Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
[ f(f-1)) is f and inverse of f]
If $\displaystyle f(A)\ \neq \ B$, $\displaystyle \ f^{-1}(S)$ may not even be defined! (As we may choose $\displaystyle S$ to be $\displaystyle B-f(A)$)Originally Posted by ummmm
Or have I missed something.
Of course I have missed something:
choose $\displaystyle S$ to be $\displaystyle B-f(A)$,
then:
$\displaystyle f^{-1}(S)\ =\ f^{-1}(B-f(A))\ =\ \emptyset$ -the empty set.
Back to the problem:
(we need only consider the case where $\displaystyle f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)
$\displaystyle f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,
so if $\displaystyle a\ \epsilon \ f^{-1}(S)$ then $\displaystyle f(a)\ \epsilon \ S$ and $\displaystyle f(a)\ \epsilon \ f(A)$
$\displaystyle \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $\displaystyle f(f^{-1}(S))\ \subseteq\ f(A)$,
i.e. $\displaystyle \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$
Now suppose $\displaystyle s \epsilon\ (S\cap f(A))$ then clearly
$\displaystyle s\ =\ f(f^{-1}(s))$ and so $\displaystyle \ s \epsilon f(f^{-1}(S))\ $
$\displaystyle \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$
hence:
$\displaystyle S\cap f(A)\ =\ f(f^{-1}(S))$
RonL
The proof on its own:
(we need only consider the case where $\displaystyle f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)
$\displaystyle f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,
so if $\displaystyle a\ \epsilon \ f^{-1}(S)$ then $\displaystyle f(a)\ \epsilon \ S$ and $\displaystyle f(a)\ \epsilon \ f(A)$
$\displaystyle \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $\displaystyle f(f^{-1}(S))\ \subseteq\ f(A)$,
i.e. $\displaystyle \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$
Now suppose $\displaystyle s \epsilon\ (S\cap f(A))$ then clearly
$\displaystyle s\ =\ f(f^{-1}(s))$ and so $\displaystyle \ s \epsilon f(f^{-1}(S))\ $
$\displaystyle \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$
hence:
$\displaystyle S\cap f(A)\ =\ f(f^{-1}(S))$
RonL