criterion of surjectivity
Proof. . Suppose that is surjective. Let be an arbitrary subset of and any element of the set . By the surjectivity, there is an in such that , and since , the element is not in , i.e. and thus . One can conclude that for all .
. Conversely, suppose the condition (1). Let again be an arbitrary subset of and any element of . We have two possibilities:
a) ; then , and by (1), . This means that there exists an element of such that .
b) ; then there exists an such that .
The both cases show the surjectivity of .
|Title||criterion of surjectivity|
|Date of creation||2013-03-22 18:04:56|
|Last modified on||2013-03-22 18:04:56|
|Last modified by||pahio (2872)|