To prove that $f : X \rightarrow Y$ is surjective under the condition that $\forall B \subset Y : f(f^{-1}(B)) = B$, let's analyze this condition carefully.

Proof

1. Given Condition: $\forall B \subset Y : f(f^{-1}(B)) = B$ This means that for any subset $B$ of $Y$, applying $f$ to the preimage of $B$ under $f$ yields $B$ itself.

2. Goal: We want to show that $f$ is surjective, i.e., for every $y \in Y$, there exists an $x \in X$ such that $f(x) = y$.

3. Contrapositive Argument: Suppose $f$ is not surjective. This would mean that there exists some $y_0 \in Y$ such that $y_0 \notin f(X)$. Define the subset $B = \{ y_0 \}$ of $Y$.

4. Analyzing $f(f^{-1}(B))$ for $B = \{ y_0 \}$:

   • Since $y_0 \notin f(X)$, there is no $x \in X$ such that $f(x) = y_0$.
   • Thus, $f^{-1}(B) = f^{-1}(\{ y_0 \}) = \emptyset$, because no element in $X$ maps to $y_0$ under $f$.

5. Applying $f$ to $f^{-1}(B)$:

   • Now we compute $f(f^{-1}(B)) = f(\emptyset)$.
   • Since the image of the empty set is empty, $f(\emptyset) = \emptyset$.

6. Contradiction:

   • According to our assumption, $f(f^{-1}(B))$ should equal $B = \{ y_0 \}$.
   • However, we found that $f(f^{-1}(B)) = \emptyset$, which contradicts $f(f^{-1}(B)) = B$.

7. Conclusion: This contradiction implies that our assumption (that $f$ is not surjective) is false. Therefore, $f$ must be surjective.

Final Answer

Hence, $f$ is surjective.

Would you like a deeper breakdown of any part of this proof, or do you have questions about the concepts?

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Relative Questions

1. How does the concept of preimages help in proving surjectivity?
2. Can this proof method be adapted to prove injectivity under different conditions?
3. What happens if the equality $f(f^{-1}(B)) = B$ holds only for certain subsets $B \subset Y$?
4. How does surjectivity relate to the concept of bijectivity in functions?
5. How does this proof change if we have a different condition on $f$ such as injectivity?

Tip

When working with preimages and images, it's helpful to examine the behavior on empty sets, as they can often reveal key insights in proofs involving mappings.