Editing Bijection (section)

==More mathematical examples==
[[File:A bijection from the natural numbers to the integers.png|thumb|A bijection from the [[natural number]]s to the [[integer]]s, which maps 2''n'' to −''n'' and 2''n'' − 1 to ''n'', for ''n'' ≥ 0.]]
* For any set ''X'', the [[identity function]] '''1'''<sub>''X''</sub>: ''X'' → ''X'', '''1'''<sub>''X''</sub>(''x'') = ''x'' is bijective.
* The function ''f'': '''R''' → '''R''', ''f''(''x'') = 2''x'' + 1 is bijective, since for each ''y'' there is a unique ''x'' = (''y'' − 1)/2 such that ''f''(''x'') = ''y''. More generally, any [[linear function]] over the reals, ''f'': '''R''' → '''R''', ''f''(''x'') = ''ax'' + ''b'' (where ''a'' is non-zero) is a bijection. Each real number ''y'' is obtained from (or paired with) the real number ''x'' = (''y'' − ''b'')/''a''.
* The function ''f'': '''R''' → (−π/2, π/2), given by ''f''(''x'') = arctan(''x'') is bijective, since each real number ''x'' is paired with exactly one angle ''y'' in the interval (−π/2,&nbsp;π/2) so that tan(''y'') = ''x'' (that is, ''y'' = arctan(''x'')). If the [[codomain]]  (−π/2,&nbsp;π/2) was made larger to include an integer multiple of π/2, then this function would no longer be onto (surjective), since there is no real number which could be paired with the multiple of π/2 by this arctan function.
* The [[exponential function]], ''g'': '''R''' → '''R''', ''g''(''x'') = e<sup>''x''</sup>, is not bijective: for instance, there is no ''x'' in '''R''' such that ''g''(''x'') = −1, showing that ''g'' is not onto (surjective). However, if the codomain is restricted to the positive real numbers <math>\R^+ \equiv \left(0, \infty\right)</math>, then ''g'' would be bijective; its inverse (see below) is the [[natural logarithm]] function ln.
* The function ''h'': '''R''' → '''R'''<sup>+</sup>, ''h''(''x'') = ''x''<sup>2</sup> is not bijective: for instance, ''h''(−1) = ''h''(1) = 1, showing that ''h'' is not one-to-one (injective). However, if the [[domain of a function|domain]] is restricted to <math>\R^+_0 \equiv \left[0, \infty\right)</math>, then ''h'' would be bijective; its inverse is the positive square root function.
*By [[Schröder–Bernstein theorem]], given any two sets ''X'' and ''Y'', and two injective functions ''f'': ''X → Y'' and ''g'': ''Y → X'', there exists a bijective function ''h'': ''X → Y''.