1. ## Inverse

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 and only if $f(X)=Y$

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

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

To me it appears that every option is correct.But then I think conditions for inverse are also to be applied .

Please explain as to why (A),(B),(C) are wrong and why (D) is correct.I am not being able to put concepts of ONE-ONE and ONTO functions in place

2. Originally Posted by pankaj
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 and only if $f(X)=Y$
counter-example: $X=\mathbb{R}, \ Y=\mathbb{R}_{+}, \ A=\{1 \},$ and $f: X \longrightarrow Y$ defined by $f(x)=x^2.$ then $f(X)=Y$ but $f^{-1}(f(A))=\{1, -1 \} \neq A.$

$\color{red} (*)$: as an exercise prove that (B) is true if and only if $f$ is one-to-one!

(C) $f(f^{-1}(B))=B$ for every $B\subset Y$
counter-example: $X=Y=\mathbb{R}, \ B=\{0,-1 \},$ and $f: X \longrightarrow Y$ defined by $f(x)=x^2.$ then $f(f^{-1}(B))=\{0 \} \neq B.$

(D) $f(f^{-1}(B))=B$ for every $B\subset Y$if and only if $f(X)=Y$
if $y \in f(f^{-1}(B)),$ then $y=f(x),$ for some $x \in f^{-1}(B).$ thus $y=f(x) \in B,$ i.e. $f(f^{-1}(B)) \subseteq B.$ this is true for any function $f.$ now if $f$ is onto and $y \in B,$ then $y=f(x),$ for some $x \in X.$

hence $f(x)=y \in B,$ which means $x \in f^{-1}(B)$ and so $y=f(x) \in f(f^{-1}(B)).$ therefore $B \subseteq f(f^{-1}(B)).$ conversely, if $f(f^{-1}(B))=B,$ for any $B \subseteq Y,$ then choosing $B=\{ y \}, \ y \in Y,$ we

have $f(f^{-1}(\{y \}))=\{y \}$ and thus $f^{-1}(\{y \}) \neq \emptyset,$ i.e. $f$ is onto.

3. Thanks
You make it look so simple.It is the language of mathematics that I am probably lacking in as I never studied maths formally

4. In one of my very first graduate school classes, I was asked to prove, in class, some things about $f^{-1}(A)$. I did the whole problem assuming f had an inverse and humilliated myself!