one-to-one function from onto function
Given an onto function from a set to a set , there exists a one-to-one function from to .
Suppose is onto, and define ; that is, is the set containing the pre-image of each singleton subset of . Since is onto, no element of is empty, and since is a function, the elements of are mutually disjoint, for if and , we have and , whence . Let be a choice function, noting that , and define by . To see that is one-to-one, let , and suppose that . This gives , but since the elements of are disjoint, this implies that , and thus . So is a one-to-one function from to . ∎
|Title||one-to-one function from onto function|
|Date of creation||2013-03-22 16:26:55|
|Last modified on||2013-03-22 16:26:55|
|Last modified by||mathcam (2727)|