I am trying to do this question based on sqtrat's answer, but I am having trouble understanding their answer based on mine.
Let $f:X\to Y$ be a map and let $B_{1}\subset Y$. Show that $f^{-1}(Y\setminus B_{1})=X\setminus f^{-1}(B_{1})$.
Let $x\in f^{-1}(Y\setminus B_{1})$. Then there exists a $y\in Y\setminus B_{1}$ such that $f^{-1}(y)=x$. Then $y\in Y$ and $y\notin B_{1}$. Then $x\in X$ and $x\notin f^{-1}(B_{1})$.
The part I do not understand: why is $f^{-1}(Y)=X$? Because the function is bijective due to it having an inverse?
The identity $f^{-1}(Y)=X$ is always true for any function $f:X \rightarrow Y$, and is in fact a trivial statement.
If $x\in X$ then $f(x) \in Y$ hence $x \in f^{-1}(Y)$. Thus, $X \subset f^{-1}(Y)$.
But certainly, we also have $f^{-1}(Y) \subset X$ by definition of the inverse image.
The equality follows.
NB: For any function, we can define the inverse image of a subset of the target space, and we use the symbol $f^{-1}$ for it. It must not be mixed with the notation $f^{-1}$ that represents the inverse function of a bijection. In order to distinguish both of them, just look at the nature of the argument. The "inverse image" symbol takes subsets of the target space as argument, whereas the inverse function of a bijection takes elements of the target space as argument.