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(C) Given $f: X \rightarrow Y$ is a function. By definition,$A_y = f^{-1}(\{y\}) = \{x \in X : f(x) = y\}$. This set $A_y$ represents the preimage of

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$f$ is any function

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(C) Given $f: X ightarrow Y$ is a function. By definition,$A_y = f^{-1}(\{y\}) = \{x \in X : f(x) = y\}$. This set $A_y$ represents the preimage of the element $y$ under the function $f$. For any two distinct elements $i, j \in Y$ where $i 
eq j$,the sets $A_i$ and $A_j$ are disjoint because a function maps each element of the domain to exactly one element in the codomain. Thus,$A_i \cap A_j = \phi$. Furthermore,the union of all preimages $A_y$ for all $y \in Y$ covers the entire domain $X$,i.e.,$\bigcup_{y \in Y} A_y = X$. These properties hold true for any function $f: X ightarrow Y$. Therefore,option $(C)$ is correct.

Let $f: X \rightarrow Y$ be a function and $A_y = f^{-1}(\{y\})$ for $y \in Y$. Then $A_i \cap A_j = \phi$ $(i \neq j)$ for all $i, j \in Y$ and $\bigcup_{y \in Y} A_y = X$,if

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