Question: I am reading a book, and it suddenly says:

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> A distribution $\nu$ is symmetric around a point $x\_0 \in \mathbb{R}$ if $h(\nu) = \nu$ where $h$ is the affine map given by $h(x) = 2x\_0 - x$. As $h^{-1}(x\_0 - A) = x\_0 + A$ for $A \subset \mathbb{R}$ ...
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What does that mean?

Answer: 
The affine map $h$ is invertible, with inverse $h^{-1}(x) = h(x)$. So $$h^{-1}(x\_0 - A) = h(x\_0 - A) = 2x\_0 - (x\_0 - A) = x\_0 + A.$$

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**Added Later:** The question has been edited so that $A \subset \mathbb{R}$ rather than $A \in \mathbb{R}$. 

Now $x\_0 - A$ denotes the set $\{x\_0 - a \mid a \in A\}$ and $x\_0 + A$ denotes the set $\{x\_0 + a \mid a \in A\}$. Given a function $f : X \to Y$, and a subset $S \subset X$, then $f(S)$ denotes the subset $\{f(s) \mid s \in S\}$ of $Y$. So,

$$h^{-1}(x\_0 - A) = \{h^{-1}(x\_0 - a) \mid a \in A\} = \{h(x\_0 - a) \mid a \in A\} = \{x\_0 + a \mid a \in A\} = x\_0 + A.$$
It's just the inverse of $h(x)$. Call it $y$:

$$y = 2x\_0 - x \to x = 2x\_0 - y = h^{-1}(y)$$
$$h^{-1}(x\_0 - A) = 2x\_0 - (x\_0 - A) = x\_0 + A.$$