0
$\begingroup$

Not the best title but I don't know how to better describe it.

So the image of a set is usually written as $f(B)$, my question is, can I use sets in the place of variables in the expression of my mapping?

The reason I'm asking is, I have to show something along the lines of

$f^{-1}(A)=B\times C$

So I have to show both sides of the inclusion, so to show this I'm thinking of inserting the (supposed)preimage in the function and see if it gives A, but I'm not sure this kind of operation is legal. If not, what would be a general way to do it?

$\endgroup$
1
$\begingroup$

What you seem to have in mind does not work properly. This because $f(f^{-1}(A))$ does not necessarily equalize $A$. We have $f(f^{-1}(A))\subseteq A$ but $f(f^{-1}(A))= A$ demands that $A$ is a subset of the image of $f$.

The equality you mention can be handled by showing that for each $x$: $$f(x)\in A\iff x\in B\times C$$

$\endgroup$
  • $\begingroup$ Hey, thanks for the answer. I tried to do it but was met with some problems. Some tips on work around would be greatly appreciated. I've edit my post. $\endgroup$ – ChuckP Nov 26 '14 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.