Suppose $A$ and $B$ are sets and $f:A\to B.$ We define $F:\mathcal P(B)\to\mathcal P(A)$ by $$F(C):=f^{-1}[C].$$ If $f$ is injective or surjective, what can we conclude about $F$?
My attempt:
- $f$ is injective.
Since $f$ is injective, for distinct $a_1,a_2\in A,$ we have $$F(f[\{a_1\}])=f^{-1}[f[\{a_1\}]]=\{a_1\}\ne\{a_2\}=f^{-1}[f[\{a_2\}]]=F(f[\{a_2\}]),$$ so, for each $A'\in\mathcal P(A),$ there is $B'=f\left(\bigcup\limits_{a\in A'}\{a\}\right)\in\mathcal P(B)$ with $A'=F(B'),$ hence, $F$ is surjective.
Let $A=B=\{1,2\}$ and $f(1)=f(2)=1.$ Then $F(\emptyset)=f^{-1}[\emptyset]=\emptyset=f^{-1}[\{2\}]=F(\{2\})$ or $F(\{1,2\})=f^{-1}[\{1,2\}]=\{1,2\}=f^{-1}[\{1\}]=F(\{1\}),$ so $F$ isn't necessarily injective.
- $f$ is surjective.
I claim $F$ is then injective. Suppose to the contrary, $f$ is surjective and $F$ isn't injective. Then, there are distinct $B_1,B_2\mathcal\in P(B)$ s. t., WLOG, $B_1\setminus B_2\ne\emptyset$ and $f^{-1}[B_1]=F(B_1)=F(B_2)=f^{-1}[B_2]\implies f^{-1}[B_1]\setminus f^{-1}[B_2]=\emptyset.$
On the other hand, as proven in this answer, $\emptyset=f^{-1}[B_1]\setminus f^{-1}[B_2]=f^{-1}[B_1\setminus B_2],$ but this is impossible, since $B_1\setminus B_2\ne\emptyset$ and $f$ is surjective. Therefore, if $f$ is surjective, $F$ is injective.
Let $A=\{1,2\},B=\{1\}$ and $f(1)=f(2)=1.$ Then $\mathcal P(B)=\{\emptyset,\{1\}\},f^{-1}[\{1\}]=\{1,2\}$ and $f^{-1}[\emptyset]=\emptyset,$ so there is no $B'\in\mathcal P(B)$ s. t. $F[B']=f^{-1}[B']=\{1\}$ or $F[B']=f^{-1}[B']=\{1\}.$
Is my deduction valid?
