes composition of $f \sqbrk \cdot : \powerset S \to \powerset T$ and $f^{-1} : \powerset T \to \powerset S$.


 * I see where you're coming from. But your notation suboptimal. If you really want to get all hung up over the confusion between a mapping $f: S \to T$ and the mapping induced by $f$ on the power set of $S$ to the power set of $T$, you can present it as $f^\to: \powerset S \to \powerset T$ and $f^\gets: \powerset T \to \powerset S$ where $\map {f^\to} X$ is defined as $f \sqbrk X$ and $\map {f^\gets} Y$ is defined as $f^{-1} \sqbrk Y$, and $\map {\paren {f^\to \circ f^\gets} } Y$ is therefore defined as $\paren {f \circ f^{-1} } \sqbrk Y$. Please see Definition:Image of Subset under Mapping and the section for clarification. --prime mover (talk) 12:35, 3 July 2022 (UTC)


 * OK, now I see what you are trying to explain with the 3 lines following where. It is fine for me. $f^\to \circ f^\gets$ is the clearest but less common.--Usagiop (talk) 12:48, 3 July 2022 (UTC)