Editing Inverse function (section)

==={{anchor|Compositional inverse}}Inverses and composition===
{{See also|Inverse element}}

Recall that if {{mvar|f}} is an invertible function with domain {{mvar|X}} and codomain {{mvar|Y}}, then

: <math> f^{-1}\left(f(x)\right) = x</math>, for every <math>x \in X</math> and <math> f\left(f^{-1}(y)\right) = y</math> for every <math>y \in Y </math>.

Using the [[composition of functions]], this statement can be rewritten to the following equations between functions:

: <math> f^{-1} \circ f = \operatorname{id}_X</math> and <math>f \circ f^{-1} = \operatorname{id}_Y, </math>

where {{math|id<sub>''X''</sub>}} is the [[identity function]] on the set {{mvar|X}}; that is, the function that leaves its argument unchanged. In [[category theory]], this statement is used as the definition of an inverse [[morphism]].

Considering function composition helps to understand the notation {{math|''f''<sup> −1</sup>}}. Repeatedly composing a function {{math|''f'': ''X''→''X''}} with itself is called [[iterated function|iteration]]. If {{mvar|f}} is applied {{mvar|n}} times, starting with the value {{mvar|x}}, then this is written as {{math|''f''<sup> ''n''</sup>(''x'')}}; so {{math|''f''<sup> 2</sup>(''x'') {{=}} ''f'' (''f'' (''x''))}}, etc. Since {{math|''f''<sup> −1</sup>(''f'' (''x'')) {{=}} ''x''}}, composing {{math|''f''<sup> −1</sup>}} and {{math|''f''<sup> ''n''</sup>}} yields {{math|''f''<sup> ''n''−1</sup>}}, "undoing" the effect of one application of {{mvar|f}}.