Difference between revisions of "Inverse"

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: <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>.
: <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]], we can rewrite this statement as follows:
Using the [[wikipedia:Function composition|composition of functions]], we can rewrite this statement as follows:


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


where <math>\operatorname{id}_X</math> is the [[identity function]] on the set <math>X</math>; that is, the function that leaves its argument unchanged. In [[Category Theory]], this statement is used as the definition of an inverse [[morphism]].
where <math>\operatorname{id}_X</math> is the [[identity function]] on the set <math>X</math>; that is, the function that leaves its argument unchanged. In [[Category Theory]], this statement is used as the definition of an inverse [[morphism]].

Revision as of 14:26, 6 August 2021

The notion of inverse in mathematical operation is simply the anti-operator.

Excerpt from Wikipedia

The following paragraph is copied from Wikipedia.

Compositional inverseInverses and composition

If is an invertible function with domain and codomain , then

, for every ; and , for every .

Using the composition of functions, we can rewrite this statement as follows:

and

where is the identity function on the set ; that is, the function that leaves its argument unchanged. In Category Theory, this statement is used as the definition of an inverse morphism.