Let $f$ be a bijection from a set $A$ to a set $B$. The inverse of $f$, noted $f^{-1}$ is the function that assigns to an element $b \in B$ the unique element $a \in A$ such that $f(a)=b$. Hence $f^{-1}(b) = a$ when $f(a) = b$.
Let $f$ be a bijection from a set $A$ to a set $B$. Let $S$ and $T$ be two subsets of $A$. a) Show that $f(S∪T)=f(S)∪f(T)$ b) Show that $f(S∩T)⊂f(S)∩f(T)$ c) Show that $f^-1(S∪T)=f^{-1}(S)∪f^{-1}(T)$
1) Suppose $y\in f(S\cup T)$ $\Rightarrow$ $\exists x\in S\cup T$ such that $f(x)=y$. Thus either $x\in S$ and/or $x\in T$. In either case, $f(x)\in f(S)\cup f(T)$, thus proving $f(S\cup T)\subset f(S)\cup f(T)$. On the other hand, suppose $y\in f(S)\cup f(T)$, then either $\in f(S)$ and/or $y\in f(T)$. In either case $\exists x\in S$ and/or $T$ such that $f(x)=y$, thus implying $f(x)\in f(S\cup t)$. Hence $f(S\cup T)=f(S)\cup f(T)$.
3.If you've really understood above two points then it's time to do 3rd on your own.
P.S.-Since you are new to MSE, hence I answered your question. But you should try to show some of your own effort while posting a question.
Hint:
