criterion of surjectivity

Theorem.  For surjectivity of a mappingf:AB,  it’s necessary and sufficient that

Bf(X)f(AX)XA. (1)

Proof.1o¯.  Suppose that  f:AB  is surjectivePlanetmathPlanetmath.  Let X be an arbitrary subset of A and y any element of the set Bf(X).  By the surjectivity, there is an x in A such that  f(x)=y, and since  yf(X),  the element x is not in X, i.e.  xAX  and thus  y=f(x)f(AX).  One can conclude that  Bf(X)f(AX)  for all  XA.

2o¯.  Conversely, suppose the condition (1).  Let again X be an arbitrary subset of A and y any element of B.  We have two possibilities:
a) yf(X); then  yBf(X), and by (1), yf(AX).  This means that there exists an element x of  AXA  such that  f(x)=y.
b) yf(X); then there exists an xXA  such that  f(x)=y.
The both cases show the surjectivity of f.

Title criterion of surjectivity
Canonical name CriterionOfSurjectivity
Date of creation 2013-03-22 18:04:56
Last modified on 2013-03-22 18:04:56
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 4
Author pahio (2872)
Entry type Theorem
Classification msc 03-00
Synonym surjectivity criterion
Related topic Function
Related topic Image
Related topic Subset