Need condition which will make functions both true

Question

Let $f:X\mapsto Y$ Let $A \subset X, B \subset Y$

What condition on $f$ can make

$A=f^{-1}(f[A])$. and $B=f(f^{-1}[B])$ both true?

For the first ,its $f$ being 1-1, the second it’s $f$ being onto. But both I am somewhat perplexed .

I know both can be solved separately. The info is on MSE.

The only other condition I can think to make it work on both if $f$ is a bijection.

If I assume $f$ is a bijection,I have to show $f$ is invertible for both ,correct?

Please don’t Downvote me if it’s a duplicate.I checked and couldn’t find it

Help?

Answer 1

The definition of "$f$ is a bijection" is that $f$ is both injective (aka "1-1") and surjective (aka "onto"). And it is an almost trivial fact that $f$ is invertible if and only if it is a bijection. So no, if you know $f$ is a bijection, it is not necessary to prove it is invertible (though you should go through that proof once just to understand why it is an almost trivial fact).

Now

• ($f$ is bijective) $\implies (f$ is injective$) \implies A = f^{-1}(f(A))$
• ($f$ is bijective) $\implies (f$ is surjective$) \implies B = f(f^{-1}(B))$

So $f$ is bijective is indeed a single condition that is sufficient for both of the implications you want.

But it is not necessary. In fact all that is required for $A = f^{-1}(f(A))$ is that $f(A)$ and $f(X\setminus A)$ are disjoint. And all that is required for $B = f(f^{-1}(B))$ is $B \subset f(X)$. Neither full injectivity nor full surjectivity is needed.