证明：先证f (A∩B )?1?f
(A )?1∩f (B )?1 事实上，因为A ∩B ?A ，故f (A∩B )?1?f
(A )?1 同理有f (A∩B )
?1?f (B )?1.故有f (A∩B )?1?f (A )∩?1f (B )?1 再证：f (A∩B )?1?f
(A )?1∩f (B )?1 任取x ∈f (A )?1∩f
(B )?1，则x=f (A )?1.从而?y ∈A 使得f (x )=y . 又x ∈f (B )
?1,由映射的定义可知，y ∈B ，y ∈A ∩B ， x ∈f (A∩B )?1 故f (A∩B )?1?f (A )?1∩f (B )?1.
于是有f (A∪B )?1=f
(A )?1∩f (B )?1.