# Does bijectivity survive set union, intersection?

Let $f :A \to B, \ g : C \to D$ be bijections then is there always a bijection $h: A\cup C \to B \cup D$ or $h : A \cap C \to B \cap D$ ?

Clearly, if $A, C$ are disjoint and $B, D$ are disjoint, then union works. I can't see how to split up the sets though since if we split up $A \cup B = A \setminus B \uplus A \cap B \uplus B \setminus A$ then this splitting is not carried over in a similar way to the codomain side.

• one thing needed for a bijection is equal cardinality. – user451844 Aug 12 '17 at 2:19

Suppose that $A = C$, then $A \cup C = A \cup A = A$. Now suppose that $B \cap D = \varnothing$. Then $| B \cup D | = |B| + |D| > |A| = |A \cup C| \implies |B \cup D| > |A \cup C|$ at least in some finite cases. A similar argument can show that no bijection $h : A \cap C \to B \cap D$ exists either as $A \cap C = A$ and $B \cap D = \varnothing$.
No. For example, assume each of $A,B,C,D$ and $5$ elements. Assume also that $A$ and $C$ are equal, but $B$ and $D$ are disjoint. Then $A \cup C$ has $5$ elements, but $B \cup D$ has $10$ elements. So there's no bijection $A \cup C \rightarrow B \cup D$, because the domain and codomain have different numbers of elements.