How to proof: if $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$ and show that it can't work conversely? Hello so I was wondering how you can prove $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$. I found someone with a similar problem on stack exchange but didn't  found it much of a proof.
I know if you have $f:A\rightarrow B$ is a function with e.g. $K\subset Y$ then the inverse image is $f^{-1}(K)=\{x\in X \mid f(x)=K \}$. Which also means that $x\in f^{-1}(K)\leftrightarrow f(x)\in K$. And that for $A\subset B$ means that $x\in A \rightarrow x\in B$.
But knowing this how can I proof this? Because I can say $x\in f^{-1}(A)\rightarrow x\in f^{-1}(B)$ So therefore $f(x)\in A \rightarrow f(x)\in B$ But now I don't know if I automatically can say that $A\subset B$ (I don't think so).
Secondly I want to show using a counterexample that $ f^{-1}(A)\subset f^{-1}(B) \rightarrow A\subset B $ Doesn't work.
If someone knows the answer please share your wisdom :)
Thanks in advance
 A: You are given $A\subseteq B$ and need to show $f^{-1}(A)\subseteq f^{-1}(B)$. You can show it by proving that every $x\in f^{-1}(A)$ also belongs to $f^{-1}(B)$.
Let $x\in f^{-1}(A)$. Then $f(x)\in A$, so $f(x)\in B$ since $A\subseteq B$. Thus $x\in f^{-1}(B)$.

For a counterexample of $f^{-1}(A)\subseteq f^{-1}(B)\implies A\subseteq B$, consider $f:\Bbb R\to\Bbb R$ with $f(x)=x^2,x\in\Bbb R$. Let $A=[-1,1],B=[0,4]$. Then $f^{-1}(A)=[-1,1],f^{-1}(B)=[-2,2]$ but $A\not\subseteq B$.
A: $\newcommand{\inv}[1]{f^{-1}[#1]}$
This is my answer.
The inverse image of the mapping $f:X \rightarrow Y$ is defined as, for any $B \subseteq Y$,
$$
f^{-1}[B] = \{x \in X~|~\exists y \in B: f(x) = y\}
$$
Therefore, for all $x\in X$, $x \in f^{-1}[B]$ is equivalent to $f(x) \in B$. We can solve your problem using this property.
Let $A \subseteq B \subseteq Y$. Then, for all $x \in X$, $x \in f^{-1}[A]$ implies $f(x) \in A \subseteq B$, which gives $x\in f^{-1}[B]$. Therefore, $f^{-1}[A] \subseteq f^{-1}[B]$.
Conversely, suppose $f^{-1}[A] \subseteq f^{-1}[B]$. In this case, we don't have much to say. However, if we assume $f$ to be subjective we can conclude $A \subseteq B$. Let $f$ be surjective. Then, for all $y \in A$, there exists $x \in X$ such that $f(x) = y$, hence $x \in \inv{A} \subseteq \inv{B}$, which gives $y = f(x) \in B$.
