Intuitive proof for a simple set theory fact To my embarrassment, while working out an algebraic geometry exercise, I became hung up on a simple set theory fact:

Proposition: Let $f : X \rightarrow Y$ be any map of sets with $A \subseteq X$ and $B \subseteq Y$. Then $f(f^{-1}(B)) \subseteq B$ and $A \subseteq f^{-1}(f(A))$.

It can be given a simple proof:

Proof: Let $b \in f(f^{-1}(B))$. Then, $b = f(a)$, where $a \in f^{-1}(B)$, i.e., $f(a) \in B$, i.e., $b \in B$. Similarly, let $a \in A$, then by definition $a \in f^{-1}(f(\{ a \})) \subseteq f^{-1}(f(A))$.

(See also here.) That said, the proposition and its proof seem somehow unintuitive to me, for reasons I find difficult to articulate. Does the proposition admit an intuitive 'summary', which makes it sound 'obvious' or otherwise something I can easily hang my hat on? Thanks.
 A: $f^{-1}(B)$ is the subset of $X$ which maps to values in $B$. If you push this set through $f$ it will give you a set with values in $B$, which is a subset of $B$. 
Similarly, $f(A)$ is the set of values in $Y$ which are the outputs of $f$ in $A$. If we pull this back through $f^{-1}$, you get the subset of $X$ whose values like in $f(A)$, which is $A$ and the parts of $X$ which are not in $A$ which take values in $f(A)$. 
Try drawing a picture, if you still don't see it - the mathematical proof is pretty much the clearest form of the idea as far as I can tell. 
A: For a slightly different perspective:
If $f$ is not onto, then it will not, in general, map to all the points in $B$. So if you take the preimage of $B$, those points will "span out" some, but not necessarily all of it: $$f\big(f^{-1}(B)\big)\subseteq B$$
If $f$ is not one-one, then it will not, in general, map only one point to each of its outputs. So if you take the image of a set $A$, there may be other points in the domain which get mapped there as well: $$A\subseteq f^{-1}\big(f(A)\big)$$
