For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.
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Proof:$\displaystyle f\text{ is injective }\Longleftrightarrow f^{-1}f(A)=A$ for every $\displaystyle A\in\text{Codom }f$
$\displaystyle \implies$:
$\displaystyle x\in A\implies f(x)\in f(A)\implies x\in f^{-1}ff(A)$ (this is always true). So, $\displaystyle A\subseteq f^{-1}f(A)$
Lemma: $\displaystyle f(x)\in f(A)\implies x\in A$
Proof: Suppose that $\displaystyle x\notin A$ and let $\displaystyle y=f(x)$. Since $\displaystyle y\in f(A)$ there exists some $\displaystyle a\in A$ such that $\displaystyle f(a)=y$. But, since $\displaystyle x\notin A$ we see that $\displaystyle x\ne a$ and so $\displaystyle f(x)=y=f(a)$ which condtradicts $\displaystyle f$'s injectivity. The conclusion follows $\displaystyle \blacksquare$
Now, then $\displaystyle x\in f^{-1}f(A)\implies f(x)\in f(A)\implies x\in A$
so $\displaystyle f^{-1}f(A)\subseteq A$
from where it follows that $\displaystyle A=ff^{-1}(A)$
$\displaystyle \Leftarrow:$ Let $\displaystyle f(x)=f(y)$ then $\displaystyle f(\{x\})=\{f(x)\}=\{f(y)\}=f(\{y\})$ and so $\displaystyle \{x\}=f^{1}f(\{x\})=f^{-1}f(\{y\})=\{y\}$ from where it follows that $\displaystyle x=y$