# Image of intersections is equal to intersections of images if and only if the inverse of the image is the set

I'm working through some exercises dealing with mappings and the properties of its image and pre-image. Let $$f: S \to T$$ be function. I want to prove that the statements

• $$f$$ is injective
• $$f(A\cap B) = f(A) \cap f(B)$$ for all $$A,B \subseteq S$$
• $$f^{-1}(f(A)) = A$$ for all $$A \subseteq S$$

are equivalent. Consulting the amazing: Overview of basic results about images and preimages one finds that it is quite easy to show that the first statement is equivalent to the other two, which should establish the equivalence of all three.

This seems to be the standard way to go about this. However, would it be possible to show that the second and third statements are equivalent?

Out of curiosity I played around with this for a bit and wasn't able to get anywhere. So this is either very easy or it takes some trick I've been unable to figure out.

If anyone could point me to a proof or give me a hint, it would be greatly appreciated!

Suppose we have $$f^{-1}(f(A)) = A$$ and want to prove $$f(A\cap B) = f(A) \cap f(B)$$.
Assume we know, that set properties holds for $$f^{-1}$$ we have $$f^{-1}(f(A\cap B)) = A\cap B = f^{-1}(f(A)) \cap f^{-1}(f(B)) = f^{-1}(f(A) \cap f(B))$$ not taking image from both sides you obtain desired.