Editing Natural transformation (section)

===Opposite group===
{{details|Opposite group}}
Statements such as
:"Every group is naturally isomorphic to its [[opposite group]]"
abound in modern mathematics. We will now give the precise meaning of this statement as well as its proof. Consider the category 
<math>\textbf{Grp}</math> of all [[group (mathematics)|group]]s with [[group homomorphism]]s as morphisms. If <math>(G, *)</math> is a group, we define 
its opposite group <math>(G^\text{op}, {*}^\text{op})</math> as follows: <math>G^\text{op}</math> is the same set as <math>G</math>, and the operation <math>*^\text{op}</math> is defined 
by <math>a *^\text{op} b = b * a</math>. All multiplications in <math>G^{\text{op}}</math> are thus "turned around". Forming the [[Opposite category|opposite]] group becomes 
a (covariant) functor from <math>\textbf{Grp}</math> to <math>\textbf{Grp}</math> if we define <math>f^{\text{op}} = f</math> for any group homomorphism <math>f: G \to H</math>. Note that 
<math>f^\text{op}</math> is indeed a group homomorphism from <math>G^\text{op}</math> to <math>H^\text{op}</math>:
:<math>f^\text{op}(a *^\text{op} b) = f(b * a) = f(b) * f(a) = f^\text{op}(a) *^\text{op} f^\text{op}(b).</math>
The content of the above statement is:
:"The identity functor <math>\text{Id}_{\textbf{Grp}}: \textbf{Grp} \to \textbf{Grp}</math> is naturally isomorphic to the opposite functor <math>{\text{op}}: \textbf{Grp} \to \textbf{Grp}</math>"
To prove this, we need to provide isomorphisms <math>\eta_G: G \to G^{\text{op}}</math> for every group <math>G</math>, such that the above diagram commutes. 
Set <math> \eta_G(a) = a^{-1}</math>.
The formulas <math>(a * b)^{-1} = b^{-1}*a^{-1}= a^{-1}*^{\text{op}} b^{-1}</math> and <math> (a^{-1})^{-1} = a</math>
show that <math>\eta_G</math> is a group homomorphism with inverse <math> \eta_{G^\text{op}}</math>. To prove the naturality, we start with a group homomorphism 
<math>f : G \to H</math> and show <math>\eta_H \circ f = f^{\text{op}} \circ \eta_G</math>, i.e. 
<math> (f(a))^{-1} = f^\text{op}(a^{-1})</math> for all <math>a</math> in <math>G</math>. This is true since 
<math>f^{\text{op}} = f</math> and every group homomorphism has the property <math>(f(a))^{-1} = f(a^{-1})</math>.