What does $f^{-1}(B)= \{ x \in X \mid f(x) \in B\}$ mean? I have encountered the expression
$$f^{-1}(B) = \{ x \in X \mid f(x) \in B\}$$
My questions are:

1) What does the $-1$ exponent mean in this context? 
  2) Is it right to say "if the set $X$ contains the element $x$ then the image of $x$ is contained in the set $B$"? 
  3) What does it mean to assign a function with this notation $\{ x \in X \mid f(x) \in B\}$? How could function be equal to this notation?

 A: In this context, $f^{-1}(B)$ is not itself a function. We read this as the inverse image of $B$ under $f$, and it is the set that contains every element in $X$ that maps into $B$ under $f$. 
A: The equation you quote is the definition of the $f^{-1}$ notation on its the left-hand side. If $f$ is a function $X\to Y$, then it defines a function called $f^{-1}:\mathcal P(Y)\to\mathcal P(X)$.
Note that this is a different $f^{-1}$ than the inverse function $f^{-1}:Y\to X$ which you can define if (but only if) $f$ is a bijection. The ambiguity usually doesn't cause confusion because you can see whether the thing $f^{-1}$ is applied to is an element of $Y$ or a subset of $Y$.
(The preimage function defined by your equation exists no matter whether $f$ is or isn't injective or surjective).
A: This is known as the preimage of a set under a function. First, some abstraction, then some concrete examples.
Let $f: X \to Y$. This means for any element $x \in X$, there exists a unique element $y \in Y$ such that $f(x) = y$. In high school algebra, you learned that the statement is equivalent to $f^{-1}(y) = x$, where $f^{-1}$ is the inverse function. In the same vein, we can define mappings of sets under functions.  If $A \subseteq X$, we say $f(A) = \{y \in Y \mid \exists x \in X \text{ s.t. } f(x) = y\}.$ Intuitively, this is the set of all values in the codomain of $f$ such that if we take every element in $A$ and map it under $f$, the value is in that set. Again we can generalize inverses. The inverse image (preimage) of $B \subseteq Y$ under $f$ is $\{ x \in X \mid f(x) \in B\}$. Intuitively, this is the largest set of elements in the domain such that the mapping of the elements is in $B$. Note the similarities and differences between $f(A)$ $f^{-1}(B)$. The set $f(A)$ is a subset of the codomain, and $f^{-1}(B)$ is a subset of the domain, yet each is explicitly determined by the action of $f$ on elements in the domain.
As an example, let $f: [-10,10] \to \mathbb{R}$ with $x \mapsto x^2$. We have

1) $f([0,10]) = [0,100]$. 
  2) $f([-10,10]) = [0,100]$. 
  3) $f(\{0\}) = \{0\}$. 
  4) $f^{-1}([0,100]) = [-10,10]$. 
  5) $f^{-1}(\mathbb{R}) = [-10,10]$. 
  6) $f^{-1}(\{0,16,25\}) = \{-5, -4, 0, 4, 5\}$.

A: $f^{-1}$ is known as the inverse mapping of $f$. $f^{-1}(B)$ is the image of inverse mapping of $f$ under $B$.
1) $-1$ means inverse like invertible matrix $A^{-1}$.
2) Yes.
3) $\{x∈X∣f(x)∈B\}$ is a set that contains all points whose image under $f$ is $B$. Generally $f^{-1}$ is not a function. It is a function if and only if $f$ is $1-1$.
