# 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.

From a ZFC perspective, this occurs because sets are defined up to equality, not up to isomorphism.

From a structuralist perspective, the reason this happens is because your question is conflating "sets" with "subsets." See Lawvere's Sets for mathematics and the nLab page on subobjects and structural set theory for more information.