the inverse image commutes with set operations

Theorem.Let $f$ be a mapping from $X$ to $Y$. If ${\{B_i\}}_{i\in I}$ isa (possibly uncountable) collection of subsets in $Y$, thenthe following relations hold for the inverse image:

1. $f^{-1}\left({\displaystyle \bigcup _{i\in I}}{B}_i\right)={\displaystyle \bigcup _{i\in I}}{f}^{-1}\left(B_i\right)$

2. $f^{-1}\left({\displaystyle \bigcap _{i\in I}}{B}_i\right)={\displaystyle \bigcap _{i\in I}}{f}^{-1}\left(B_i\right)$

If $A$ and $B$ are subsets in $Y$, then we also have:

1. For the set complement,

   $${\left(f^{-1}(A)\right)}^{\mathrm{\complement }}=f^{-1}(A^{\mathrm{\complement }}).$$

2. For the set difference,

   $$f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B).$$

3. For the symmetric difference,

   $$f^{-1}(A△B)=f^{-1}(A)△f^{-1}(B).$$

Proof.For part (1), we have

$f^{-1}\left({\displaystyle \bigcup _{i\in I}}{B}_i\right)$$=$$\{x\in X\mid f(x)\in {\displaystyle \bigcup _{i\in I}}{B}_i\}$

$=$$\{x\in X\mid f(x)\in B_i\text{for some}i\in I\}$

$=$${\displaystyle \bigcup _{i\in I}}\{x\in X\mid f(x)\in B_i\}$

$=$${\displaystyle \bigcup _{i\in I}}{f}^{-1}\left(B_i\right).$

Similarly, for part (2), we have

$f^{-1}\left({\displaystyle \bigcap _{i\in I}}{B}_i\right)$$=$$\{x\in X\mid f(x)\in {\displaystyle \bigcap _{i\in I}}{B}_i\}$

$=$$\{x\in X\mid f(x)\in B_i\text{for all}i\in I\}$

$=$${\displaystyle \bigcap _{i\in I}}\{x\in X\mid f(x)\in B_i\}$

$=$${\displaystyle \bigcap _{i\in I}}{f}^{-1}\left(B_i\right).$

For the set complement, suppose $x\notin f^{-1}(A)$. This is equivalent to$f(x)\notin A$, or $f(x)\in A^{\mathrm{\complement }}$, which is equivalent to$x\in f^{-1}(A^{\mathrm{\complement }})$. Since the set difference $A\setminus B$ can bewritten as $A\cap B^c$, part (4) follows from parts (2) and(3). Similarly, since $A△B=(A\setminus B)\cup (B\setminus A)$,part (5) follows from parts (1) and(4). $\mathrm{\square }$