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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?

UnknownW
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2 Answers2

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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.

platty
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  • I think you mean the function will not be a bijection. Non-empty intersection does not imply undefined function even by the answer you have posed. Please fix it. – Bertrand Wittgenstein's Ghost Dec 02 '18 at 03:33
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    @BertrandWittgenstein'sGhost Nope, I meant what I said. If there is $x \in A \cap B$ such that $F_1(x) \neq F_2(x)$, then the function $F$ will not be well-defined, as we would have both $F(x) = F_1(x)$ and $F(x) = F_2(x)$. It certainly does not imply that the function is always undefined if the intersection is nonempty, but in order to guarantee that the function is well-defined, you want this to hold. – platty Dec 02 '18 at 03:37
  • I still do not see the logic, your answer and the comment imply the function is not surjective. Which is fine, but not necessarily undefined. – Bertrand Wittgenstein's Ghost Dec 02 '18 at 03:42
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    Nothing I wrote implies that the function will necessarily not be surjective, as it is quite possible that the function is actually bijective (let $A = C = {1,2}$ and $B = D = {2,3}$ and take all three functions to be the identity). My point is that, if we do not add the constraint that $A \cap B$, you cannot even guarantee that $F$ is well-defined. Is there some aspect of this argument in particular which you do not understand? – platty Dec 02 '18 at 03:45
  • You said, $F_1(x)\neq F_2(x)$, which one do you choose evaluting. What I am saying is that indeterminacy does not imply that it is undefined. There is a big leap between the premise and conclusion. What I am saying is given the conditions we can deduce the function is not a bijection, but you are saying the function is not defined altogether. Your argument is not supported by what you are writing. – Bertrand Wittgenstein's Ghost Dec 02 '18 at 04:07
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    By definition, a function maps each element do the domain to exactly one element of the codomain. Ergo, in the chance that $F_1(x) \neq F_2(x)$, the function $F$ is not well-defined. – platty Dec 02 '18 at 04:09
  • You are absolutely correct. My god, that totally skipped my mind lol. Sorry about that! – Bertrand Wittgenstein's Ghost Dec 02 '18 at 04:20
  • +1 :) for this answer. – Bertrand Wittgenstein's Ghost Dec 02 '18 at 04:21
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If $y \in A\cap B$ then which do you choose to define $F(y)$ as? $F_1(y)$ or $F_2(y)$? If you define it as $F_1(y)$ then it's possible that nothing will be mapped to $F_2(y)$ and vice versa so it may not be surjective. (And if $C\cap D = \emptyset$ then definitely on of $F_1(y)$ or $F_2(y)$ will not be mapped to.)

Likewise if $z \in C\cap D$ then it's possible there is an $a \in A$ and a $b \in B$ so that $F_1(a) = F_2(b) = z$. So $F$ may not be one to one.

fleablood
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