# the inverse image commutes with set operations

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

1. (1)

$\displaystyle f^{-1}\big{(}\bigcup_{i\in I}B_{i}\big{)}=\bigcup_{i\in I}f^{-1}% \big{(}B_{i}\big{)}$

2. (2)

$\displaystyle f^{-1}\big{(}\bigcap_{i\in I}B_{i}\big{)}=\bigcap_{i\in I}f^{-1}% \big{(}B_{i}\big{)}$

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

1. (3)

For the set complement,

 $\big{(}f^{-1}(A)\big{)}^{\complement}=f^{-1}(A^{\complement}).$
2. (4)

For the set difference,

 $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B).$
3. (5)

For the symmetric difference,

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

Proof. For part (1), we have

 $\displaystyle f^{-1}\big{(}\bigcup_{i\in I}B_{i}\big{)}$ $\displaystyle=$ $\displaystyle\Big{\{}x\in X\mid f(x)\in\bigcup_{i\in I}B_{i}\Big{\}}$ $\displaystyle=$ $\displaystyle\left\{x\in X\mid f(x)\in B_{i}\ \mbox{for some}\ i\in I\right\}$ $\displaystyle=$ $\displaystyle\bigcup_{i\in I}\left\{x\in X\mid f(x)\in B_{i}\right\}$ $\displaystyle=$ $\displaystyle\bigcup_{i\in I}f^{-1}\big{(}B_{i}\big{)}.$

Similarly, for part (2), we have

 $\displaystyle f^{-1}\big{(}\bigcap_{i\in I}B_{i}\big{)}$ $\displaystyle=$ $\displaystyle\big{\{}x\in X\mid f(x)\in\bigcap_{i\in I}B_{i}\big{\}}$ $\displaystyle=$ $\displaystyle\left\{x\in X\mid f(x)\in B_{i}\ \mbox{for all}\ i\in I\right\}$ $\displaystyle=$ $\displaystyle\bigcap_{i\in I}\left\{x\in X\mid f(x)\in B_{i}\right\}$ $\displaystyle=$ $\displaystyle\bigcap_{i\in I}f^{-1}\big{(}B_{i}\big{)}.$

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

Title the inverse image commutes with set operations TheInverseImageCommutesWithSetOperations 2013-03-22 13:35:24 2013-03-22 13:35:24 matte (1858) matte (1858) 11 matte (1858) Proof msc 03E20 SetDifference