1. ## help, i'd be impressed

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]

2. Originally Posted by ummmm
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 $f(A)\ \neq \ B$, $\ f^{-1}(S)$ may not even be defined! (As we may choose $S$ to be $B-f(A)$)

Or have I missed something.

Of course I have missed something:

choose $S$ to be $B-f(A)$,

then:

$f^{-1}(S)\ =\ f^{-1}(B-f(A))\ =\ \emptyset$ -the empty set.

Back to the problem:

(we need only consider the case where $f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)

$f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,

so if $a\ \epsilon \ f^{-1}(S)$ then $f(a)\ \epsilon \ S$ and $f(a)\ \epsilon \ f(A)$

$\Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $f(f^{-1}(S))\ \subseteq\ f(A)$,

i.e. $\ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$

Now suppose $s \epsilon\ (S\cap f(A))$ then clearly
$s\ =\ f(f^{-1}(s))$ and so $\ s \epsilon f(f^{-1}(S))\$

$\Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$

hence:

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

RonL

3. ## i think f(a)=b

i think f(a) has to equal b because there if a function from a to b.

4. ## anyone?

anyone?

5. B is merely the codomain for f
f(A) may be a proper subset of B

All that is necessary is that if x and y are in A and f(x) and f(y) are distinct, then x and y are distinct.

6. The proof on its own:

(we need only consider the case where $f^{-1}(S)\ \neq\ \emptyset$ as the result holds trivally otherwise)

$f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \}$,

so if $a\ \epsilon \ f^{-1}(S)$ then $f(a)\ \epsilon \ S$ and $f(a)\ \epsilon \ f(A)$

$\Rightarrow\ f(f^{-1}(S))\ \subseteq\ S$ and $f(f^{-1}(S))\ \subseteq\ f(A)$,

i.e. $\ f(f^{-1}(S))\ \subseteq\ S\cap f(A)$

Now suppose $s \epsilon\ (S\cap f(A))$ then clearly
$s\ =\ f(f^{-1}(s))$ and so $\ s \epsilon f(f^{-1}(S))\$

$\Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))$

hence:

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

RonL