Let X and Y be nonempty set, let f : X \rightarrow Y be an injective mapping and let A,B \subset X. Show that f(A \cap B) = f(A) \cap f(B).

We are tasked with showing that for an injective mapping \( f: X \to Y \) and subsets \( A, B \subseteq X \), the following holds:

\[
f(A \cap B) = f(A) \cap f(B)
\]

### Step 1: Show \( f(A \cap B) \subseteq f(A) \cap f(B) \)
We will start by showing that \( f(A \cap B) \subseteq f(A) \cap f(B) \).

- Take any element \( y \in f(A \cap B) \). By definition, this means there exists some \( x \in A \cap B \) such that \( f(x) = y \).
- Since \( x \in A \cap B \), it follows that \( x \in A \) and \( x \in B \).
- Therefore, \( f(x) \in f(A) \) and \( f(x) \in f(B) \), which means \( y \in f(A) \cap f(B) \).

Thus, every element of \( f(A \cap B) \) is also in \( f(A) \cap f(B) \), so:
\[
f(A \cap B) \subseteq f(A) \cap f(B)
\]

### Step 2: Show \( f(A) \cap f(B) \subseteq f(A \cap B) \)
Now, we will show the reverse inclusion \( f(A) \cap f(B) \subseteq f(A \cap B) \).

- Take any element \( y \in f(A) \cap f(B) \). By definition, this means there exist elements \( x_1 \in A \) and \( x_2 \in B \) such that:
  \[
  f(x_1) = y \quad \text{and} \quad f(x_2) = y
  \]
- Since \( f \) is injective, it follows that \( x_1 = x_2 \), because injectivity implies that \( f(x_1) = f(x_2) \) only if \( x_1 = x_2 \).
- Therefore, there exists some \( x \in A \cap B \) such that \( f(x) = y \).

Thus, every element of \( f(A) \cap f(B) \) is also in \( f(A \cap B) \), so:
\[
f(A) \cap f(B) \subseteq f(A \cap B)
\]

### Conclusion
Since we have shown both \( f(A \cap B) \subseteq f(A) \cap f(B) \) and \( f(A) \cap f(B) \subseteq f(A \cap B) \), it follows that:
\[
f(A \cap B) = f(A) \cap f(B)
\]

This completes the proof.

Do you have any questions or would you like further details?

### 5 Related Questions:
1. How does this proof change if the function \( f \) is not injective?
2. What is the significance of injectivity in set theory and mappings?
3. Can you provide an example where \( f(A \cap B) \neq f(A) \cap f(B) \) for a non-injective function?
4. What happens if \( A \) or \( B \) is the empty set in this scenario?
5. How can this result be extended to other set operations, like union or difference?

### Tip:
Injective functions (one-to-one) preserve distinctness, making them useful when proving equalities involving set images under mappings.

Understand how to prove that f(A ∩ B) = f(A) ∩ f(B) for an injective function f, where A and B are subsets of a set X. This problem covers concepts in set theory and injective functions, including the properties of mappings and subset intersections. Suitable for college or university-level students.

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