# Show that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$

Show that $f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)$ but not necessarily

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

Let $S=A\cup B$

I know that $f^{-1}(S)=\{x:f(x)\in S\}$ assuming that that $f$ is one to one. Is this true $\{x:f(x)\in S\}=\{x:f(x) \in A\}\cup\{x:f(x)\in B\}$?

Why doesn't the intersection work?

Sources : ♦ 2nd Ed, $\;$ P219 9.60(d), $\;$ Mathematical Proofs by Gary Chartrand,
♦ P214, $\;$ Theorem 12.4.#4, $\;$ Book of Proof by Richard Hammack,
♦ P257-258, $\;$ Theorem 5.4.2.#2(b), $\;$ How to Prove It by D Velleman.

• $f^{-1}[S]=\{x:f(x)\in S\}$ by definition; this has nothing to do with whether $f$ is one-to-one. – Brian M. Scott Apr 10 '13 at 23:52
• then f^-1 isnt a function right if f isnt one to one – sarah Apr 10 '13 at 23:53
• If $f:A\to B$, then $f^{-1}:P(B)\to P(A)$ is a function on the power sets. – Berci Apr 10 '13 at 23:55
• That’s right, but it has no bearing on this problem. The statement that $f^{-1}[A\cup B]=f^{-1}[A]\cup f^{-1}[B]$ is true for all $A,B$, and $f$. – Brian M. Scott Apr 10 '13 at 23:55
• I think of $f^{-1}(S)$ informally as "stuff that lands inside $S$ (when I hit it with $f$)". You're being asked to show that "x lands inside $A\cup B$" iff "x lands inside $A$ or $B$", which is obviously true. You're also being asked to disprove the claim that "x lands inside $A\cap B$" iff "x lands inside both $A$ and $B$", but this claim is obviously true too, as Cameron Buie says. – Billy Apr 11 '13 at 0:25

\begin{align}f^{-1}[A\cap B] &:= \{x\in\text{dom}(f):f(x)\in A\cap B\}\\ &= \{x\in\text{dom}(f):f(x)\in A\text{ and }f(x)\in B\}\\ &= \{x\in\text{dom}(f):f(x)\in A\}\cap\{x\in\text{dom}(f):f(x)\in B\}\\ &=: f^{-1}[A]\cap f^{-1}[B].\end{align}
You'll proceed similarly to show that $$f^{-1}[A\cup B] = f^{-1}[A]\cup f^{-1}[B],$$ trading "and" for "or".
On the other hand, while we have $$f[A\cup B]=f[A]\cup f[B]$$ and $$f[A\cap B]\subseteq f[A]\cap f[B],$$ we don't generally have equality in the last case, unless $$f$$ is one-to-one. Pick any constant function on your personal favorite set of two or more elements, then choose non-empty two disjoint subsets $$A$$ and $$B$$ for an example where the inclusion is strict.