Let $f: A\rightarrow B$ be function and $X\subseteq A$.
The image of set X is defined as a set $f[X] = \{b\in B \:|\: \exists a\in X: f(a) = b\}$. Inverse image of set $Y\subseteq B$ is defined as $f^{-1} = \{a\in A\:|\:f(a)\in Y\}$.

Let $f : X\rightarrow Y$ and $C, D \subseteq Y$. Fill and prove formulas: $$f[f^{-1}[C]] \: ? \: C$$ $$f[f^{-1}[C]] = \: ?$$

I have read few articles about functions, images and inverse images and found that these two formulas should probably be written the following way: $$f[f^{-1}[C]] \subseteq C \text{ for all functions}$$ $$f[f^{-1}[C]] = C \text{ for surjective (onto) functions}$$ I could not find any proofs of them and it is what I am looking for.


By definition $x\in f^{-1}(C)\implies f(x)\in C$. Thus $f(f^{-1}(C))\subseteq C$.

Now if $f$ is surjective, then for each $c\in C$, $f^{-1}(c)\not=\emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(\bigcup_{c\in C}c)=f(\bigcup_{c\in C}f^{-1}(c))=\bigcup_{c\in C} f(f^{-1})(c)=\bigcup_{c\in C}c=C$.


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