I was solving a problem of set theory and applications then suddenly I faced this problem:
Let $ f: X \to Y $ be a function, and let $ B \subseteq Y $. Prove that:
\begin{align*}
f^{-1}(Y-B) = X – f^{-1}(B)
\end{align*}
I proved that:
\begin{align*}
f^{-1}(Y-B) = f^{-1}(Y) – f^{-1}(B)
\end{align*}
Now I think It is enough to show that $ f^{-1}(Y) = X $.
A friend of mine told me that the equality $ f^{-1}(Y) = X $ is always true, but I have doubt. What does that mean? Is it really true that the pre-image of a co-domain is always the domain?! if not what is the counterexample?
Best Answer
Let $x\in f^{-1}(Y-B)$ then $\exists$some $y\in Y-B$ such that $y=f(x)$ Now $f(x)\notin B$ so $x\notin f^{-1}(B)$ so $x \in X-f^{-1}(B)$.Now conversely$x\notin f^{-1}(B)$ implies $f(x)\notin B$ so $f(x)\in Y-B$ thus $x\in f^{-1}(Y-B)$.Trivial case where $f^{-1}(Y-B)$ is empty is not discussed.I think you can do it from above. And for your doubt,check the definition, $f^{-1}(Y)=\{x\in X : f(x) \in Y\}$ which is obviuously $X$,because for any element of $X$,the property is true.