Saturated set

Let ${\displaystyle f:X\to Y}$ be a map. Given any subset ${\displaystyle S\subseteq X,}$ define its image under ${\displaystyle f}$ to be the set:

$${\displaystyle f(S):=\{f(s)~:~s\in S\}}$$

and define its preimage or inverse image under ${\displaystyle f}$ to be the set:

$${\displaystyle f^{-1}(S):=\{x\in X~:~f(x)\in S\}.}$$

Given ${\displaystyle y\in Y,}$ the fiber of ${\displaystyle f}$ over ${\displaystyle y}$ is defined to be the preimage:

$${\displaystyle f^{-1}(y):=f^{-1}(\{y\})=\{x\in X~:~f(x)=y\}.}$$

Any preimage of a single point in ${\displaystyle f}$'s codomain ${\displaystyle Y}$ is referred to as a fiber of ${\displaystyle f.}$

A set ${\displaystyle C}$ is called ${\displaystyle f}$-saturated and is said to be saturated with respect to ${\displaystyle f}$ if ${\displaystyle C}$ is a subset of ${\displaystyle f}$'s domain ${\displaystyle X}$ and if any of the following equivalent conditions are satisfied:[1]

1. ${\displaystyle C=f^{-1}(f(C)).}$
2. There exists a set ${\displaystyle S}$ such that ${\displaystyle C=f^{-1}(S).}$
   • Any such set ${\displaystyle S}$ necessarily contains ${\displaystyle f(C)}$ as a subset and moreover, it will also necessarily satisfy the equality ${\displaystyle f(C)=S\cap \operatorname {Im} f,}$ where ${\displaystyle \operatorname {Im} f:=f(X)}$ denotes the image of ${\displaystyle f.}$
3. If ${\displaystyle c\in C}$ and ${\displaystyle x\in X}$ satisfy ${\displaystyle f(x)=f(c),}$ then ${\displaystyle x\in C.}$
4. If ${\displaystyle y\in Y}$ is such that the fiber ${\displaystyle f^{-1}(y)}$ intersects ${\displaystyle C}$ (that is, if ${\displaystyle f^{-1}(y)\cap C\neq \varnothing }$), then this entire fiber is necessarily a subset of ${\displaystyle C}$ (that is, ${\displaystyle f^{-1}(y)\subseteq C}$).
5. For every ${\displaystyle y\in Y,}$ the intersection ${\displaystyle C\cap f^{-1}(y)}$ is equal to the empty set ${\displaystyle \varnothing }$ or to ${\displaystyle f^{-1}(y).}$