Show $f(f^{-1}(B_0))\subset B_0$ and that equality holds if $f$ is surjective Let $f: A\to B$. Let $A_0\subset A$ and $B_0\subset B$. Show $f(f^{-1}(B_0))\subset B_0$ and that equality holds if $f$ is surjective.
Attempt:
I already did the first part. It is showing that equality holds for surjectivity that is troubling me.
Assume $f$ is surjective but $f(f^{-1}(B_0))\subsetneq B_0$.
$\implies$ there exists $f(a)\in B_0$ s.t $a\notin A_0$, but if $f$ is surjective this implies for all $f(x)\in B_0$ there exists $x\in A_0$ s.t. $f(x) = b\in B_0$ But this is a contradiction because we assumed $f$ was surjective. So $f(f^{-1}(B_0))\subset B_0$. So equality would hold??? I have a fealing this is completely bumbled up....
 A: $A_0$ doesn't seem to play a role here (the domain of $f$ is stated to be $A$, and nowhere else is $A_0$ referred to; was $A_0$ relevant to a different part of this problem?). Otherwise, you seem to have the right idea, but perhaps you are trying to do too many steps simultaneously. Try slowing down and carefully using the definitions (of images and pre-images, set inclusion/equality, and of surjectivity).
We already know $B_0\supseteq f(f^{-1}(B_0))$, to show they are equal, we must additionally show $B_0\subseteq f(f^{-1}(B_0))$.
You can also argue directly. Choose any $b \in B_0$. Then, there is $a \in A$ such that $f(a) = b$, and $a \in f^{-1}(B_0)$ (by definition of pre-image). Thus, $b = f(a) \in f(f^{-1}(B_0))$.
The assumption for contradiction is "There exists $b \in B_0$ so that $b \not\in f(f^{-1}(B_0))$" (i.e. for every $a \in f^{-1}(B_0)$, $f(a) \neq b$; though we really don't even need to state this consequence, as we will see). 
EDIT: and to complete the argument by contradiction, because $f$ is surjective, there is some $a_1 \in A$ so that $f(a_1) = b$. Then $a_1 \in f^{-1}(B_0)$ and so $b = f(a_1) \in f(f^{-1}(B_0))$, which directly contradicts our assumption. (and we have basically made the direct argument anyway)
A: $A_0$ doesn't have anything to do with this.
Since you already proved that $f(f^{-1}(B_0)) \subset B_0$, the other inclusion remains to prove. Let $b\in B_0$. Surjectivity implies that there exists $a\in A$ s.t. $f(a) = b$. Hence $a\in f^{-1}(B_0)$. So $b = f(a) \in f(f^{-1}(B_0))$.
