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.

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: $$A\subseteq f^{-1}\big(f(A)\big)$$
