Question

Let $f:X\to Y$ be a function and $A_y=\left\{\frac{f^{-1}(y)}{y\in Y}\right\}$. Then $A_i\cap A_j=\varphi (i\ne j)\forall i,j\in Y$ and ${\bigcup }_{y\in Y}{A}_y=X$, if

• A. f is onto function only
• B. f is one-one function only
• C. f is any function
• D. X and Y are finite sets only

Answer 1

Answer: (C)

It is given that function $f:X\to Y$ and $A_y=\left\{\frac{f^{-1}(y)}{y\in Y}\right\}$
$\because$ Inverse of the function exists, so function $f$ is one-one and onto.
then $A_i\cap A_j=\varphi ,(i\ne j)\forall i,j\in Y$
and $\bigcup A_y=X$ for every function $f$.