surjection and axiom of choice
In this entry, we show the statement that
AC implies (*).
Remark. The function is easily seen to be an injection: if , then .
(*) implies AC.
(**) given a set of pairwise disjoint non-empty sets, there is a choice function
seemingly weaker than AC, turns out to be equivalent to AC, and we will prove this fact first.
Obviously AC implies (**). Conversely, assume (**). Let be a collection of non-empty sets. We assume . For each , define a set . Since , . In addition, iff (true since elements of and elements of have distinct second coordinates). So the collection is a set consisting of pairwise disjoint non-empty sets. By (**), there is a function such that for every . Now, define two functions and by and Then, for any , we have . Since , its first coordinate is an element of . Therefore , and hence is the desired choice function. ∎
Proof of Propositon 2.
Let be a collection of pairwise disjoint non-empty sets. Each element of belongs to a unique set in . Then the function taking each element of to the set it belongs in , is a well-defined function. It is clearly surjective. Hence, by assumption, there is a function such that (a right inverse of ). For each , , which is the same as saying that is an element of by the definition of . ∎
Remark. In the category of sets, AC is equivalent to saying that every epimorophism is a split epimorphism. In general, a category is said to have the axiom of choice if every epimorphism is a split epimorphism.
|Title||surjection and axiom of choice|
|Date of creation||2013-03-22 18:44:37|
|Last modified on||2013-03-22 18:44:37|
|Last modified by||CWoo (3771)|