Every subset of A is f-saturated Let $f:A\rightarrow B$ be a function such that $\forall X\subseteq A[ f^{-1}[f[X]]=X]$ (In other words, every subset of $A$ is $f$-saturated). Does the property of the function $f$ have a name ?
I understand that there might not be a name (In this case I will delete my question). In case there is a term, I don't want to miss it.
Thank you
 A: It's common to say that a function $f$ with such property is injective.
Statement: The function $f$ is injective if, and only if, for every subset $X$ of $A$ the equality $f^{-1}[f[X\textbf{]}\textbf{]}=X$ holds.
Proof: If $B=\varnothing$, then so $A=\varnothing$ and the statement is trivially true. If $A=\varnothing$ it also is trivially true. Suppose $A\neq \varnothing \neq B$.
$\Longrightarrow:$ Suppose $f$ is injective and let $X\subseteq A$. We wish to prove that $f^{-1}[f[X\textbf{]}\textbf{]}=X$. If $X=\varnothing$ it's easy. Suppose $X\neq \varnothing$.


*

*$\subseteq:$ Let $x\in f^{-1}[f[X\textbf{]}\textbf{]}\ \bigl(=\{a\in A:f(a)\in f[X\textbf{]}\}\bigr)$. So $f(x)\in f[X\textbf{]}\ \bigl(=\{f(a): a\in X\}\bigr)$. Since $f(x)\in f[X\textbf{]}\Longrightarrow (\exists a\in X)(f(a)=f(x))$, from $f$ being injective follows that $x=a$, for some $a\in X$, which implies that $x\in X$.

*$\supseteq:$ Let $x\in X$. From $$x\in X\Longrightarrow x\in X \wedge f(x)\in f[X\textbf{]} \Longrightarrow x\in\{a\in A:f(a)\in f[X\textbf{]}\}=f^{-1}[f[X\textbf{]}\textbf{]}$$
follows that $f^{-1}[f[X\textbf{]}\textbf{]}\supseteq X$. (Note that the injectivity wasn't used here which makes this inclusion true wether $f$ is injective or not).


Therefore $f^{-1}[f[X\textbf{]}\textbf{]}=X$.
$\Longleftarrow$: Proving the contrapositive suppose that $f$ isn't injective. Then there are $x,y\in A$ such that $f(x)=f(y) \wedge x\neq y$. However $f^{-1}[f[\{x\}\textbf{]}\textbf{]}=\{a\in A: f(a)=f(x)\}\supseteq \{x,y\}\neq \{x\}$, which means that the equality $f^{-1}[f[X\textbf{]}\textbf{]}=X$ fails for $X=\{x\}$.
