I am currently proving the following set theory question for a real analysis course:
Given a function $f : S \to T$ and $A \subset S$ establish the following: $A \subset f^{-1}(f(A))$, with equality for all $A$ iff $f$ is injective.
I do not need help with the details of this proof, however, I am struggling to understand how I should write and format it. Is this question really asking to show $A = f^{-1}(f(A))$ iff $f$ is injective?
If so, do I assume $f$ is injective for both set inclusions ($A \subseteq f^{-1}(f(A))$ and $f^{-1}(f(A)) \subseteq A$ to show equality) or do I simply show $A = f^{-1}(f(A))$ iff $f$ is injective? I just want to make sure I am answering this question in full.
Best Answer
Prove the first theorem alone: $A\subseteq f^{-1}(f(A)).$
Then, instead of the stated theorem, show the equivalent, and more easily shown:
$$\left(\exists A: A\neq f^{-1}(f(A))\right)\iff (f\text{ is not injective})$$
This is equivalent because $P\iff Q$ is equivalent to $\lnot P\iff \lnot Q,$ and $\lnot \forall A: P(A)$ is equivalent to $\exists A:\lnot P(A).$
First assume $f$ is not injective. Then $f(x)=f(y)$ for some $x\neq y.$ Then let $A=\{x\}.$ You see that $y\in f^{-1}(f(A))$ so $f^{-1}(f(A))\neq A.$
Assume $A\neq f^{-1}(f(A)).$ Now, $A\subsetneq f^{-1}(f(A))$ so there must be a $y\in f^{-1}(f(A))$ such that $y\notin A.$ From there, prove that $f$ is not injective.