How to proof inverse function of unions are equal to union of inverse functions.

Question

I want to proof this equation below. $$ f^{-1}(\bigcup B_{j})=\bigcup(f^{-1}B_{j}) $$ I know it's true but I don't have idea how to proof and explain. Please help.

Answer 1

Essential is: $$x\in f^{-1}(A)\iff f(x)\in A$$

So the following statements are equivalent:

• $x\in f^{-1}(\bigcup_{j\in J}B_j)$
• $f(x)\in\bigcup_{j\in J}B_j$
• $f(x)\in B_{j_0}$ for some $j_0\in J$
• $x\in f^{-1}(B_{j_0})$ for some $j_0\in J$
• $x\in\bigcup_{j\in J}f^{-1}(B_j)$

Answer 2

Let $x \in f^{-1}(\bigcup B_j)$. This means that there exists $y \in \bigcup B_j$ such that $f(x)=y$. For some $i$ in the index set of $\bigcup$, $y \in B_i \subset \bigcup B_j$. This gives that $ x \in f^{-1}(B_j)$, so we also get $x \in \bigcup f^{-1}(B_j)$.

Conversely, assume $x \in \bigcup f^{-1}(B_j)$. There must exist an $i$ such that $x \in f^{-1} (B_i)$. Again we have $B_i \subset \bigcup B_j$, so $x \in f^{-1}(\bigcup B_j)$.

Therefore $\bigcup f^{-1}(B_j) = f^{-1}(\bigcup B_j)$.