Editing Surjective function (section)

===Surjections as right invertible functions===
The function {{Nowrap|''g'' : ''Y'' → ''X''}} is said to be a [[Inverse function#Left and right inverses|right inverse]] of the function {{Nowrap|''f'' : ''X'' → ''Y''}} if {{Nowrap begin}}''f''(''g''(''y'')) = ''y''{{Nowrap end}} for every ''y'' in ''Y'' (''g'' can be undone by ''f'').  In other words, ''g'' is a right inverse of ''f'' if the [[function composition|composition]] {{Nowrap|''f'' <small>o</small> ''g''}} of ''g'' and ''f'' in that order is the [[identity function]] on the domain ''Y'' of ''g''. The function ''g'' need not be a complete [[inverse function|inverse]] of ''f'' because the composition in the other order, {{Nowrap|''g'' <small>o</small> ''f''}}, may not be the identity function on the domain ''X'' of ''f''. In other words, ''f'' can undo or "''reverse''" ''g'', but cannot necessarily be reversed by it.

Every function with a right inverse is necessarily a surjection. The proposition that every surjective function has a right inverse is equivalent to the [[axiom of choice]].

If {{Nowrap|''f'' : ''X'' → ''Y''}} is surjective and ''B'' is a [[subset]] of ''Y'', then {{Nowrap begin}}''f''(''f''<sup> −1</sup>(''B'')) = ''B''{{Nowrap end}}. Thus, ''B'' can be recovered from its [[preimage]] {{Nowrap|''f''<sup> −1</sup>(''B'')}}.

For example, in the first illustration in the [[#Gallery|gallery]], there is some function ''g'' such that ''g''(''C'') = 4. There is also some function ''f'' such that ''f''(4) = ''C''.  It doesn't matter that ''g'' is not unique (it would also work if ''g''(''C'') equals 3); it only matters that ''f''  "reverses" ''g''.