TITLE: Inverse image of a union equals the union of the inverse images
QUESTION [19 upvotes]: I just wonder if my following solution is true.
Let $X,Y$ be sets, let $f:X\to Y$ be a function, let $\{Y_i\}_{i\in I}$ be a family of subsets of $Y$. (Note: I use equalities instead of mutual containment)
$$\begin{align}f^{-1}\left[\bigcup_{i\in I} Y_i\right]
 &= \{x \in X: \mbox{there exists an}\quad i \in I\mbox{ such that } y \in Y_i,f(x)=y\}
 \\&=\bigcup_{i \in  I} f^{-1}\left[Y_i\right]
\end{align}$$
I initially do not know how to get from the left to right, but when I put both sets in set notation, they turn out to be the same, hence the one line proof. Something go ultimately wrong?

REPLY [33 votes]: The statement is true, and your argument is essentially right, but I would say that you are skipping steps to achieve that identification.  (Also, you should not have those $\infty$'s on top of the union symbol). I would instead add:
$$\begin{align*}
f^{-1}\left[\bigcup_{i\in I}Y_i\right] &= \left\{ x\in X\;\left|\; f(x)\in\bigcup_{i\in I}Y_i\right\}\right.\\
&=\Biggl\{x\in X\;\Biggm|\; \exists i\in I\text{ such that }f(x)\in Y_i\Biggr\}\\
&= \bigcup_{i\in I}\{x\in X\mid f(x)\in Y_i\}\\
&= \bigcup_{i\in I}f^{-1}[Y_i].
\end{align*}$$
The first equality is by definition of inverse image; the second by definition of the union; the third is by definition of union; and the fourth by definition of inverse image.