Question

Let $f: X \rightarrow Y$ be a function and $A_{y}=\left\{\frac{f^{-1}(y)}{y \in Y}\right\}$. Then $A_{i} \cap A_{j}=\phi(i \neq j) \forall i, j \in Y$ 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 $f: X \rightarrow Y$ 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 $A_{i} \cap A_{j}=\phi,(i \neq j) \forall i, j \in Y$
and $\bigcup A_{y}=X$ for every function $f$.