define: $ A \approx B $ $ \iff $ exists bijection $ f:A \to B $
let $B,C $ be such $ B\cap C=\emptyset $
I want to prove that $ A^{(B \cup C)} \approx A^B\times A^C $
$ A^{(B \cup C)} $ = the set of all the functions from $ B\cup C $ to $ A $
I dont have any idea of how to define the bijection so it would satisfy the terms.
Also, Im trying to learn this subject independently, so any tips about how to approach such questions will be helpful.
For any $x\in A^{(B\cup C)}$, define $f:A^{(B\cup C)}\rightarrow A^B\times A^C$ as $$fx=(x_1,x_2),\text{ where }x_1(b):=x(b),x_2(c):=x(c),\quad\forall b\in B,\forall c\in C.$$ Then $f$ is injection :
If $x\neq y$, then there exists $a\in B\cup C$ s.t. $x(a)\neq y(a)$, hence (let $a\in B$ without loosing generality) $(fx)(a,c)=(x(a),x(c))\neq (y(a),y(c))=(fy)(a,c)$, i.e. $fx\neq fy$.
On the other hand, for any $(x,y)\in A^B\times A^C$, let $g:A^B\times A^C\rightarrow A^{(B\cup C)}$ as $$g(x,y)(a)=\bigg\{\begin{array}&x(a)\quad\text{if }a\in B\\y(a)\quad\text{if }a\in C. \end{array}$$ Then again $g$ is injection :
If $(x,y)\neq (x',y')$, then either $x\neq x'$ or $y\neq y'$, so let $x\neq x'$ without loosing generality, then there exists $a\in B$ s.t. $x(a)\neq x'(a)$ so $g(x,y)(a)=x(a)\neq x'(a)=g(x',y')(a)$.
Therefore by Schröder-Bernstein theorem two sets have bijection.