1. ## another mapping question

Let $f$ be a function from $X$ onto $Y$ and let $B \subset Y$. Show that $f(f^{-1}(B))=B$.

I was thinking of something along the lines of

let $b \in B \ \& \ x \in X$ then $f^{-1}(b) = x \longrightarrow f(f^{-1}(b)) = f(x) = b$, since it's onto and it's pretty much by definition. I'm not quite sure that this is correct.

2. $z \in f\left( {f^{ - 1} (B)} \right)\; \Rightarrow \;\left( {\exists t \in f^{ - 1} (B)} \right)\left[ {z = f(t) \in B} \right]\;\; \Rightarrow \;f\left( {f^{ - 1} (B)} \right) \subseteq B$

By surjectivity $b \in B\; \Rightarrow \;\left( {\exists s \in X} \right)\left[ {f(s) = b} \right]$.
It follows: $\left[ {f(s) = b} \right]\; \Rightarrow \;s \in f^{ - 1} (B)\; \Rightarrow \;b = f(s) \in f\left( {f^{ - 1} (B)} \right)$

3. Plato:

I am sorry to bother you again but can you do this problem in predicate calculus,i would like to see the face of my lecturer to show him a proof in predicate calculus.

Thanks