Answer
See explanation
Work Step by Step
To prove \(F\bigl(F^{-1}(B)\bigr) = B\) for all subsets \(B \subseteq Y\), given that \(F : X \to Y\) is onto (surjective), proceed as follows:
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### 1. Show \(F\bigl(F^{-1}(B)\bigr) \subseteq B\)
Let \(y \in F\bigl(F^{-1}(B)\bigr)\). By definition, there exists some \(x \in X\) such that
\[
x \in F^{-1}(B) \quad \text{and} \quad F(x) = y.
\]
Since \(x \in F^{-1}(B)\), it follows that \(F(x) \in B\). Hence \(y \in B\). Because \(y\) was arbitrary, we conclude
\[
F\bigl(F^{-1}(B)\bigr) \subseteq B.
\]
---
### 2. Show \(B \subseteq F\bigl(F^{-1}(B)\bigr)\)
Now let \(y \in B\). Because \(F\) is onto, there exists an \(x \in X\) such that \(F(x) = y\). Since \(y \in B\), that same \(x\) satisfies \(x \in F^{-1}(B)\). Therefore,
\[
y = F(x) \in F\bigl(F^{-1}(B)\bigr).
\]
Thus every \(y \in B\) is in \(F\bigl(F^{-1}(B)\bigr)\), giving us
\[
B \subseteq F\bigl(F^{-1}(B)\bigr).
\]
---
### 3. Conclude equality
Combining both inclusions,
\[
F\bigl(F^{-1}(B)\bigr) \;\subseteq\; B
\quad\text{and}\quad
B \;\subseteq\; F\bigl(F^{-1}(B)\bigr),
\] we obtain
\[
F\bigl(F^{-1}(B)\bigr) \;=\; B.
\]
This completes the proof.
