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Let $f: X \rightarrow Y$ be a function and $A_y=\left\{\frac{f^{-1}(y)}{y \in Y}\right\}$. Then $A_i \cap A_j=\dot{\phi}(i \neq j) \forall i, j \in Y$ and $\bigcup_{y \in Y} A_y=X$, if
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The correct answer is:
$f$ is any function
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 $f$.
Hence, ${ }^Y$ option (c) is correct.
$A_y=\left\{\frac{f^{-1}(y)}{y \in Y}\right\}$
$\because$ 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 $f$.
Hence, ${ }^Y$ option (c) is correct.
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