Question

Suppose $X$ and $Y$ are two sets and $f: X \rightarrow Y$ is a function. For a subset $A$ of $X$, define $f(A)$ to be the subset $\{f(a): a \in A\}$ of $Y$. For a subset $B$ of $Y$, define $f^{-1}(B)$ to be the subset $\{x \in X: f(x) \in B\}$ of $X$. Then which of the following statements is true?

$(A) f^{-1}(f(A))=A$ for every $A \subset X$

(B) $f^{-1}(f(A))=$ A for every $A \subset X$ if only if $f(X)=Y$

(C) $f\left(f^{-1}(B)\right)=B$ for every $B \subset Y$

(D) $f\left(f^{-1}(B)\right)=B$ for every $B \subset Y$ if only if $f(X)=Y$

Answer 1

For option A

$f^{-1}(f(A))=A \cup A^{\prime} \neq A$

Hence $A$ is wrong

For option B 

$f(X)=Y \quad \Rightarrow \quad f$ is onto but it will not effect on mapping of function

Hence $B$ is wrong

For option A \& B other explaination can be given else if $Y$ is a singleton set then the function $f$ is constant function and hence is trivially onto (unless $X=\phi$ ). But in such a case, even if A consists of just one point, $f(A)$ is entire set $Y$ and so $^{-1}(f(A))$ is the entire set $X$, which could be much bigger than $A$. So $A$ and $B$ are wrong even if $f(X)=Y$

For option C

$f(X) \subset Y(\text { range } \neq \text { co-domain })$

$f(X)$ is a proper subset of $Y$ (so that $f$ is not onto), then for $B=Y$ option $C$ is wrong because $f^{-1}(Y)=X$ but $f\left(f^{-1}(Y)\right)=f(X) \neq Y$.

For option D

If $B=Y$, then $f\left(f^{-1}(Y)\right)$ is the range of the function $f$. If this is equal to $Y$, then function must be onto, thus $f(X)=Y$ is necessary condition

Hence $D$ is correct