# 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.

• If $U_i \not \subseteq Y$, we cannot make sense of $f^{-1}(U_i)$. – user21436 May 5 '12 at 12:11
• @AsafKaragila: Would it necessarily work if $f$ is continuous? What is $f$ is some general surjective map? – Bumbry May 5 '12 at 12:12
• What ever sets and functions you consider, you always have that the preimage of a union of subsets of the target set is the union of the preimages, same goes for intersections. The only thing that can fail is that the direct image of an intersection in the domain of the function may be smaller than the intersection of their images (in the target space.) – Olivier Bégassat May 5 '12 at 12:14

## 1 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.

• I assumed that $U_i$ is a subset of $Y$ and not of $X$. – Manos May 5 '12 at 12:13
• Thanks! You have assumed correctly. – Bumbry May 5 '12 at 12:37
• Then please edit your question so it actually says what you now say it says. – Gerry Myerson May 5 '12 at 12:41
• @GerryMyerson: done – Bumbry May 5 '12 at 13:43