Editing Inverse function (section)

===Symmetry===
There is a symmetry between a function and its inverse. Specifically, if {{mvar|f}} is an invertible function with domain {{mvar|X}} and codomain {{mvar|Y}}, then its inverse {{math|''f''<sup> −1</sup>}} has domain {{mvar|Y}} and image {{mvar|X}}, and the inverse of {{math|''f''<sup> −1</sup>}} is the original function {{mvar|f}}. In symbols, for functions {{math|''f'':''X'' → ''Y''}}  and {{math|''f''<sup>−1</sup>:''Y'' → ''X''}},<ref name=Wolf72 />

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

This statement is a consequence of the implication that for {{mvar|f}} to be invertible it must be bijective. The [[involution (mathematics)|involutory]] nature of the inverse can be concisely expressed by<ref>{{harvnb|Smith|Eggen|St. Andre|2006|loc=pg. 141 Theorem 3.3(a)}}</ref> 
:<math>\left(f^{-1}\right)^{-1} = f.</math>

[[Image:Composition of Inverses.png|thumb|right|240px|The inverse of {{math| ''g'' ∘ ''f'' }} is {{math| ''f''<sup> −1</sup> ∘ ''g''<sup> −1</sup>}}.]]
The inverse of a composition of functions is given by<ref>{{harvnb|Lay|2006|loc=p. 71, Theorem 7.26}}</ref> 
:<math>(g \circ f)^{-1} = f^{-1} \circ g^{-1}.</math>
Notice that the order of {{mvar|g}} and {{mvar|f}} have been reversed; to undo {{mvar|f}} followed by {{mvar|g}}, we must first undo {{mvar|g}}, and then undo {{mvar|f}}.

For example, let {{math|1= ''f''(''x'') = 3''x''}} and let {{math|1= ''g''(''x'') = ''x'' + 5}}. Then the composition {{math| ''g'' ∘ ''f''}} is the function that first multiplies by three and then adds five,

: <math>(g \circ f)(x) = 3x + 5.</math>

To reverse this process, we must first subtract five, and then divide by three,

: <math>(g \circ f)^{-1}(x) = \tfrac13(x - 5).</math>

This is the composition
{{math| (''f''<sup> −1</sup> ∘ ''g''<sup> −1</sup>)(''x'')}}.