I am not sure how to go about this problem, can someone please help.
Problem
let $\,f:X\rightarrow Y$ and $\wp$ is the power set.
$\forall \, B_1,B_2 \in \wp(Y),\,\, f^{-1}(B_1\,\cup\,B_2) = f^{-1}(B_1)\,\,\cup\,\,f^{-1}(B_2) $,
Equivalently $F^* : \wp(Y), \, \cup \rightarrow \wp(X), \, \cup$, is a homomorphism.
Solution attempt :
I am not sure if I should just construct 2 arbitrary sets, say $\{ x_1,x_2\}$ and $\{y_1,y_2\}$, and possibly complete the proof by applying unions to those two sets? I also know this identity but not sure if it pans out :
$B_1 \, \cup B_2 = B1 + B2 - B1 \cap B_2 \implies f^{-1}(B_1)+f^{-1}(B_2)-f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$
However, without knowing beforehand that $f$ a homomorphism, I don't think I can prove the converse, $\,\,f^{-1}(B_1\cup B_2)=f^{-1}(B_1)+f^{-1}(B_2)-f^{-1}(B_1\cap B_2)$
I was also given two identities, they are probably needed in the proof, but no idea how :
Image : We have map $f_* : \wp(X) \rightarrow \wp(Y) \text{ given by } $ $f_*(A)=\{y \in Y : y=f(a) \text{for some } a \in A\}$In conventional notation, $f_*(A)$ is just denoted by $f(A)$. It is called the image $f$ of $A$
Inverse Image We also have $f^* : \wp (Y)\rightarrow \wp (X)$ given by $ f^*(B)=\{x \in X : f(X) \in B\}.$ In conventional notation $f^*(B)$ is denoted by $f^{-1}(B)$ It is always defined, whether or not $f$ has an inverse. It is called the inverse image of $f$ of $B$. In the case that $B=\{b\}$ we denote $f^{-1}(\{b\})$ by $f^{-1}(b)$
Note that by the definition of the inverse image $f^{-1}(B)= \{x\in X\mid f(x)\in B\}$, it follows that for $x\in X$: $$x\in f^{-1}(B)\ \ \mbox{ iff }\ \ f(x)\in B.$$
Now, what you have to prove is: $f^{-1}(B_1\cup B_2)= f^{-1}(B_1)\cup f^{-1}(B_2)$, i.e. that for $x\in X$: $$x\in f^{-1}(B_1\cup B_2)\ \ \mbox{ iff }\ \ x\in f^{-1}(B_1)\cup f^{-1}(B_2).$$ So, let $x\in X$. We have: \begin{align} x\in f^{-1}(B_1\cup B_2) && \mbox{iff} && f(x)\in B_1\cup B_2 && \mbox{by the definition of the inverse image}\\ && \mbox{iff} && f(x)\in B_1\ \ \mbox{or}\ \ f(x)\in B_2 && \mbox{by the definition of union}\\ && \mbox{iff} && x\in f^{-1}(B_1)\ \ \mbox{or}\ \ x\in f^{-1}(B_2) && \mbox{by the definition of the inverse image}\\ && \mbox{iff} && x\in f^{-1}(B_1)\cup f^{-1}(B_2) && \mbox{by the definition of union} \end{align}