In this video by 0:55, it says
Lemma: Let $A,B,C,D$ be sets with $A\cap B=\emptyset$ and $C\cap D=\emptyset$. Suppose that $F_1:A\to C$ and $F_2:B\to D$ are both bijections. Define $F:A\cup B\to C\cup D$ by
$$
F(x)=\begin{cases}
F_1(x) & \text{ if } x\in A \\
F_2(x) & \text{ if } x\in B
\end{cases}
$$
Then $F$ is a bijection.
To me, it is easy to prove it by looking at the $F|_A$ and $F|_B$ seperately, and then, by construction, $F$ is bijective by collecting the elements of $A$ and $B$. The question is, why do the intersections $A\cap B$ and $C\cap D$ have to be empty, in particular the last one?
Best Answer
If the intersection $A \cap B$ is nonempty, the function may not be well-defined — if $x\in A \cap B$ and $F_1(x) \neq F_2(x)$, which one do you choose when evaluating $F(x)$?
If the intersection $C \cap D$ is nonempty, the function may not be an injection. For $y \in C \cap D$, there must be $x_1 \in A$ and $x_2 \in B$ such that $F_1(x_1) = y$ and $F_2(x_2) = y$. But then $F(x_1) = F(x_2)$; if $x_1 \neq x_2$ (which may very well be true), then $F$ is not an injection.