I would like to know if my proof for the next argument is correct:
Let ${\displaystyle f\colon X\to Y} $ be a function, let ${\displaystyle G\colon P(X)\to P(Y)} $ be a function.
Suppose ${\displaystyle G[B]=F^{-1}[B] }$
so,
- G is Surjective iff F is Injective
- G is Injective iff F is Surjective
-----my final answer-----
- ${\displaystyle \to \colon}$ suppose $G$ is surjective we will prove that $F$ is injective.
let ${\displaystyle a,b \in X}$ , suppose ${\displaystyle f(a)=f(b)}$ we will show that ${\displaystyle a=b}$.
${\displaystyle a,b \in X}$ , so ${\displaystyle {\{a}\},{\{b}\} \subseteq X}$ so , ${\displaystyle {\{a}\},{\{b}\} \in P(X)}$
$G$ is surjective so there exist ${\displaystyle A,B \in P(Y)}$ so that $G(A)= {\{a}\} $ and $G(B)= {\{b}\} $ w.l.o.g
${\displaystyle G[B]=F^{-1}[B]={\{b}\} }$ and ${\displaystyle G[A]=F^{-1}[A] ={\{a}\}}$ ,
${\displaystyle A,B \in P(Y)}$ so ${\displaystyle A,B \subseteq Y}$ and ${\displaystyle f\colon X\to Y}$ , so ${\displaystyle F^{-1}[A],F^{-1}[B] \subseteq X}$
$f$ is a function so if the preimage of $A$ and $B$ contains only one element then $A and B$ contains only one element
${\displaystyle F^{-1}[B]={\{b}\}}$ then ${\displaystyle f(b) \in B}$ so, ${\{f(b)}\} = B$
${\displaystyle F^{-1}[A]={\{a}\}}$ then ${\displaystyle f(a) \in A}$ so, ${\{f(a)}\} = A$ , $f(a)=f(b)$ so, $A=B$ so $G(A)=G(B)$ so, $a=b$
${\displaystyle \leftarrow \colon}$ suppose $F$ is injective we will prove that $G$ is surjective.
let ${\displaystyle B \in P(X)}$ we will show that there exist ${\displaystyle A\in P(Y)}$ so that $G(A)=B$.
${\displaystyle B \in P(X)}$ so ${\displaystyle B \subseteq X}$ , let $A$ be:
${\displaystyle A= {\{f(b) \in Y | b \in B }\}}$ , we can easily see that $A=Im(B)$ .
${\displaystyle lemma \colon}$ $f$ is injective so ${\displaystyle F^{-1}[F[B]]=B (F[B]=A})$
so ${\displaystyle F^{-1}[A]=G(A)}$ so ${\displaystyle G(A)=B}$ thus, $G$ is surjective
- ${\displaystyle \rightarrow \colon}$ suppose G is Injective we need to prove that F is Surjective.
let ${\displaystyle y \in Y}$ so ${\displaystyle {\{y}\} \subseteq Y}$ so, ${\displaystyle F^{-1}[{\{y}\}]=G({\{y}\})}$,
${\displaystyle G({\{y}\}) \in P(X)}$ so, ${\displaystyle G({\{y}\}) \subseteq X}$, so ${\displaystyle F^{-1}[{\{y}\}] \subseteq X}$
${\displaystyle F^{-1}[B]={\{x \in X| f(x) \in B}\}}$
thus, ${\displaystyle F^{-1}[∅]=∅}$ , so ${\displaystyle G(∅)=∅}$ , G is injective so there exist ${\displaystyle x \in G({\{y}\}}$ so, ${\displaystyle x \in F^{-1}[{\{y}\}]}$ so, ${\displaystyle f(x) \in {\{y}\}}$
only ${\displaystyle y \in {\{y}\}}$ thus, $f(x)=y$ thus, F in surjective.
${\displaystyle \leftarrow \colon}$ suppose f is surjective we will show that G is injective
let ${\displaystyle A',A \in P(Y)}$ so, ${\displaystyle A',A \subseteq Y}$, suppose $G(A')=G(A)$ , we will show that $A'=A$
${\displaystyle F^{-1}[A']=G(A')}$ and ${\displaystyle F^{-1}[A]=G(A)}$
${\displaystyle \subseteq \colon }$ let ${\displaystyle a \in A}$ , f is surjective so there is ${\displaystyle b \in X}$ so that, $f(b)=a$ so, ${\displaystyle b \in F^{-1}[A]}$ , so, ${\displaystyle b \in G(A)}$,
$G(A)=G(A')$ so, ${\displaystyle b \in G(A')}$ so, ${\displaystyle b \in F^{-1}[A']}$ so, ${\displaystyle f(b) \in A'}$ , $f(b) = a $ , so ${\displaystyle a \in A'}$ , so ${\displaystyle A \subseteq A'}$
same for ${\displaystyle A' \subseteq A}$ , so $A'=A$ thus, G is injective.
Lemma: if F is injective then ${\displaystyle F^{-1}[F[B]] = B}$ (the preimage of the image is the image)
suppose f is injective:
${\displaystyle F^{-1}[F[B]] \subseteq B \colon}$ let ${\displaystyle a \in F^{-1}[F[B]]}$ then ${\displaystyle f(a) \in [F[B]]}$ then there exist ${\displaystyle a' \in B}$ so that $f(a')=f(a)$,
$f$ is injective therefore, $a'=a$ ,so ${\displaystyle a \in B}$
${\displaystyle B \subseteq F^{-1}[F[B]] \colon}$ let ${\displaystyle a \in B}$ then ${\displaystyle f(a) \in [F[B]}$ then, ${\displaystyle a \in F^{-1}[F[B]]}$
thus, ${\displaystyle B = F^{-1}[F[B]]}$.
I would like to know if the proof is correct or not, thank you:)