functions - Union of preimages and preimage of union

Is it possible to have a map $f:X\to Y$ from a topological space $X$ to a set $Y$ and some subsets of $Y$ namely $U_i,i\in I$ such that $\bigcup_{i\in I} f^{-1}(U_i)$ is not equal to $f^{-1}\left(\bigcup_{i\in I}U_i\right)$ ?

I can't think of a case that this is true, but of course this doesn't mean anything! Thanks.

Answer

It is a general fact that for any mapping of sets $f: X \rightarrow Y$,

$\bigcup f^{-1}(U_i) = f^{-1}\left(\bigcup U_i\right)$ and $\bigcap f^{-1}(U_i)=f^{-1}\left( \bigcap U_i\right)$. Try proving this by elementary set
theory, i.e. take an element of $\bigcup f^{-1}(U_i)$ and show that it is an element of $f^{-1}\left(\bigcup U_i\right)$ and conversely.