When do we have $A \subset B$ imples $f^{-1}(A) \subset f^{-1}(B)$? Im not sure if continuity of the function $f$ is enough to have the above. Or is it the monotonicity of $f$?
 A: All you need is that $f$ is a function. If $x\in f^{-1}A$, then $f(x)\in A$, so $f(x)\in B$ as well.
Carl Mummert pointed out that if $R$ is any relation between $X$ and $Y$ then the claim still holds, where $R^{-1}A=\{x\in X: xRa\text{ for some }a\in A\}$. This is because if $A\subset B$, then $x\in R^{-1}A$ implies $xRa$ for some $a\in A$ implies $x\in R^{-1}B$.
A: Here is an alternative proof, where we start with the most complex side and expand the definitions, and then work towards $\;A \subseteq B\;$:
\begin{align}
& f^{-1}[A] \;\subseteq\; f^{-1}[B] \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \langle \forall x :: x \in f^{-1}[A] \;\Rightarrow\; x \in f^{-1}[B] \rangle \\
\equiv & \qquad \text{"basic property of $\;^{-1}[\quad]\;$, twice"} \\
& \langle \forall x :: f(x) \in A \;\Rightarrow f(x) \in B \rangle \\
\Leftarrow & \qquad \text{"logic: generalize $\;f(x)\;$ to any $\;y\;$ -- to get rid of $\;f\;$"} \\
& \langle \forall y :: y \in A \;\Rightarrow y \in B \rangle \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& A \subseteq B
\end{align}
