Answer
Yes, the inclusion
\[
F\bigl(F^{-1}(C)\bigr) \;\subseteq\; C
\] is **always valid** for any function \(F\) and any subset \(C \subseteq Y\).
Work Step by Step
For a function \(F: X \to Y\) and any subset \(C \subseteq Y\), determine whether
\[
F\bigl(F^{-1}(C)\bigr) \;\subseteq\; C
\] is always true.
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## Proof
1. **Definition of \(F^{-1}(C)\).**
By definition,
\[
F^{-1}(C) \;=\; \{\,x \in X : F(x) \in C\}\,.
\] So if \(x\in F^{-1}(C)\), it means \(F(x)\) is in \(C\).
2. **Apply \(F\) to \(F^{-1}(C)\).**
An element \(y \in F\bigl(F^{-1}(C)\bigr)\) means \(y = F(x)\) for some \(x \in F^{-1}(C)\).
3. **Show \(y \in C\).**
Since \(x \in F^{-1}(C)\), we have \(F(x) \in C\). But \(y = F(x)\). Therefore \(y \in C\).
Hence, every element of \(F(F^{-1}(C))\) lies in \(C\). This proves
\[
F\bigl(F^{-1}(C)\bigr) \;\subseteq\; C.
\]
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## Conclusion
For any function \(F : X \to Y\) and any subset \(C\) of \(Y\),
\[
F\bigl(F^{-1}(C)\bigr) \;\subseteq\; C
\] always holds. This is a fundamental property of inverse images under functions.
