Answer
See explanation
Work Step by Step
Below are standard proofs of the two statements, using the fact that \(F\colon X \to Y\) is **one-to-one** (injective).
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## (a) \(F^{-1}\bigl(F(A)\bigr) = A\) for all \(A \subseteq X\)
Recall that
\[
F^{-1}(B) \;=\;\{\,x \in X : F(x)\in B\}
\]
for any \(B \subseteq Y\).
1. **Show \(\subseteq\).**
Take any \(x \in F^{-1}(F(A))\). Then \(F(x) \in F(A)\). By definition of \(F(A)\), there is some \(a \in A\) such that \(F(a) = F(x)\). Since \(F\) is injective, \(F(a) = F(x)\) implies \(a = x\). Thus \(x \in A\). Hence,
\[
F^{-1}(F(A)) \;\subseteq\; A.
\]
2. **Show \(\supseteq\).**
Take any \(x \in A\). Then \(F(x) \in F(A)\). By definition of preimage, \(x \in F^{-1}(F(A))\). Thus,
\[
A \;\subseteq\; F^{-1}(F(A)).
\]
Combining both inclusions, we get
\[
F^{-1}\bigl(F(A)\bigr) \;=\; A.
\]
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## (b) \(F(A_{1}\,\cap\,A_{2}) \;=\; F(A_{1}) \,\cap\,F(A_{2})\) for all \(A_{1}, A_{2} \subseteq X\)
1. **Show \(\subseteq\).**
Suppose \(y \in F(A_{1} \cap A_{2})\). Then there exists some \(x \in A_{1} \cap A_{2}\) with \(F(x) = y\). Since \(x\in A_{1}\) and \(x\in A_{2}\), it follows \(y \in F(A_{1})\) and \(y \in F(A_{2})\). Hence \(y \in F(A_{1}) \cap F(A_{2})\). Therefore,
\[
F(A_{1} \cap A_{2}) \;\subseteq\; F(A_{1}) \cap F(A_{2}).
\]
2. **Show \(\supseteq\).**
Suppose \(y \in F(A_{1}) \cap F(A_{2})\). Then \(y \in F(A_{1})\) and \(y \in F(A_{2})\). So there exist \(x_{1} \in A_{1}\) and \(x_{2} \in A_{2}\) such that
\[
F(x_{1}) = y \quad\text{and}\quad F(x_{2}) = y.
\]
Because \(F\) is injective, \(F(x_{1}) = F(x_{2})\) implies \(x_{1} = x_{2}\). Call this common element \(x\). Then \(x \in A_{1}\) and \(x \in A_{2}\), i.e. \(x \in A_{1} \cap A_{2}\). Thus \(y = F(x) \in F(A_{1} \cap A_{2})\). Hence,
\[
F(A_{1}) \cap F(A_{2}) \;\subseteq\; F(A_{1} \cap A_{2}).
\]
Combining both inclusions, we get
\[
F(A_{1} \cap A_{2}) \;=\; F(A_{1}) \cap F(A_{2}).
\]
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## Conclusion
Because \(F\) is injective:
1. For every subset \(A \subseteq X\),
\[
F^{-1}\bigl(F(A)\bigr) = A.
\]
2. For every pair of subsets \(A_{1}, A_{2} \subseteq X\),
\[
F(A_{1}\,\cap\,A_{2}) = F(A_{1}) \,\cap\, F(A_{2}).
\]
