# criterion of surjectivity

Theorem. For surjectivity of a mapping $f:A\to B$, it’s necessary and sufficient that

$B\setminus f(X)\subseteq f(A\setminus X)\mathit{\hspace{1em}}\forall X\subseteq A.$ | (1) |

Proof. ${1}^{\underset{\xaf}{o}}$. Suppose that $f:A\to B$ is surjective^{}. Let $X$ be an arbitrary subset of $A$ and $y$ any element of the set $B\setminus f(X)$. By the surjectivity, there is an $x$ in $A$ such that $f(x)=y$, and since $y\notin f(X)$, the element $x$ is not in $X$, i.e. $x\in A\setminus X$ and thus $y=f(x)\in f(A\setminus X)$. One can conclude that $B\setminus f(X)\subseteq f(A\setminus X)$ for all $X\subseteq A$.

${2}^{\underset{\xaf}{o}}$. 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) $y\notin f(X)$; then $y\in B\setminus f(X)$, and by (1), $y\in f(A\setminus X)$. This means that there exists an element $x$ of $A\setminus X\subseteq A$ such that $f(x)=y$.

b) $y\in f(X)$; then there exists an $x\in X\subseteq A$ 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 |