How possible for function to be $A \subset f^{-1} \left( f(A) \right) \forall f$? While trying to understand concept of measurable function I read on wiki more about function inverse and found interesting fact about them. 

For every function $f$, subset $A$ of the domain and subset $B$ of the
  codomain we have $A \subset f^{−1}(f(A))$ and $f(f^{−1}(B))\subset B$. 
If $f$ is injective we have $A = f^{−1}(f(A))$ and if ''f'' is
  surjective we have $f(f^{−1}(B)) = B$.



I have made a sketch, and have some questions:
1) do we need to have mapping from all elements of A to other set? (Otherwise I get $f^{−1}(f(A)) \subset A$, see picture 2)
2) can inverse of surjective function have 2 elements at the domain? 
3)$A \subset f^{−1}(f(A))$ does not work in general as you see! Maybe I do some restricted operations? It only works if I have 2 elements: 1 from set and 1 out of the set mapping to the same element in codomain.
 A: 1: Yes it does have to be such that $f:A\to B$ is defined for all elements in $A$ because otherwise you have a subset $A^\ast\subset A$ and the function is instead defined for this subset, so it's really is $f:A^\ast\to B$
2: Yes, for example $f:\mathbb{R}\to\mathbb{R}_{0\leq}$ with $f(x)=x^2$ is surjective, and $f^{-1}(1)=\{-1,1\}$ as both gives us the number $1$
3: is is always valid because $f(A)$ has already mapped all elements of $A$ to other elements in B, so when you inverse it you get it back, but I think you might have typoed as it should be $A\subseteq f^{-1}(f(A))$ as it may very well be equal, as if when $f$ is injective.
A: Let $f\colon X\to Y$ be a map and assume $A\subset X$.
We want to prove that $A\subset f^{-1}(f(A))$.
Note that, for an element $x\in X$ and $B\subset Y$, the statement $x\in f^{-1}(B)$ is equivalent to $f(x)\in B$.
So what we have to verify is that, for $x\in A$, $f(x)\in f(A)$, which is true by definition of $f(A)$. So the statement
$$
A\subset f^{-1}(f(A))
$$
holds for all subsets $A$ of $X$.
Similarly, if $B\subset Y$, we want to show that $f(f^{-1}(B))\subset B$. So, let $y\in f(f^{-1}(B))$; by definition, there is $x\in f^{-1}(B)$ such that $y=f(x)$. However, saying $x\in f^{-1}(B)$ is the same as saying $f(x)\in B$. Therefore $y=f(x)\in B$. Hence the statement
$$
f(f^{-1}(B))\subset B
$$
holds for all subsets $B$ of $Y$.
In the second diagram you don't have a function, because you lack a definition for $f(5)$.
In the fourth diagram, $A=\{1,2,3\}$, so $f(A)=\{f(1),f(2),f(3)\}=\{a,c,d\}$. Next $f^{-1}(f(A))=f^{-1}(\{a,c,d\})=\{1,2,3,4\}$ which confirms the general statement.
For your final questions.
1) Yes, there must be a definition of $f$ on all the elements of the domain
2) I can't understand what you're asking
3) The statement is valid in general, as I proved above.
