Inverse of a function on two sets.

Question

I understand that $f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)$, but what is $f^{-1}(A\cap B)$?

Is it necessarily $f^{-1}(A)\cap f^{-1}(B)$?

Answer 1

Well, we have for $A,B\subset \Omega$ and a function $f:\Omega\rightarrow \Omega'$ \begin{align} f^{-1}(A\cap B):=\{x\in\Omega:f(x)\in(A\cap B)\}&= \{x\in\Omega:f(x)\in A\; \wedge \; f(x)\in B\} \\ &=\{x\in\Omega:f(x)\in A\} \cap \{x\in \Omega: f(x)\in B\}\\ &= f^{-1}(A)\cap f^{-1}( B) \end{align} so yes, it is exactly the same and therefore necessary.

Answer 2

Yes: $$ \eqalign{x \in& f^{-1}(A \cap B)\cr \iff &f(x) \in A \cap B \cr \iff &(f(x) \in A) \ \text{and}\ (f(x) \in B)\cr \iff &(x \in f^{-1}(A))\ \text{and}\ (x \in f^{-1}(B)) \cr \iff & x \in f^{-1}(A) \cap f^{-1}(B)}$$

Answer 3

If $f(x)$ is in A and also in B , then, in particular, $f(x)$ is in $A$ so $x$ is in $f^{-1}(A)$. Also, $f(x)$ is in $B$, so $x$ is in $f^{-1}(B)$. That is $x\in f^{-1}(A)\cap f^{-1}(B)$. Conversely, if $x$ is in both $f^{-1}(A)$ and $f^{-1}(B)$, then, in particular, $x$ is in $f^{-1}(A)$. That is $f(x)\in A$. Also, $x$ is in $f^{-1}(B)$, so $f(x)\in B$. That is $f(x)$ is in A and also in B.