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Please verify if this proof is rigorous and correct. And if not provide feedback of what is wrong/lacking.

def1 $ A \subseteq B: \Leftrightarrow \forall x \in A \Rightarrow x \in B$

Given a function $f: X \rightarrow Y$, and subsets $A$ of $X$ and $B$ of $Y$ we define

def2 $f(A)=\{y \in Y \mid y=f(x)$ whenever $x \in A\}$ and

$f^{-1}(B)=\{x \in X \mid f(x) \in B\}$

Prov that:

$f^{-1}(G) \subseteq f^{-1}(H)$ whenere $G \subseteq H$

Let $ x \in f^{-1}(G)$, then by def 2 there exist $f(x) \in G$ by def 1 and the fact that $G \subseteq H$ this implies that $f(x) \in H$. Then it follows by def 2 that $x \in f^{-1}(H)$. therefore $f^{-1}(g) \subseteq f^{-1}(f)$

David C. Ullrich
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    Yes its correct. – Surb Feb 24 '22 at 11:31
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    If you want to make it really right you should note that "Let $ x \in f^{-1}(G)$, then by def 2 there exist $f(x) \in G$" doesn't quite make sense - in fact there's no $\exists$ in def 2. You meant just "Let $ x \in f^{-1}(G)$, then by def 2 $f(x) \in G$". – David C. Ullrich Feb 24 '22 at 13:31

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