The equation $f(A \cap B) = f(A) \cap f(B)$ relates to functions and set theory. It states that the image of the intersection of two sets $A$ and $B$ under a function $f$ is equal to the intersection of the images of $A$ and $B$ under the same function.

This property holds true under specific conditions. It is true if and only if the function $f$ is injective (one-to-one). An injective function is a function where different elements in the domain map to different elements in the codomain.

Explanation:

1. Injective Function (One-to-One): If $f$ is injective, no two different elements in the domain have the same image in the codomain. Therefore, if $x$ belongs to $A \cap B$, then $f(x)$ belongs to both $f(A)$ and $f(B)$, and so $f(x)$ belongs to $f(A) \cap f(B)$.

2. Non-Injective Function: If $f$ is not injective, it is possible for elements that are not in $A \cap B$ to map to the same image under $f$, which can cause the equation $f(A \cap B) = f(A) \cap f(B)$ to fail.

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Related Questions:

1. What are the properties of injective functions?
2. How does the surjectivity of a function relate to set operations?
3. Can $f(A \cap B) = f(A) \cap f(B)$ hold if $f$ is surjective but not injective?
4. How does the equation change if we consider $f(A \cup B)$ instead of $f(A \cap B)$?
5. What is the significance of bijective functions in set theory?

Tip:

When analyzing function-related set identities, always check whether the function is injective, surjective, or bijective, as these properties often determine the validity of such identities.