Answer
Yes. The equality
\[
F^{-1}(C - D) \;=\; F^{-1}(C) \;-\; F^{-1}(D)
\] holds for **all** functions \(F\) and **all** subsets \(C, D \subseteq Y\).
Work Step by Step
For a function \(F: X \to Y\) and subsets \(C, D \subseteq Y\), determine whether the following equality always holds:
\[
F^{-1}(C - D) \;=\; F^{-1}(C) \;-\; F^{-1}(D).
\]
---
## Proof
Recall the definitions:
\(C - D = \{\,y \in Y : y \in C \text{ and } y \notin D\}\).
\(F^{-1}(S) = \{\,x \in X : F(x) \in S\}\) for any \(S \subseteq Y\).
\(A - B = \{\,x \in A : x \notin B\}\).
We must show two set inclusions to establish equality:
1. **\(\subseteq\) Direction**
- Take any \(x \in F^{-1}(C - D)\).
- By definition, \(F(x) \in C - D\), meaning \(F(x) \in C\) **and** \(F(x) \notin D\).
- Consequently, \(x \in F^{-1}(C)\) and \(x \notin F^{-1}(D)\).
- Hence \(x \in F^{-1}(C) - F^{-1}(D)\).
- This shows \(F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D)\).
2. **\(\supseteq\) Direction**
- Take any \(x \in F^{-1}(C) - F^{-1}(D)\).
- Then \(x \in F^{-1}(C)\) (so \(F(x) \in C\)) and \(x \notin F^{-1}(D)\) (so \(F(x) \notin D\)).
- Therefore \(F(x)\) lies in \(C\) but not in \(D\), i.e., \(F(x) \in C - D\).
- Hence \(x \in F^{-1}(C - D)\).
- This shows \(F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D)\).
Since both inclusions hold, the two sets are equal:
\[
F^{-1}(C - D) \;=\; F^{-1}(C) \;-\; F^{-1}(D).
\]
---
## Conclusion
The property
\[
F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D)
\]
is **always true** for any function \(F: X \to Y\) and any subsets \(C, D \subseteq Y\).
